Made Easy Chemistry

Why are copper and chromium exceptions to electron configuration?

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Electronic Configuration of Elements

Mendeleev noticed the recurrence of properties of elements as the atomic weight increased, and he invented the Periodic Table of Element, which is a useful tool for organizing and correlating chemical and physical properties of chemical elements. Today, the most popular Periodic Table form is shaped by results of quantum theory. Quantum theory rationalized the existence of and arrangement of all elements in today's Periodic Table. It has also been applied to explain their chemical properties.

Electronic Configuration of Elements

Skills to develop

Explain the rules for filling electrons in atomic orbitals -- Pauli exclusion principle and Hund's rule

Fill electrons in atomic orbitals--Aufbau process

Explain the arrangement of elements in terms of quantum numbers

Explain the systematic variation of element properties

Energy Levels in Many-Electron Atoms

In order to fill the electrons in various atomic orbitals, we need to know how the energy levels vary as the nuclear charge increases. For hydrogen-like atoms, the approximate energy levels are as indicated below:

The shielding effect and electron-electron interactions cause the energy levels of subshells such as 2s & 2p to be different from those of H-like atoms. This is done by treating the electron shield cores as a proton but the core has an effective nuclear charge Z.
For the H-like atoms, energy levels for 2s, 2p stay the same, but the separation between 2s and 2p energy levels increases as the atomic number (Z) increases. Similar situations happen for 3s, 3p, and 3d energy levels. The energy diagrams of H, Li & K are used to illustrate this point. The color diagram is from a Hyperion website discussing quantum numbers and structure of atoms

Understand how the energy level vary is the key to the Aufbau process, because Electrons tend to occupy the lowest energy level available. But before we talk about the Aufbau process, we need to be aware of the Pauli exclusion principle and the Hund's rule.

The Pauli Exclusion Principle

The Pauli exclusion principle suggests that only two electrons with opposite spin can occupy an atomic orbital. Stated another way, no two electrons have the same 4 quantum numbers n, l, m, s. Pauli's exclusion principle can be stated in some other ways, but the idea is that energy states have limit room to accommodate electrons. A state accepts two electrons of different spins.
In applying this rule, You should realize that an atomic orbital is an energy state.

Hund's RULE

Hund's rule suggests that electrons prefer parallel spins in separate orbitals of subshells. This rule guides us in assigning electrons to different states in each sub-shell of the atmic orbitals. In other words, electrons fill each and all orbitals in the subshell before they pair up with opposite spins. Pauli exclusion principle and Hund's rule guide us in the aufbau process, which is figuring out the electron configurations for all elements.

The Aufbau Procedure

The aufbau procedure (filling order of atomic orbitals) is used to work out the electron confiturations of all atoms. However, modification should be made by applying Hund's rule to be discussed in the next section. The aufbau procedure is based on a rough energy levels diagram of many-electron atoms as shown below:



     7s |~            5d |-----   5f |_______    
                                     Actanides


     6s |_  6p |~~~   5d |=====   4f |=======    
                                     Lanthanides


     5s |_  5p |~~~   4d |-----

                            Note the trend 5s 4d 5p 
                            & develope a pattern

     4s |_  4p |~~~   3d |-----
                         Transition elements

     3s |_  3p |---


     2s |_  2p |---



     1s |-

The fine difference between energy levels cannot be adquately shown.
You've learned various techniques to work out the electronic configurations of elements. Here is yet another form of an energy level diagram:

                                          7p _ _ _
    7s _  5f - - - - - - -  6d ~ ~ ~ ~ ~
                                          6p _ _ _
    6s _  4f - - - - - - -  5d ~ ~ ~ ~ ~

    5s _  4d - - - - -  5p ~ ~ ~

    4s _  3d - - - - -  4p ~ ~ ~

    3s _  3p - - -

    2s _  2p - - -

    1s _

In order to master the technique of the aufbau procedure, you should apply the Hund's rule and Pauli exclusion principle to work out the electronic configuration for closed shells of inert gases. After you worked out, you may compare your result with the following. Do not try to remember it, it is important to know how to work it out.

Z=   2      10     18       36         54           86
    1s²  2s²2p⁶ 3s²3p⁶ 4s²3d¹⁰4p⁶ 5s²4d¹⁰5p⁶ 6s²4f¹⁴5d¹⁰6p⁶
    He      Ne     Ar       Kr         Xe           Rn

Block of elements by highest occupied atomic orbitals

Block of elements by last filled atomic orbitals
1s 2s 3s 4s 5s 6s 7s	4f - - - - - 4f 5f - - - - - 5f	3d - - - 3d 4d - - - 4d 5d - - - 5d 6d - - - 6d	2p - 2p 3p - 3p 4p - 4p 5p - 5p 6p - 6p 7p - 7p

The highest atomic orbitals occupied by electrons determine the properties of the elements. According to this scheme, the periodic table can be divided into s, p, d, and f blocks as seen in the table on the right. This table shows the filling order of atomic orbitals as
1s 2s 2p 3s 3p 4s 3d 4p 5s 4d 5p 6s 4f 5d 6p 7s 5f 6d 7p The s- and p-blocks of elements are called main group elements. The d-block elements are called transition elements The f-block elements are called the inner transition elements.
In an ordinary periodic table, the s, p, and d block elements are in the main body of the Periodic Table, whereas the f block elements are placed below the main body. If we placed them on the same period where they belong, the Periodic Table would be too long for the screen to accommodate. Thus, we keep the Periodic Table in the usual (long) form.

Special Electronic Configurations

When two electrons occupy the same orbital, they not only have different spins (Pauli exclusion principle), the pairing raises the energy slightly. On the other hand, a half filled subshell and a full filled subshell lower the energy, gaining some stability. Bearing this in mind, you will be able to understand why we have the following special electronic configurations.

Cr [Ar]4s¹ 3d⁵	<=All s and d subshells are half full
Cu [Ar]4s¹ 3d¹⁰	<=Prefers a filled d subshell, leaving s with 1
Nb [Kr]5s¹ 4d⁴	<=5s and 4d energy levels are close
Mo [Kr]5s¹ 4d⁵	similar to Cr above
Tc [Kr]5s² 4d⁵	(not special, but think of Hund's rule)
Ru [Kr]5s¹ 4d⁷	<= Only 1 5s electron
Rh [Kr]5s¹ 4d⁸	<= in both
Pd [Kr]5s⁰ 4d¹⁰	<= Note filled 4d and empty 5s
Ag [Kr]5s¹ 4d¹⁰	<= partial filled 5s, but filled d

For CHEM120 students, you are not required to remember the special ones, but you should take notice of the electronic configurations of Cr and Cu to realize the gaining of stability due to half and full filled subshells.

Confidence Building Questions

For the H-like atom, which subshell has the highest energy level?
4f, 3d, 2p, 1s
Skill:
Describe the energy levels of hydrogen atoms.
For an element with atomic number 20, which is the last or highest occupied subshell of atomic orbitals?
1s 2s 2p 3s 3p 3d 4s 5s
Discussion:
The element is Calcium, Ca.
How many electrons are required to fill all the following sub shells?
1s 2s 2p 3s 3p 4s 3d 4p
Discussion:
The element is Krypton, Kr.
Which of the following two electronic configuration is more stable?
a [Ar]4s¹ 3d⁵
b [Ar]4s² 3d⁴
Discussion:
Check the electronic configuration for Cu and Cr.
Which of the following two electronic configurations is more stable?
a [Ar]4s² 3d⁹
b [Ar]4s¹ 3d¹⁰
Discussion:
This is the electronic configuration for copper.
Choose the electronic configuration for palladium, Pd (Z = 46).
a [Kr]5s¹ 4d⁷
b [Kr]5s¹ 4d⁸
c [Kr]5s⁰ 4d¹⁰
d [Kr]5s¹ 4d¹⁰

Electron Configuration Anomalies

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Electron Configuration Anomalies

Some of the elements have electron configurations that differ slightly from what our general procedure would lead us to predict. Because a few of these elements are important elements, it is useful to know their actual electron configurations. Six of these are listed on the table below.

Unusual Electron Configurations

Element

Predicted Electron Configuration

Actual Electron Configuration

copper, Cu

[Ar] 3d⁹ 4s²

[Ar] 3d¹⁰4s¹

silver, Ag

[Kr] 4d⁹ 5s²

[Kr] 4d¹⁰ 5s¹

gold, Au

[Xe] 4f¹⁴ 5d⁹ 6s²

[Xe] 4f¹⁴ 5d¹⁰ 6s¹

palladium, Pd

[Kr] 4d⁸ 5s²

[Kr] 4d¹⁰

chromium, Cr

[Ar] 3d⁴ 4s²

[Ar] 3d⁵ 4s¹

molybdenum, Mo

[Kr] 4d⁴ 5s²

[Kr] 4d⁵ 5s¹

Filling Electron Shells

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Filling Electron Shells

When an atom or ion receives electrons into its orbitals, the orbitals and shells fill up in a particular manner.

Aufbau principle

You may consider an atom as being "built up" from a naked nucleus by gradually adding to it one electron after another, until all the electrons it will hold have been added. Much as one fills up a container with liquid from the bottom up, the orbitals of an atom are filled from the lowest energy orbitals to the highest energy orbitals.
Orbitals with the lowest principal quantum number () have the lowest energy and will fill up first. Within a shell, there may be several orbitals with the same principal quantum number. In that case, more specific rules must be applied. For example, the three p orbitals of a given shell all occur at the same energy level. So, how are they filled up? ans: all the three p orbitals have same energy so while filling the p orbitals we can fill any one of the Px, Py or Pz first. it is a convention that we chose to fill Px first ,then Py and then Pz for our simplicity. Hence you can opt for filling these three orbitals from right to left also.

Hund's Rule

According to Hund's rule, orbitals of the same energy are each filled with one electron before filling any with a second. Also, these first electrons have the same spin.

This rule is sometimes called the "bus seating rule". As people load onto a bus, each person takes his own seat, sitting alone. Only after all the seats have been filled will people start doubling up.

Pauli Exclusion principle

No two electrons can have all four quantum numbers the same. What this translates to in terms of our picture of orbitals is that each orbital can only hold two electrons, one "spin up" (+½) and one "spin down" (-½).

This animation demonstrates the Aufbau principle, Hund's Rule, and the Pauli Exclusion principle.

Orbital Order

1s, 2s, 2p, 3s, 3p, 4s, 3d, 4p, 5s, 4d, 5p, 6s, 4f, 5d, 6p, 7s, 5f, 6d, 7p, 8s.
Although this looks confusing, there is an easy way to remember.

Understanding the above rules and diagrams will allow you to determine the electron configuration of almost any atom or ion.

How to Write the Electron Configuration of an Isolated Atom

Wikipedia has related information at Electron configuration

Electron-configuration notation is relatively straightforward. An isolated Calcium atom, for example, would have configuration of 1s²2s²2p⁶3s²3p⁶4s² in its ground state. Other configurations like 1s²2s²2p⁶3s²3p⁶4s¹4p¹ are possible, but these excited states have a higher energy. They are not stable and generally only exist for a brief moment.
The ground state configuration for Ca could be abbreviated by using the preceding noble gas (the elements found all the way on the right of the periodic table) as [Ar]4s², where Ar is argon.
Noble gases have very stable configurations, and are extremely reluctant to lose or gain electrons. Noble gas atoms are also the only ones regularly found as isolated atoms in the ground state. Atoms of other elements all undergo bonding under the conditions that we live under and this affects the orbitals that the outermost electrons are in. In that sense the electron configurations for the other elements are somewhat hypothetical: to encounter an isolated atom of, say, tungsten (W), we would have to first vaporize a metal that boils at 5800K. However, knowing atomic configurations is useful because it does help us to understand how and why they bond, i.e. why and how they change the configuration of their outer valence electrons.

Rule of Stability

A subshell is particularly stable if it is half full or full. Given two configurations, the atom would "choose" the more stable one.
Example: In the following configuration, Cu: [Ar]4s²3d⁹, copper's d shell is just one away from stability, and therefore, one electron from the s shell jumps into the d shell: [Ar]4s¹3d¹⁰. This way, the d shell is full, and is therefore stable, and the s shell is half full, and is also stable.

Another example: Chromium has a configuration of [Ar]4s¹3d⁵, although you would expect to see four d electrons instead of five. This is because an s electron has jumped into the d orbital, giving the atom two half-full shells—much more stable than a d orbital with only four electrons.
The stability rule applies to atoms in the same group as chromium and copper.
If one of these atoms has been ionized, that is, it loses an electron, it will come from the s orbital rather than the d orbital. For instance, the configuration of Cu⁺ is [Ar]4s⁰3d¹⁰. If more electrons are removed, they will come from the d orbital.

Magnetism

The spin of an electron creates a magnetic field (albeit ridiculously weak), so unpaired electrons create a small magnetic field. Paired electrons have opposite spin, so the magnetic fields cancel each other out, leading to diamagnetism.

Magnetism is a well-known effect. Chances are, you have magnets on your refrigerator. As you already know, only certain elements are magnetic. Electron configurations help to explain why.
Diamagnetism is actually a very weak repulsion to magnetic fields. All elements have diamagnetism to some degree. It occurs when there are paired electrons.
Paramagnetism is an attraction to external magnetic fields. It is also very weak. It occurs whenever there is an unpaired electron in an orbital.
Both diamagnetism and paramagnetism are responses of spins acting independently from each other. This leads to rather weak repulsion and attraction respectively. However, when they are located in a solid they may also interact with each other and respond collectively and that can lead to rather different properties:
Ferromagnetism is the permanent magnetism that we encounter in our daily lives. It occurs when all the unpaired spins in a solid couple and tend to align themselves in the same direction, leading to a strong attraction when exposed to a magnetic field. This only occurs at room temperature with three elements: iron (Fe), nickel (Ni), and cobalt (Co). Gadolinium (Gd) is a borderline case. It loses its ferromagnetism at 20^oC; above that temperature the spins start to act alone. However, there are many alloys and compounds that exhibit strong ferromagnetic coupling. The strongest one is Nd₂Fe₁₄B
Antiferromagnetism is also a permanent magnetism in which unpaired spins align, but they do so in opposite directions. The result is that the material does not react very strongly to a magnetic field at all. Chromium (Cr) is an example.
'Ferrimagnetism is a combination of ferro- and antiferromagetism. Unpaired spins align partly in opposite directions, but the compensation is not complete. This is why the material is still attracted strongly to a magnetic field. Magnetite Fe₃O₄ is such a substance. It was the first material studied for its magnetic properties and may well be the one sitting on your fridge.

Order of Filling of Electron States

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As the periodic table of the elements is built up by adding the necessary electrons to match the atomic number, the electrons will take the lowest energy consistent with the Pauli exclusion principle. The maximum population of each shell is determined by the quantum numbers and the diagram at left is one way to illustrate the order of filling of the electron energy states.
For a single electron, the energy is determined by the principal quantum n number and that quantum number is used to indicate the "shell" in which the electrons reside. For a given shell in multi-electron atoms, those electrons with lower orbital quantum number l will be lower in energy because of

The electron configuration for any element may be found by clicking on that element in the periodic table. The first exception to the above scheme that is encountered is chromium, where the fifth 3d electron state is occupied instead of the second 4s state

Shapes of Atomic Orbitals

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s Sublevels are Spherically Shaped

All ℓ = 0 electron waves are s waves, or waves from the s sublevel, and they all describe electrons in s orbitals. As suggested in the previous section, all electron waves from the s sublevel have the same overall shape, regardless of the value of n, regardless of their size, and regardless of the number of nodes they contain. s orbitals always correspond to spherical waves. The quantum numbers n = 1 and ℓ = 0 describe a small spherical wave with no nodes, the quantum numbers n = 2 and ℓ = 0 describe a larger spherical wave with a single node, and the quantum numbers n = 3 and ℓ = 0 describe an even larger spherical wave with two nodes. These waves all look slightly different, as shown in Figure 6.16.

Figure 6.16: Various s orbitals. All of these orbitals have ℓ = 0, but they have different values for n. The first orbital has n = 1, and thus is small and has no nodes. The second orbital has n = 2, and thus is larger and has one node. The third orbital has n = 3, and thus is even larger and has two nodes.

Figure : Notice that the amount of electron density (here represented by the intensity of the blue color) doesn't depend on direction. It does, however, depend on distance from the center of the atom.

Nevertheless, they are all spherical, because they all have ℓ = 0. Their shapes don't change – only their sizes and the number of nodes that they contain.
Now if you think back to an earlier lesson, you might remember something special about the different orientations of a spherical wave. Do you remember what happened when we rotated the spherical wave so that it pointed in different directions? It ended up looking the same, didn't it! No matter which way you rotate a sphere, it always looks the same. So how many different m_l values do you expect for a spherical wave? One, of course! Now that you know spherical waves all have ℓ = 0, you can use your rules for m_l to figure out exactly how many different m_l are allowed. If you look back to Example 4 in the previous lesson, you'll see that we actually did that calculation. It turned out that there was only one allowable value for m_l, and that was m_l = 0. In other words, there is only one orientation of a spherical wave. It all makes sense!
So what does a spherical wave really mean? It means that your probability of finding an electron at any particular distance from the center of the atom only depends on the distance, and not on the direction. You can see this in Figure 6.17.

When the Azimuthal Quantum Number is 1, then m_l Can Only Be -1, 0 or +1

All ℓ = 1 electron waves are p waves, or waves from the p sublevel, and they describe electrons in what are known as p orbitals. Unlike s orbitals, p orbitals are not spherical, so they can have different orientations in space. Now that you know all p orbitals have ℓ = 1, you should be able to figure out exactly how many different p orbital orientations exist by using your rules for m_l. (m_l is the quantum number associated with the orientation of a particular orbital). Let's figure it out.

Example 1

How many different p orbital orientations are possible?

Solution:
ℓ = 1
From now on, whenever you're told an electron is in a p orbital, you're expected to know that electron has the quantum number ℓ = 1.
The question asks how many p orbital orientations are possible, but what it's really asking is how many different m_l values are allowed when ℓ = 1. We've already done this type of problem.
1. Find the minimum value of m_l.

The minimum value of m_l is always −ℓ.

minimum m_l = −1

2. Find the maximum value of m_l.

The maximum value of m_l is always +ℓ

maximum m_l = +1

3. List all of the integers (no decimals!) starting from the minimum value of m_l and ending with the maximum value of m_l.

m_l = −1, 0, 1

In this case m_l can equal −1, 0 or 1, so there are a total of 3 allowed values for m_l, and thus 3 possible p orbital orientations.

The p Orbitals are Often Described as Dumb-bell Shaped

Even though you know that there are three possible orientations for p orbitals, you can't really predict their shape unless you know a lot more about mathematics, physics, and wave functions. When scientists use the wave function to draw the shape of an electron's p orbital, though, they always end up with is something that looks a lot like a dumb-bell. Not only that, the three different p orbitals (one with m_l = −1, another with m_l = 0, and the third with m_l = 1) turn out to be perpendicular to each other. In other words, if one p orbital points along the x-axis, another p orbital points along the y-axis, and the third points along the z-axis. Scientists typically label these three orbitals p_x, p_y, and p_z respectively. The below figure shows each of the three p orbitals separately, and then all three together on the same atom.

Sometimes we get so caught up thinking about electron wave functions, and electron orbitals, that we forget entirely about the atom itself. Remember, electron standing waves form because electrons get trapped inside an atom by the positive charge on the atom's nucleus. As a result, s orbitals, and p orbitals and even d and f orbitals always extend out from the atom's nucleus. Don't get so caught up in orbitals that you forget where they are and why they exist.
As with s orbitals, p orbitals can be big or small, depending on the value of n, and they can also have more or less nodes, also depending on the value of n. Notice, however, that unlike the s orbital, which can have no nodes at all, a p orbital always has at least one node. Take a look at the p orbital figure above again. Can you spot the node in each of the p orbitals? Since all p orbitals have at least one node, there are no p orbitals with n = 1. In fact, the first principal quantum number, n, for which p orbitals are allowed is n = 2. Of course you could have figured that out for yourself, right? No? Well, here’s a hint – remember the rules for predicting which values of ℓ are allowed for any given value of n. In the last lesson, you learned that ℓ must be no less than 0, but also, no greater than n − 1. For the n = 1 energy level, then, the maximum allowed value for ℓ is:

maximum ℓ = n − 1

maximum ℓ = 1 − 1

maximum ℓ = 0

As a result, only s orbitals (ℓ = 0) are allowed. For the n = 2 energy level, though, the maximum allowed value for ℓ is

maximum ℓ = n − 1

maximum ℓ = 2 − 1

maximum ℓ = 1

which means p orbitals (ℓ = 1) are allowed as well. So now you see that the restrictions on ℓ are actually there to make sure that all n = 1 wave functions have no nodes, all n = 2 wave functions have 1 node, all n = 3 wave functions have 2 nodes, all… well, you get the picture.
One interesting property of p orbitals that is different from s orbitals is that the total amount of electron density changes with both the distance from the center of the atom and the direction. Take a look at Figure 6.18. Notice how the electron density is different depending on which direction you travel from the center of the atom out. In the particular p orbital shown, the probability of finding the electron is greater as you head straight up from the center of the atom than it is as you head straight to the left or to the right of the atom. It turns out that this dependence on direction is very important when it comes to studying how different atoms interact and form bonds. We'll talk more about that in a later chapter.

Figure : For p orbitals, the amount of electron density, and thus the probability of finding an electron, depends on both the distance from the center of the atom and the direction.

The d Orbitals and f Orbitals are Not Easily Visualized

Did you notice how the p orbitals looked a lot fancier than the simple spherical s orbitals? Well, you can imagine that if p orbitals with ℓ = 1 are fancy, then d orbitals, with ℓ = 2 are even fancier, and f orbitals, with ℓ = 3 are just plain crazy! Most people can visualize p orbitals, but d orbitals and f orbitals are actually rather difficult to imagine. Most d orbitals are butterfly shaped, although one has an unusual shape that looks like a doughnut surrounding a Q-tip! The 5 possible d orbitals are shown in Figure 6.19 (and you could have figured out that there were 5, right?). Don't worry too much about why one of the d orbitals is different. Again, it takes a lot of complex math to understand where the different d orbital pictures came from, and you won't have to worry about that unless you decide to go on and study quantum chemistry at the university level. As for f orbitals, they are even hard to draw, not to mention the fact that there are a total of 7 of them (Figure 6.20)! We won't be too concerned with d orbitals in this course, although they do become very important if you want to study certain metals like those found in the center of the periodic table. Similarly, f orbitals aren't all that important when it comes to common chemicals like hydrogen, or oxygen, or even copper. They do become important, though when you want to study some of the most important radioactive elements like uranium and plutonium!

Figure : The probability patterns for the d orbitals.

Figure : The probability patterns for the f orbitals.

Quantum Numbers and Atomic Orbitals

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Quantum Numbers and Atomic Orbitals

By solving the Schrödinger equation (Hy = Ey), we obtain a set of mathematical equations, called wave functions (y), which describe the probability of finding electrons at certain energy levels within an atom.
A wave function for an electron in an atom is called an atomic orbital; this atomic orbital describes a region of space in which there is a high probability of finding the electron. Energy changes within an atom are the result of an electron changing from a wave pattern with one energy to a wave pattern with a different energy (usually accompanied by the absorption or emission of a photon of light).
Each electron in an atom is described by four different quantum numbers. The first three (n, l, m_l) specify the particular orbital of interest, and the fourth (m_s) specifies how many electrons can occupy that orbital.

Principal Quantum Number (n): n = 1, 2, 3, …, ∞
Specifies the energy of an electron and the size of the orbital (the distance from the nucleus of the peak in a radial probability distribution plot). All orbitals that have the same value of n are said to be in the same shell (level). For a hydrogen atom with n=1, the electron is in its ground state; if the electron is in the n=2 orbital, it is in an excited state. The total number of orbitals for a given n value is n².

Angular Momentum (Secondary, Azimunthal) Quantum Number (l): l = 0, ..., n-1.
Specifies the shape of an orbital with a particular principal quantum number. The secondary quantum number divides the shells into smaller groups of orbitals called subshells (sublevels). Usually, a letter code is used to identify l to avoid confusion with n:

l	0	1	2	3	4	5	...
Letter	s	p	d	f	g	h	...

The subshell with n=2 and l=1 is the 2p subshell; if n=3 and l=0, it is the 3s subshell, and so on. The value of l also has a slight effect on the energy of the subshell; the energy of the subshell increases with l (s < p < d < f).

Magnetic Quantum Number (m_l): m_l = -l, ..., 0, ..., +l.
Specifies the orientation in space of an orbital of a given energy (n) and shape (l). This number divides the subshell into individual orbitals which hold the electrons; there are 2l+1 orbitals in each subshell. Thus the s subshell has only one orbital, the p subshell has three orbitals, and so on.

Spin Quantum Number (m_s): m_s = +½ or -½.
Specifies the orientation of the spin axis of an electron. An electron can spin in only one of two directions (sometimes called up and down).
The Pauli exclusion principle (Wolfgang Pauli, Nobel Prize 1945) states that no two electrons in the same atom can have identical values for all four of their quantum numbers. What this means is that no more than two electrons can occupy the same orbital, and that two electrons in the same orbital must have opposite spins.
Because an electron spins, it creates a magnetic field, which can be oriented in one of two directions. For two electrons in the same orbital, the spins must be opposite to each other; the spins are said to be paired. These substances are not attracted to magnets and are said to be diamagnetic. Atoms with more electrons that spin in one direction than another contain unpaired electrons. These substances are weakly attracted to magnets and are said to be paramagnetic.

Table of Allowed Quantum Numbers

n	l	*m_l*	Number of orbitals	Orbital Name	Number of electrons
1	0	0	1	1s	2
2	0	0	1	2s	2
	1	-1, 0, +1	3	2p	6
3	0	0	1	3s	2
	1	-1, 0, +1	3	3p	6
	2	-2, -1, 0, +1, +2	5	3d	10
4	0	0	1	4s	2
	1	-1, 0, +1	3	4p	6
	2	-2, -1, 0, +1, +2	5	4d	10
	3	-3, -2, -1, 0, +1, +2, +3	7	4f	14

Writing Electron Configurations

The distribution of electrons among the orbitals of an atom is called the electron configuration. The electrons are filled in according to a scheme known as the Aufbau principle ("building-up"), which corresponds (for the most part) to increasing energy of the subshells:

1s, 2s, 2p, 3s, 3p, 4s, 3d, 4p, 5s, 4d, 5p, 6s, 4f, 5d, 6p, 7s, 5f

It is not necessary to memorize this listing, because the order in which the electrons are filled in can be read from the periodic table in the following fashion:

Or, to summarize:

In electron configurations, write in the orbitals that are occupied by electrons, followed by a superscript to indicate how many electrons are in the set of orbitals (e.g., H 1s¹)
Another way to indicate the placement of electrons is an orbital diagram, in which each orbital is represented by a square (or circle), and the electrons as arrows pointing up or down (indicating the electron spin). When electrons are placed in a set of orbitals of equal energy, they are spread out as much as possible to give as few paired electrons as possible (Hund's rule).

examples will be added at a later date

In a ground state configuration, all of the electrons are in as low an energy level as it is possible for them to be. When an electron absorbs energy, it occupies a higher energy orbital, and is said to be in an excited state.

Properties of Monatomic Ions

The electrons in the outermost shell (the ones with the highest value of n) are the most energetic, and are the ones which are exposed to other atoms. This shell is known as the valence shell. The inner, core electrons (inner shell) do not usually play a role in chemical bonding.
Elements with similar properties generally have similar outer shell configurations. For instance, we already know that the alkali metals (Group I) always form ions with a +1 charge; the "extra" s¹ electron is the one that's lost:

IA	Li	1s²2s¹	Li+	1s²
	Na	1s²2s²2p⁶3s¹	Na+	1s²2s²2p⁶
	K	1s²2s²2p⁶3s²3p⁶4s¹	K+	1s²2s²2p⁶3s²3p⁶

The next shell down is now the outermost shell, which is now full — meaning there is very little tendency to gain or lose more electrons. The ion's electron configuration is the same as the nearest noble gas — the ion is said to be isoelectronic with the nearest noble gas. Atoms "prefer" to have a filled outermost shell because this is more electronically stable.

The Group IIA and IIIA metals also tend to lose all of their valence electrons to form cations.

IIA	Be	1s²2s²	Be²⁺	1s²
	Mg	1s²2s²2p⁶3s²	Mg²⁺	1s²2s²2p⁶
IIIA	Al	1s²2s²2p⁶3s²3p¹	Al3+	1s²2s²2p⁶

The Group IV and V metals can lose either the electrons from the p subshell, or from both the s and p subshells, thus attaining a pseudo-noble gas configuration.

IVA	Sn	[Kr]4d¹⁰5s²5p²	Sn²⁺	[Kr]4d¹⁰5s²
			Sn⁴⁺	[Kr]4d¹⁰
	Pb	[Xe]4f¹⁴5d¹⁰6s²6p²	Pb²⁺	[Xe]4f¹⁴5d¹⁰6s²
			Pb⁴⁺	[Xe]4f¹⁴5d¹⁰
VA	Bi	[Xe]4f¹⁴5d¹⁰6s²6p³	Bi³⁺	[Xe]4f¹⁴5d¹⁰6s²
			Bi⁵⁺	[Xe]4f¹⁴5d¹⁰

The Group IV - VII non-metals gain electrons until their valence shells are full (8 electrons).

IVA	C	1s²2s²2p²	C4-	1s²2s²2p⁶
VA	N	1s²2s²2p³	N3-	1s²2s²2p⁶
VIA	O	1s²2s²2p⁴	O2-	1s²2s²2p⁶
VIIA	F	1s²2s²2p⁵	F-	1s²2s²2p⁶

The Group VIII noble gases already possess a full outer shell, so they have no tendency to form ions.

VIIIA	Ne	1s²2s²2p⁶
	Ar	1s²2s²2p⁶3s²3p⁶

Transition metals (B-group) usually form +2 charges from losing the valence s electrons, but can also lose electrons from the highest d level to form other charges.

B-group	Fe	1s²2s²2p⁶3s²3p⁶3d⁶4s²	Fe²⁺	1s²2s²2p⁶3s²3p⁶3d⁶
			Fe³⁺	1s²2s²2p⁶3s²3p⁶3d⁵

Atomic Sub Shells

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Atomic Sub Shells

Atoms have electrons arranged in definite energy levels. The electrons within the first quantum shell are closest to the nucleus and have the lowest energies.
Subsequent energy levels further away from the nucleus have higher energies.
The shells are labelled by assigning each one a principal number, n, where n = 1, 2, 3…
Evidence for electrons being arranged in energy levels is given by atomic spectroscopy and successive ionisation energy values.

The energy levels are evident due to the large increase in ionisation energies after the removal of certain electrons.
The number of electrons allowed in one energy level can be summarised by the formula 2n² where n = energy level quantum number.
Within energy levels in atoms, there are also sub energy levels, or orbitalss.
The orbitals are labelled s, p, d and f. In the first energy level, there is only one orbital (s). The second shell has two orbitals (s and p), the third three (s, p, d) and the fourth has four (s, p, d, f).
The different orbitals can hold a different number of electrons:
- 1 s orbital can hold 2
- 3 p orbitals can hold 6
- 5 d orbitals can hold 10
- 7 f orbitals can hold 14

This explains why the total number of electrons in each energy level is 2n².

The sub shells can be summarised in the diagram below, notice that the n=3 and n=4 sub shells overlap.

Atomic Orbitals

The s, p, d and f s orbitals compose of regions of space where the electrons can be found.

Electrons within the same orbitals have the same amount of energy.
The electrons do not lie within a specific place within an orbital, but can be found anywhere around the orbital.
Each atomic orbital can contain 2 electrons, which can spin in two directions , the electrons within the same orbital spin in opposite directions.
To describe accurately the state of an electron, you need to be able to state:
1. The electron shell it is in (1, 2, 3, 4….)
2. The orbital it is in within the electron shell (s, p, d, f)
3. Its spin.

Filling the orbitals

The arrangement of electrons within an atom, is referred to as the electronic configuration of an atom.
The orbitals are filled within a definite order to produce the lowest energy arrangement available.
The lowest energy levels are always filled first.
Where there is more than one orbital within the same energy, the separate orbitals are filled separately with the same spin (think of bus analogy, i.e. people will aim to sit on their own seat on the bus, but when the seats are full, they will be forced to sit on a seat with someone else!). Then the electrons pair up in orbitals.

Difference between shells, subshells and orbitals

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Here's a graphic I use to explain the difference :

All electrons that have the same value for n (the principle quantum number) are in the same shell
Within a shell (same n), all electrons that share the same l (the angular momentum quantum number, or orbital shape) are in the same sub-shell

When electrons share the same n, l, and ml, we say they are in the same orbital (they have the same energy level, shape, and orientation)

So to summarize:

same n - shell
same n and l - sub-shell
same n, l, and ml - orbital

Now, in the other answer, there is some discussion about spin-orbitals, meaning that each electron would exist in its own orbital. For practical purposes, you don't need to worry about that - by the time those sorts of distinctions matter to you, there won't be any confusion about what people mean by "shells" and "sub-shells." For you, for now, orbital means "place where up to two electrons can exist," and they will both share the same

l, and

ml values, but have opposite spins (

ms).

Orbitals that have the same value of the principal quantum number n form a shell. Orbitals within a shell are divided into subshells that have the same value of the angular quantum number l. Chemists describe the shell and subshell in which an orbital belongs with a two-character code such as 2p or 4f. The first character indicates the shell (n = 2 or n = 4). The second character identifies the subshell. By convention, the following lowercase letters are used to indicate different subshells.

s: l = 0

p: l = 1

d: l = 2

f: l = 3

What is called an orbital might differ according to the context. With orbitals in the context of shells and subshells one usually means atomic orbitals, i.e. two-electron eigenstates of an atom's Hamilton operator which are characterized by the three quantum numbers: the principal quantum number

n, the angular quantum number

l and the magnetic quantum number

m. But often the word orbital is also used for spin-orbitals, i.e. one-electron eigenstates of the system's one-electron Hamilton operator which are characterized not only by

l and

m but also by the spin quantum number

ms which can be either

+12 or

−12.

Nucleons Calculation

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Nucleons

The nuclei of atoms are made up of protons and neutrons. These two components of the nucleus are referred to as nucleons. The electrons occupy the space outside the nucleus. Since an atom is electrically neutral, the number of protons in the nucleus is exactly equal to the number of electrons. This number is the atomic number given by the symbol Z.
Atomic number represents the number of protons in an atom. As atoms are electrically neutral, an atom contains as many electrons as it has protons. The total number of protons and neutrons present in one atom of an element is known as its mass number.Mass number = number of protons + number of neutrons
It can also be said that:Mass number = atomic number + number of neutrons
Atomic number and mass number are represented on the symbol of an element as follows:

Example:

Represent the atom of sodium whose atomic mass is 23 and atomic number is 11. Calculate the number of protons, electrons and neutrons.

Atomic number Z -11Atomic mass A - 23No. of protons = Z = 11 No. of electrons = 11No. of neutrons = A - Z23 - 11 = 12

Ionization Energy

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Ionization Energy Definition

The electrons are raised to higher energy levels by absorption of energy from an external source. If this process is continued, a stage comes when the electron goes completely out of the influence of the nucleus and a positive ion is produced. The electron in an atom is attracted by the positively charged nucleus.
In order to remove an electron from an atom, energy has to be supplied to it to overcome the attractive force. This energy is referred to as ionization energy or ionization potential.
"The amount of energy required to remove the most loosely bound electron or the outermost electron from an isolated gaseous atom of an element in its lowest energy state or the ground state to produce a cation is known as ionization potential or ionization energy of that element."

A_(g) + Energy A⁺_(g) + electron

It is generally represented as I or IP and is measured in electron volts (eV) or kilo calories (K calories) per gram of atom.

1eV atom = 23.06 kcal / mole = 96.3 kj/mole

Ionization potential is a very important property which gives an idea about the tendency of an atom to form a gaseous positive ion. The smaller the value of ionization energy, the easier it is to remove the electron from the atom.

The process by which the element loses an electron, to convert itself into a cation is called its ionization. This process is an endothermic process, since the energy is supplied to effect it.

Ionization Energy Trend

In a group, as we move from top to bottom, ionization potential decreases. As we move down the group, with the increase in atomic number, new shells ass on and the size of the atom gradually increases. The distance between the nucleus and the outermost shell increases and the screening effect also increases.

As a result, the attractive force of the nucleus on the outermost electron decreases. Thereby, the amount of energy required to remove an electron from the atom decreases. Ionization potential, thus, decreases down a group.

Element	kJ/mol	eV/atom
H	1312.0	13.595
Li	520.1	5.390
Na	459.2	5.138
K	418.7	4.339
Rb	403.0	4.176
Cs	375.7	3.893

The element with highest ionization potential in the periodic table is Helium (2372.1 kj/mol), while the element with the lowest ionization potential is Caesium (375.7 kj/mol).

First Ionization Energy

First ionization energy of an atom is the energy required to remove the first electron from the outermost shell of an atom.

Once the first electron has been removed from the gaseous atom, it is possible to remove second and successive electrons from positive ions one after the other.

Ionization Energy Periodic Table

In a period, as we move from left to right, the ionization energy increases.
As we move along a period, the nuclear charge increases and the electrons are added into the same shell. Thereby the effective nuclear charge increases and size decreases. Therefore, the energy required to remove an electron increases.

Elements	Li	Be	B	C	N	O	F	Ne
Ionization Potential	520	900	802	1085	1398	1315	1680	2080

However, some irregularities in the general trend have been noticed. These are due to half - filled and completely filled configurations which have extra stability.

To illustrate this, let us consider the ionization energies beryllium and boron. The ionization energy of boron is slightly less than that of beryllium even though boron has a greater nuclear charge. This can be understood by comparing the electronic configurations of Be (1s², 2s² ) and B (1s², 2s², sp¹).

In the case of Beryllium, the electron removed during ionization is an s- electron whereas the electron removed in the case of boron is p-electron. The penetration of a 2s electron to the nucleus is more than that of a 2p electron, hence the 2p - electron of boron is more shielded than the 2s electron of Beryllium. Therefore, ionization energy of Beryllium is more than that of Boron.

Second Ionization Energy

Once the first ionization electron is removed from the gaseous atom, it is possible to remove second and successive electrons from positive ions one after the other. Removal of a second electron from an already ionized gaseous atom is called as second ionization energy.

A_(g)+ IE₁ → A⁺_(g) + e^- [First Ionization]
A⁺_(g)+ IE₁ → A²⁺_(g) + e^- [Second ionization]

The amount of energies required to remove most loosely bound electron from unipositive, dipositive, tripositive .... ions of the element in gaseous state are called second, third, fourth, etc ionization energies respectively.

The second, third and fourth, etc ionization energies are collectively called as successive ionization energies.

It is also seen that IE₃ > IE₂ > IE₁.

Ionization Energy Chart

Ionization chart of II A group elements are as follows.

Element	Number of inner shells	eV/atom
Be	1	9.3
Ma	2	7.6
Ca	3	6.1
Sr	4	5.7
Ba	5	5.2

Factors Affecting Ionization Potential

The magnitude of Ionization potential depends upon:

1. Charge in the nucleus

The greater the charge on the nucleus, the more difficult it will be to remove an outermost electron from that atom. Thus, greater the nuclear charge of an atom, greater will be its ionization potential. Halogens have highest ionization potential in their respective periods.

2. Atomic radius

With the increase in the atomic radius, the ionization potential decreases. This is because of the fact that in case of larger atoms the attraction between the nucleus and the outer most electron is less and hence it is easier to remove an electron from a larger atom than from a smaller one.

This can be seen from the fact that as we move down a group, the size increases, while ionization potential decreases.

3. Completely and partially filled in shells

The atoms having a completely filled or partially filled outer most shell are comparatively more stable than the atoms with incompletely filled outer shells. This, the ionization potential of such atoms are relatively higher than expected normally from their positions in the periodic table.

Successive Ionization Energies

The amount of energy necessary for removing the subsequent electrons of a gaseous atom is better known as successive ionization energies. These are basically termed as 1st , 2nd, 3rd or 4th …… ionization energy and these depends completely on the removal of the 1st, 2nd, 3rd electrons respectively.

2nd ionization energy amount are higher compared to 1st because after the first electron is removed the atom changes into a positive monovalent ion.
The nuclear charge remains same although there is a decrease in electron number and this ultimately leads to a situation where the remaining electrons getting held more tightly by the nucleus which ultimately makes it very difficult to remove the second electron.
The value of the 2nd ionization energy (IE2) is more than the 1st ionization energy (IE1). Removing 2nd electron would ultimately lead to forming a di-positive ion which results in a stronger attraction between the nucleus and remaining electrons. Higher value of ionization energy results from this.

Factors Affecting Ionization Potential

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Factors Affecting Ionization Potential

The magnitude of Ionization potential depends upon:

1. Charge in the nucleus

2. Atomic radius

3. Completely and partially filled in shells

You can then have as many successive ionisation energies as there are electrons in the original atom.
The first four ionisation energies of aluminium, for example, are given by

		1st I.E. = 577 kJ mol^-1
		2nd I.E. = 1820 kJ mol^-1
		3rd I.E. = 2740 kJ mol^-1
		4th I.E. = 11600 kJ mol^-1

In order to form an Al³⁺_(g) ion from Al_(g) you would have to supply:
577 + 1820 + 2740 = 5137 kJ mol^-1
That's a lot of energy. Why, then, does aluminium form Al³⁺ ions?
It can only form them if it can get that energy back from somewhere, and whether that's feasible depends on what it is reacting with.
For example, if aluminium reacts with fluorine or oxygen, it can recover that energy in various changes involving the fluorine or oxygen - and so aluminium fluoride or aluminium oxide contain Al³⁺ ions.
If it reacts with chlorine, it can't recover sufficient energy, and so solid anhydrous aluminium chloride isn't actually ionic - instead, it forms covalent bonds.

Why doesn't aluminium form an Al⁴⁺ ion? The fourth ionisation energy is huge compared with the first three, and there is nothing that aluminium can react with which would enable it to recover that amount of extra energy.
Why do successive ionisation energies get larger?
Once you have removed the first electron you are left with a positive ion. Trying to remove a negative electron from a positive ion is going to be more difficult than removing it from an atom. Removing an electron from a 2+ or 3+ (etc) ion is going to be progressively more difficult.
Why is the fourth ionisation energy of aluminium so large?
The electronic structure of aluminium is 1s²2s²2p⁶3s²3p_x¹. The first three electrons to be removed are the three electrons in the 3p and 3s orbitals. Once they've gone, the fourth electron is removed from the 2p level - much closer to the nucleus, and only screened by the 1s² (and to some extent the 2s²) electrons.
Using ionisation energies to work out which group an element is in
This big jump between two successive ionisation energies is typical of suddenly breaking in to an inner level. You can use this to work out which group of the Periodic Table an element is in from its successive ionisation energies.
Magnesium (1s²2s²2p⁶3s²) is in group 2 of the Periodic Table and has successive ionisation energies:

Here the big jump occurs after the second ionisation energy. It means that there are 2 electrons which are relatively easy to remove (the 3s² electrons), while the third one is much more difficult (because it comes from an inner level - closer to the nucleus and with less screening).
Silicon (1s²2s²2p⁶3s²3p_x¹3p_y¹) is in group 4 of the Periodic Table and has successive ionisation energies:

Here the big jump comes after the fourth electron has been removed. The first 4 electrons are coming from the 3-level orbitals; the fifth from the 2-level.
The lesson from all this:
Count the easy electrons - those up to (but not including) the big jump. That is the same as the group number.
Another example:
Decide which group an atom is in if it has successive ionisation energies:

The ionisation energies are going up one or two thousand at a time for the first five. Then there is a huge jump of about 15000. There are 5 relatively easy electrons - so the element is in group 5.
Exploring the patterns in more detail
If you plot graphs of successive ionisation energies for a particular element, you can see the fluctuations in it caused by the different electrons being removed.
Not only can you see the big jumps in ionisation energy when an electron comes from an inner level, but you can also see the minor fluctuations within a level depending on whether the electron is coming from an s or a p orbital, and even whether it is paired or unpaired in that orbital.
Chlorine has the electronic structure 1s²2s²2p⁶3s²3p_x²3p_y²3p_z¹.
This graph plots the first eight ionisation energies of chlorine. The green labels show which electron is being removed for each of the ionisation energies.

If you put a ruler on the first and second points to establish the trend, you'll find that the third, fourth and fifth points lie above the value you would expect. That is because the first two electrons are coming from pairs in the 3p levels and are therefore rather easier to remove than if they were unpaired.
Again, if you put a ruler on the 3rd, 4th and 5th points to establish their trend, you'll find that the 6th and 7th points lie well above the values you would expect from a continuation of the trend. That is because the 6th and 7th electrons are coming from the 3s level - slightly closer to the nucleus and slightly less well screened.
The massive jump as you break into the inner level at the 8th electron is fairly obvious!

Warning! People sometimes get confused with these graphs because they forget that they are removing electrons from the atom. For example, the first point refers to the first electron being lost - from a 3p orbital. Basically, you start from the outside of the atom and work in towards the middle. If you start from the 1s orbital and work outwards, you are doomed to failure!

To plot any more ionisation energies for chlorine needs a change of vertical scale. The seventeenth ionisation energy of chlorine is nearly 400,000 kJ mol^-1, and the vertical scale has to be squashed to accommodate this.

This is now a "log graph" - plotted by finding the logarithm of each ionisation energy (press the "log" button on your calculator). This doesn't simply squash the vertical scale. It distorts it as well, to such an extent that the only useful thing the graph now shows is the major jumps where the next electron to be removed comes from an inner level. The distortion is so great in the first 8 ionisation energies, for example, that the patterns shown by the previous graph are completely (and misleadingly) destroyed.

Electron Affinity

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Electron Affinity

Ionization energies measure the tendency of a neutral atom to resist the loss of electrons. It takes a considerable amount of energy, for example, to remove an electron from a neutral fluorine atom to form a positively charged ion.

F(g) F⁺(g) + e^-

H^o = 1681.0 kJ/mol

The electron affinity of an element is the energy given off when a neutral atom in the gas phase gains an extra electron to form a negatively charged ion. A fluorine atom in the gas phase, for example, gives off energy when it gains an electron to form a fluoride ion.

F(g) + e^- F^-(g)

H^o = -328.0 kJ/mol

Electron affinities are more difficult to measure than ionization energies and are usually known to fewer significant figures. The electron affinities of the main group elements are shown in the figure below.

Several patterns can be found in these data.

Electron affinities generally become smaller as we go down a column of the periodic table for two reasons. First, the electron being added to the atom is placed in larger orbitals, where it spends less time near the nucleus of the atom. Second, the number of electrons on an atom increases as we go down a column, so the force of repulsion between the electron being added and the electrons already present on a neutral atom becomes larger.
Electron affinity data are complicated by the fact that the repulsion between the electron being added to the atom and the electrons already present on the atom depends on the volume of the atom. Among the nonmetals in Groups VIA and VIIA, this force of repulsion is largest for the very smallest atoms in these columns: oxygen and fluorine. As a result, these elements have a smaller electron affinity than the elements below them in these columns as shown in the figure below. From that point on, however, the electron affinities decrease as we continue down these columns.

At first glance, there appears to be no pattern in electron affinity across a row of the periodic table, as shown in the figure below.

When these data are listed along with the electron configurations of these elements, however, they make sense. These data can be explained by noting that electron affinities are much smaller than ionization energies. As a result, elements such as helium, beryllium, nitrogen, and neon, which have unusually stable electron configurations, have such small affinities for extra electrons that no energy is given off when a neutral atom of these elements picks up an electron. These configurations are so stable that it actually takes energy to force one of these elements to pick up an extra electron to form a negative ion.

Electron Affinities and Electron Configurations for the First 10 Elements in the Periodic Table

Element	Electron Affinity (kJ/mol)	Electron Configuration
H	72.8	1s¹
He	<0 td="">	1s²
Li	59.8	[He] 2s¹
Be	<0 td="">	[He] 2s²
B	27	[He] 2s² 2p¹
C	122.3	[He] 2s² 2p²
N	<0 td="">	[He] 2s² 2p³
O	141.1	[He] 2s² 2p⁴
F	328.0	[He] 2s² 2p⁵
Ne	<0 td="">	[He] 2s² 2p⁶

Consequences of the Relative Size of Ionization Energies and Electron Affinities

Students often believe that sodium reacts with chlorine to form Na⁺ and Cl^- ions because chlorine atoms "like" electrons more than sodium atoms do. There is no doubt that sodium reacts vigorously with chlorine to form NaCl.

2 Na(s) + Cl₂(g) 2 NaCl(s)

Furthermore, the ease with which solutions of NaCl in water conduct electricity is evidence for the fact that the product of this reaction is a salt, which contains Na⁺ and Cl^- ions.

NaCl(s)

H₂O

Na⁺(aq) + Cl^-(aq)

The only question is whether it is legitimate to assume that this reaction occurs because chlorine atoms "like" electrons more than sodium atoms.
The first ionization energy for sodium is one and one-half times larger than the electron affinity for chlorine.

Na: 1st IE = 495.8 kJ/mol

Cl: EA = 328.8 kJ/mol

Thus, it takes more energy to remove an electron from a neutral sodium atom than is given off when the electron is picked up by a neutral chlorine atom. We will obviously have to find another explanation for why sodium reacts with chlorine to form NaCl. Before we can do this, however, we need to know more about the chemistry of ionic compounds.

Ionization Energy

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The First Ionization Energy

The energy needed to remove one or more electrons from a neutral atom to form a positively charged ion is a physical property that influences the chemical behavior of the atom. By definition, the first ionization energy of an element is the energy needed to remove the outermost, or highest energy, electron from a neutral atom in the gas phase.
The process by which the first ionization energy of hydrogen is measured would be represented by the following equation.

H(g) H⁺(g) + e^-

H^o = -1312.0 kJ/mol

Practice Problem 3:Use the Bohr model to calculate the wavelength and energy of the photon that would have to be absorbed to ionize a neutral hydrogen atom in the gas phase.

The magnitude of the first ionization energy of hydrogen can be brought into perspective by comparing it with the energy given off in a chemical reaction. When we burn natural gas, about 800 kJ of energy is released per mole of methane consumed.

CH₄(g) + 2 O₂(g) CO₂(g) + 2 H₂O(g)

H^o = -802.4 kJ/mol

The thermite reaction, which is used to weld iron rails, gives off about 850 kJ of energy per mole of iron oxide consumed.

Fe₂O₃(s) + 2 Al(s) Al₂O₃(s) + 2 Fe(l)

H^o = -851.5 kJ/mol

The first ionization energy of hydrogen is half again as large as the energy given off in either of these reactions.

Patterns in the First Ionization Energies

The first ionization energy for helium is slightly less than twice the ionization energy for hydrogen because each electron in helium feels the attractive force of two protons, instead of one.

He(g) He⁺(g) + e^-

H^o = 2372.3 kJ/mol

It takes far less energy, however, to remove an electron from a lithium atom, which has three protons in its nucleus.

Li(g) Li⁺(g) + e^-

H^o = 572.3 kJ/mol

This can be explained by noting that the outermost, or highest energy, electron on a lithium atom is in the 2s orbital. Because the electron in a 2s orbital is already at a higher energy than the electrons in a 1s orbital, it takes less energy to remove this electron from the atom.
The first ionization energies for the main group elements are given in the two figures below.

Two trends are apparent from these data.

In general, the first ionization energy increases as we go from left to right across a row of the periodic table.
The first ionization energy decreases as we go down a column of the periodic table.

The first trend isn't surprising. We might expect the first ionization energy to become larger as we go across a row of the periodic table because the force of attraction between the nucleus and an electron becomes larger as the number of protons in the nucleus of the atom becomes larger.
The second trend results from the fact that the principal quantum number of the orbital holding the outermost electron becomes larger as we go down a column of the periodic table. Although the number of protons in the nucleus also becomes larger, the electrons in smaller shells and subshells tend to screen the outermost electron from some of the force of attraction of the nucleus. Furthermore, the electron being removed when the first ionization energy is measured spends less of its time near the nucleus of the atom, and it therefore takes less energy to remove this electron from the atom.

Exceptions to the General Pattern of First Ionization Energies

The figure below shows the first ionization energies for elements in the second row of the periodic table. Although there is a general trend toward an increase in the first ionization energy as we go from left to right across this row, there are two minor inversions in this pattern. The first ionization energy of boron is smaller than beryllium, and the first ionization energy of oxygen is smaller than nitrogen.

These observations can be explained by looking at the electron configurations of these elements. The electron removed when a beryllium atom is ionized comes from the 2s orbital, but a 2p electron is removed when boron is ionized.

Be: [He] 2s²

B: [He] 2s² 2p¹

The electrons removed when nitrogen and oxygen are ionized also come from 2p orbitals.

N: [He] 2s² 2p³

O: [He] 2s² 2p⁴

But there is an important difference in the way electrons are distributed in these atoms. Hund's rules predict that the three electrons in the 2p orbitals of a nitrogen atom all have the same spin, but electrons are paired in one of the 2p orbitals on an oxygen atom.

Hund's rules can be understood by assuming that electrons try to stay as far apart as possible to minimize the force of repulsion between these particles. The three electrons in the 2p orbitals on nitrogen therefore enter different orbitals with their spins aligned in the same direction. In oxygen, two electrons must occupy one of the 2p orbitals. The force of repulsion between these electrons is minimized to some extent by pairing the electrons. There is still some residual repulsion between these electrons, however, which makes it slightly easier to remove an electron from a neutral oxygen atom than we would expect from the number of protons in the nucleus of the atom.

Practice Problem 4:Predict which element in each of the following pairs has the larger first ionization energy.
(a) Na or Mg
(b) Mg or Al
(b) Mg or Al
(c) F or Cl

Second, Third, Fourth, and Higher Ionization Energies

By now you know that sodium forms Na⁺ ions, magnesium forms Mg²⁺ ions, and aluminum forms Al³⁺ ions. But have you ever wondered why sodium doesn't form Na²⁺ ions, or even Na³⁺ ions? The answer can be obtained from data for the second, third, and higher ionization energies of the element.
The first ionization energy of sodium, for example, is the energy it takes to remove one electron from a neutral atom.

Na(g) + energy Na⁺(g) + e^-

The second ionization energy is the energy it takes to remove another electron to form an Na²⁺ ion in the gas phase.

Na⁺(g) + energy Na²⁺(g) + e^-

The third ionization energy can be represented by the following equation.

Na²⁺(g) + energy Na³⁺(g) + e^-

The energy required to form a Na³⁺ ion in the gas phase is the sum of the first, second, and third ionization energies of the element.

First, Second, Third, and Fourth Ionization Energies
of Sodium, Magnesium, and Aluminum (kJ/mol)

It doesn't take much energy to remove one electron from a sodium atom to form an Na⁺ ion with a filled-shell electron configuration. Once this is done, however, it takes almost 10 times as much energy to break into this filled-shell configuration to remove a second electron. Because it takes more energy to remove the second electron than is given off in any chemical reaction, sodium can react with other elements to form compounds that contain Na⁺ ions but not Na²⁺ or Na³⁺ ions.
A similar pattern is observed when the ionization energies of magnesium are analyzed. The first ionization energy of magnesium is larger than sodium because magnesium has one more proton in its nucleus to hold on to the electrons in the 3s orbital.

Mg: [Ne] 3s²

The second ionization energy of Mg is larger than the first because it always takes more energy to remove an electron from a positively charged ion than from a neutral atom. The third ionization energy of magnesium is enormous, however, because the Mg²⁺ ion has a filled-shell electron configuration.
The same pattern can be seen in the ionization energies of aluminum. The first ionization energy of aluminum is smaller than magnesium. The second ionization energy of aluminum is larger than the first, and the third ionization energy is even larger. Although it takes a considerable amount of energy to remove three electrons from an aluminum atom to form an Al³⁺ ion, the energy needed to break into the filled-shell configuration of the Al³⁺ ion is astronomical. Thus, it would be a mistake to look for an Al⁴⁺ ion as the product of a chemical reaction.

Practice Problem 5:Predict the group in the periodic table in which an element with the following ionization energies would most likely be found.

1st IE = 786 kJ/mol

2nd IE = 1577

3rd IE = 3232

4th IE = 4355

5th IE = 16,091

6th IE = 19,784

Practice Problem 6:Use the trends in the ionization energies of the elements to explain the following observations.
(a) Elements on the left side of the periodic table are more likely than those on the right to form positive ions.
(b) The maximum positive charge on an ion is equal to the group number of the element

ELECTRONIC STRUCTURES

2015-07-10T01:18:00.000+06:00

ELECTRONIC STRUCTURES This page explores how you write electronic structures for atoms using s, p, and d notation. It assumes that you know about simple atomic orbitals - at least as far as the way they are named, and their relative energies. If you want to look at the electronic structures of simple monatomic ions (such as Cl^-, Ca²⁺ and Cr³⁺), you will find a link at the bottom of the page.

The electronic structures of atoms Relating orbital filling to the Periodic Table

UK syllabuses for 16 - 18 year olds tend to stop at krypton when it comes to writing electronic structures, but it is possible that you could be asked for structures for elements up as far as barium. After barium you have to worry about f orbitals as well as s, p and d orbitals - and that's a problem for chemistry at a higher level. It is important that you look through past exam papers as well as your syllabus so that you can judge how hard the questions are likely to get.
This page looks in detail at the elements in the shortened version of the Periodic Table above, and then shows how you could work out the structures of some bigger atoms.

The first period
Hydrogen has its only electron in the 1s orbital - 1s¹, and at helium the first level is completely full - 1s².
The second period
Now we need to start filling the second level, and hence start the second period. Lithium's electron goes into the 2s orbital because that has a lower energy than the 2p orbitals. Lithium has an electronic structure of 1s²2s¹. Beryllium adds a second electron to this same level - 1s²2s².
Now the 2p levels start to fill. These levels all have the same energy, and so the electrons go in singly at first.

B		1s²2s²2p_x¹
C		1s²2s²2p_x¹2p_y¹
N		1s²2s²2p_x¹2p_y¹2p_z¹

The next electrons to go in will have to pair up with those already there.

O		1s²2s²2p_x²2p_y¹2p_z¹
F		1s²2s²2p_x²2p_y²2p_z¹
Ne		1s²2s²2p_x²2p_y²2p_z²

You can see that it is going to get progressively tedious to write the full electronic structures of atoms as the number of electrons increases. There are two ways around this, and you must be familiar with both.
Shortcut 1: All the various p electrons can be lumped together. For example, fluorine could be written as 1s²2s²2p⁵, and neon as 1s²2s²2p⁶.
This is what is normally done if the electrons are in an inner layer. If the electrons are in the bonding level (those on the outside of the atom), they are sometimes written in shorthand, sometimes in full. Don't worry about this. Be prepared to meet either version, but if you are asked for the electronic structure of something in an exam, write it out in full showing all the p_x, p_y and p_z orbitals in the outer level separately.
For example, although we haven't yet met the electronic structure of chlorine, you could write it as 1s²2s²2p⁶3s²3p_x²3p_y²3p_z¹.
Notice that the 2p electrons are all lumped together whereas the 3p ones are shown in full. The logic is that the 3p electrons will be involved in bonding because they are on the outside of the atom, whereas the 2p electrons are buried deep in the atom and aren't really of any interest.
Shortcut 2: You can lump all the inner electrons together using, for example, the symbol [Ne]. In this context, [Ne] means the electronic structure of neon - in other words: 1s²2s²2p_x²2p_y²2p_z² You wouldn't do this with helium because it takes longer to write [He] than it does 1s².
On this basis the structure of chlorine would be written [Ne]3s²3p_x²3p_y²3p_z¹.
The third period
At neon, all the second level orbitals are full, and so after this we have to start the third period with sodium. The pattern of filling is now exactly the same as in the previous period, except that everything is now happening at the 3-level.
For example:

		short version
Mg	1s²2s²2p⁶3s²	[Ne]3s²
S	1s²2s²2p⁶3s²3p_x²3p_y¹3p_z¹	[Ne]3s²3p_x²3p_y¹3p_z¹
Ar	1s²2s²2p⁶3s²3p_x²3p_y²3p_z²	[Ne]3s²3p_x²3p_y²3p_z²

The beginning of the fourth period
At this point the 3-level orbitals aren't all full - the 3d levels haven't been used yet. But if you refer back to the energies of the orbitals, you will see that the next lowest energy orbital is the 4s - so that fills next.

K		1s²2s²2p⁶3s²3p⁶4s¹
Ca		1s²2s²2p⁶3s²3p⁶4s²

There is strong evidence for this in the similarities in the chemistry of elements like sodium (1s²2s²2p⁶3s¹) and potassium (1s²2s²2p⁶3s²3p⁶4s¹)
The outer electron governs their properties and that electron is in the same sort of orbital in both of the elements. That wouldn't be true if the outer electron in potassium was 3d¹.
s- and p-block elements

The elements in Group 1 of the Periodic Table all have an outer electronic structure of ns¹ (where n is a number between 2 and 7). All Group 2 elements have an outer electronic structure of ns². Elements in Groups 1 and 2 are described as s-block elements.
Elements from Group 3 (the boron group) across to the noble gases all have their outer electrons in p orbitals. These are then described as p-block elements.

d-block elements
We are working out the electronic structures of the atoms using the Aufbau ("building up") Principle. So far we have got to calcium with a structure of 1s²2s²2p⁶3s²3p⁶4s².
The 4s level is now full, and the structures of the next atoms show electrons gradually filling up the 3d level. These are known as d-block elements.

Once the 3d orbitals have filled up, the next electrons go into the 4p orbitals as you would expect.
d-block elements are elements in which the last electron to be added to the atom using the Aufbau Principle is in a d orbital.
The first series of these contains the elements from scandium to zinc, which at GCSE you probably called transition elements or transition metals. The terms "transition element" and "d-block element" don't quite have the same meaning, but it doesn't matter in the present context.

d electrons are almost always described as, for example, d⁵ or d⁸ - and not written as separate orbitals. Remember that there are five d orbitals, and that the electrons will inhabit them singly as far as possible. Up to 5 electrons will occupy orbitals on their own. After that they will have to pair up.
d⁵ means
d⁸ means
Notice in what follows that all the 3-level orbitals are written together - with the 4s electrons written at the end of the electronic structure.

Sc		1s²2s²2p⁶3s²3p⁶3d¹4s²
Ti		1s²2s²2p⁶3s²3p⁶3d²4s²
V		1s²2s²2p⁶3s²3p⁶3d³4s²
Cr		1s²2s²2p⁶3s²3p⁶3d⁵4s¹

Whoops! Chromium breaks the sequence. In chromium, the electrons in the 3d and 4s orbitals rearrange so that there is one electron in each orbital. It would be convenient if the sequence was tidy - but it's not!

Mn	1s²2s²2p⁶3s²3p⁶3d⁵4s²	(back to being tidy again)
Fe	1s²2s²2p⁶3s²3p⁶3d⁶4s²
Co	1s²2s²2p⁶3s²3p⁶3d⁷4s²
Ni	1s²2s²2p⁶3s²3p⁶3d⁸4s²
Cu	1s²2s²2p⁶3s²3p⁶3d¹⁰4s¹	(another awkward one!)
Zn	1s²2s²2p⁶3s²3p⁶3d¹⁰4s²

And at zinc the process of filling the d orbitals is complete.
Filling the rest of period 4
The next orbitals to be used are the 4p, and these fill in exactly the same way as the 2p or 3p. We are back now with the p-block elements from gallium to krypton. Bromine, for example, is 1s²2s²2p⁶3s²3p⁶3d¹⁰4s²4p_x²4p_y²4p_z¹.

Summary
Writing the electronic structure of an element from hydrogen to krypton

Use the Periodic Table to find the atomic number, and hence number of electrons.

Fill up orbitals in the order 1s, 2s, 2p, 3s, 3p, 4s, 3d, 4p - until you run out of electrons. The 3d is the awkward one - remember that specially. Fill p and d orbitals singly as far as possible before pairing electrons up.

Remember that chromium and copper have electronic structures which break the pattern in the first row of the d-block.

Writing the electronic structure of big s- or p-block elements

First work out the number of outer electrons. This is quite likely all you will be asked to do anyway.
The number of outer electrons is the same as the group number. (The noble gases are a bit of a problem here, because they are normally called group 0 rather then group 8. Helium has 2 outer electrons; the rest have 8.) All elements in group 3, for example, have 3 electrons in their outer level. Fit these electrons into s and p orbitals as necessary. Which level orbitals? Count the periods in the Periodic Table (not forgetting the one with H and He in it).
Iodine is in group 7 and so has 7 outer electrons. It is in the fifth period and so its electrons will be in 5s and 5p orbitals. Iodine has the outer structure 5s²5p_x²5p_y²5p_z¹.
What about the inner electrons if you need to work them out as well? The 1, 2 and 3 levels will all be full, and so will the 4s, 4p and 4d. The 4f levels don't fill until after anything you will be asked about at A'level. Just forget about them! That gives the full structure: 1s²2s²2p⁶3s²3p⁶3d¹⁰4s²4p⁶4d¹⁰5s²5p_x²5p_y²5p_z¹.
When you've finished, count all the electrons to make sure that they come to the same as the atomic number. Don't forget to make this check - it's easy to miss an orbital out when it gets this complicated.
Barium is in group 2 and so has 2 outer electrons. It is in the sixth period. Barium has the outer structure 6s².
Including all the inner levels: 1s²2s²2p⁶3s²3p⁶3d¹⁰4s²4p⁶4d¹⁰5s²5p⁶6s².
It would be easy to include 5d¹⁰ as well by mistake, but the d level always fills after the next s level - so 5d fills after 6s just as 3d fills after 4s. As long as you counted the number of electrons you could easily spot this mistake because you would have 10 too many.

Inside the Atom

2015-07-10T01:14:00.000+06:00

Whats inside the atom?

Watch video -01

Watch Video- 02

Using radioactivity safely

2015-07-10T01:06:00.001+06:00

	Using radioactivity safely

		Ionising radiations are hazardous and it is best to avoid being exposed to them. However, some exposure is unavoidable (from natural as well as man-made sources) so an understanding of the hazards is essential. It means we can put them to good use and maintain safety.

	Radiation hazards

		Ionising radiations cause damage when they pass through a cell. The most direct method is if the radiation ionises the DNA in a cell. This may alter the function of the cell or kill it. There are also indirect methods in which water molecules are ionised producing free radicals (such as H⁺ and OH^-). These can then react with and damage parts of the cell. For low doses, the damage can and might cause physiological problems at a later date (even years later) such as cancer or genetic diseases. These can be caused by even the smallest dose – there is no such thing as a safe dose. However, although the outcome doesn't change as the dose decreases, the chances are reduced. It's similar to smoking-related diseases – lung cancer is no better or worse if you smoked 5 or 40 cigarettes a day. But the chance of contracting it increases with your intake. For high doses, the effects are more predictable. The unit for dose is the gray. Doses above 1 gray cause burns and other skin problems. A whole body dose of 5 grays can be fatal and 20 grays is always fatal. However, whilst we are all exposed to small doses, you won't be exposed to these higher doses in normal circumstances.

	Are all radiations equally hazardous?

		Low doses of the different types of ionising radiations are hazardous in different ways. Alpha radiation is almost harmless from the outside. It is absorbed by the outer layer of dead skin and doesn't reach any living cells. However, if an alpha emitter is ingested (either by inhaling it or eating something containing a radioactive substance), then there is a chance that an alpha particle will be emitted inside your body. It will then cause lots of ionisation within living cells. The equivalent dose from alpha radiation is reckonned to be twenty times that of beta and gamma because of the amount of ionisation it causes. To protect ourselves from alpha radiation, we need to ensure that no sources of alpha radiation contaminate our food, water or air supply. Beta radiation can cause skin problems but is, again, more of a hazard if a beta emitter is ingested. Gamma radiation has similar effects whether its source is inside or outside the body. In each case, we need to consider the source of radiation rather than the radiation itself. For example, the isotope radon-222 gives out alpha radiation. Where possible, we should avoid radon gas because this isotope will be passing in and out with each breath.

Picture 2.8 A smoke detector uses a small source of alpha radiation.

	Uses of alpha

		Smoke detectors The most effective smoke detectors use a source of alpha radiation – americium-241. The alpha radiation is aimed into a gap between two electrodes. There is a voltage across these electrodes. The alpha radiation ionises the air in the gap and allows a current to flow between the electrodes. When smoke finds its way between the electrodes, the air ions attract the smoke particles and stop flowing so freely. The resulting drop in current is detected by the circuitry and sets off the alarm. Luminous paint The old stye luminous paint is no longer used. However, you can still get second-hand watches that have luminous numerals painted on them. This paint is a mixture of radium salt and a fluorescing substance. The radium emits alpha particles which cause the fluorescent material to glow. Although this is perfectly safe for owners of the watches, the factory workers (who used to lick their paint brushes to get a good point) ingested the radium and suffered badly. Hence it was banned in the 1950s.

	Uses of beta

		Therapeutic Isotopes The main use of beta radiation is in the treatment of cancers. The radiation kills the cancer cells. It is sufficiently penetrating to allow it to be held in a container (if necessary) and to penetrate a number of cells. Yet it is sufficiently short range to limit its action to a local area, like a tumour. The beta radiation penetrates a few millimetres of tissue.Basic radio-therapy involves sealing a radioactive isotope, such as technetium-99, into a catheter or a probe. This is inserted into the patient and left in the area of the tumour. Technetium has a half life of six hours. This means it releases a lot of beta radiation in a short time. Another version is to put the radioactive istope in tiny metal spheres which are injected into a tumour. New systemic methods of radiotherapy are being developed. These involve modifying monoclonal antibodies so that they are carrying a radioactive isotope - often yttrium-90. These antibodies attach themselves to the cells of specific tissues, including tumours. In this way, the radiotherapy (in this case called radiotherapy) is extremely accurate in the tissue that it treats.

	Uses of gamma

		Medical investigations Sources of gamma radiation are used to see inside the body. Unlike X-rays, which have to go through the body, a gamma source can be injected into the body and its radiation produces pictures of specific areas. For example, technetium methylene diphosphonate is a chemical that will build up in bones. This can be made with radioactive technetium-99, which emits gamma radiation. The radiation can be used to expose photographic plates or be picked up by a scanner outside the body giving a very detaile picture of the bone structure. Thallium-201 is a potassium analogue that is taken up by tumours and emits gamma radiation. Again, using photographic plates, radiologists can get very accurate pictures of tumours – because they effectively emit their own radiation. Thallium-201 is also used to evaluate how well regions of the heart are pumping after a coronary artery bypass surgery. The thallium acts as a tracer, giving out gamma radiation. The radiation is captured in special detectors and an image is built up. By using 3 detectors placed around a body, it is possible to create a 3D image. This technique is called Emission Computed Tomography (ECT). In the case of the heart, the method is refined and is known as Single Photon ECT (SPECT). Industrial tracers Gamma sources can be used to detect leaks in pipes and to trace pollution. Scientists can label possible sources of the leak or pollution by introducing a short lived gamma emitter. If they detect gamma radiation later on, they can work out the source of the leak. Similarly if they're not sure where a leak is in a buried pipe, they can add a gamma source to the inflow and search for a build up of radiation. Sterilisation Gamma-emitting radioisotopes, such as cobalt-60, can be used to sterilise pharmaceuticals and cosmetics. Their gamma radiation kills bacteria without needing to heat up these heat-sensitive products. This technique can also be used in hospitals for medical equipment and is useful because instruments can be sterilised without opening the packages. Crack detection Gamma radiation can be used to search for cracks in welds and pipes and give an early warning of failure where they are wearing thin – useful for aeroplane safety checks. This works in a similar way to X-rays – the gamma radiation penetrates the metal a little but will show up as hot spots where there is a fault. However, unlike X-rays, the gamma source can be put in awkward places – it doesn't involve carrying around a high energy piece of machinery.

Using radioactivity safely

2015-07-10T01:05:00.003+06:00

	Using radioactivity safely

		Ionising radiations are hazardous and it is best to avoid being exposed to them. However, some exposure is unavoidable (from natural as well as man-made sources) so an understanding of the hazards is essential. It means we can put them to good use and maintain safety.

	Radiation hazards

		Ionising radiations cause damage when they pass through a cell. The most direct method is if the radiation ionises the DNA in a cell. This may alter the function of the cell or kill it. There are also indirect methods in which water molecules are ionised producing free radicals (such as H⁺ and OH^-). These can then react with and damage parts of the cell. For low doses, the damage can and might cause physiological problems at a later date (even years later) such as cancer or genetic diseases. These can be caused by even the smallest dose – there is no such thing as a safe dose. However, although the outcome doesn't change as the dose decreases, the chances are reduced. It's similar to smoking-related diseases – lung cancer is no better or worse if you smoked 5 or 40 cigarettes a day. But the chance of contracting it increases with your intake. For high doses, the effects are more predictable. The unit for dose is the gray. Doses above 1 gray cause burns and other skin problems. A whole body dose of 5 grays can be fatal and 20 grays is always fatal. However, whilst we are all exposed to small doses, you won't be exposed to these higher doses in normal circumstances.

	Are all radiations equally hazardous?

		Low doses of the different types of ionising radiations are hazardous in different ways. Alpha radiation is almost harmless from the outside. It is absorbed by the outer layer of dead skin and doesn't reach any living cells. However, if an alpha emitter is ingested (either by inhaling it or eating something containing a radioactive substance), then there is a chance that an alpha particle will be emitted inside your body. It will then cause lots of ionisation within living cells. The equivalent dose from alpha radiation is reckonned to be twenty times that of beta and gamma because of the amount of ionisation it causes. To protect ourselves from alpha radiation, we need to ensure that no sources of alpha radiation contaminate our food, water or air supply. Beta radiation can cause skin problems but is, again, more of a hazard if a beta emitter is ingested. Gamma radiation has similar effects whether its source is inside or outside the body. In each case, we need to consider the source of radiation rather than the radiation itself. For example, the isotope radon-222 gives out alpha radiation. Where possible, we should avoid radon gas because this isotope will be passing in and out with each breath.

Picture 2.8 A smoke detector uses a small source of alpha radiation.

	Uses of alpha

		Smoke detectors The most effective smoke detectors use a source of alpha radiation – americium-241. The alpha radiation is aimed into a gap between two electrodes. There is a voltage across these electrodes. The alpha radiation ionises the air in the gap and allows a current to flow between the electrodes. When smoke finds its way between the electrodes, the air ions attract the smoke particles and stop flowing so freely. The resulting drop in current is detected by the circuitry and sets off the alarm. Luminous paint The old stye luminous paint is no longer used. However, you can still get second-hand watches that have luminous numerals painted on them. This paint is a mixture of radium salt and a fluorescing substance. The radium emits alpha particles which cause the fluorescent material to glow. Although this is perfectly safe for owners of the watches, the factory workers (who used to lick their paint brushes to get a good point) ingested the radium and suffered badly. Hence it was banned in the 1950s.

	Uses of beta

		Therapeutic Isotopes The main use of beta radiation is in the treatment of cancers. The radiation kills the cancer cells. It is sufficiently penetrating to allow it to be held in a container (if necessary) and to penetrate a number of cells. Yet it is sufficiently short range to limit its action to a local area, like a tumour. The beta radiation penetrates a few millimetres of tissue.Basic radio-therapy involves sealing a radioactive isotope, such as technetium-99, into a catheter or a probe. This is inserted into the patient and left in the area of the tumour. Technetium has a half life of six hours. This means it releases a lot of beta radiation in a short time. Another version is to put the radioactive istope in tiny metal spheres which are injected into a tumour. New systemic methods of radiotherapy are being developed. These involve modifying monoclonal antibodies so that they are carrying a radioactive isotope - often yttrium-90. These antibodies attach themselves to the cells of specific tissues, including tumours. In this way, the radiotherpay (in this case called radioimmunotherapy) is extremely accurate in the tissue that it treats.

	Uses of gamma

		Medical investigations Sources of gamma radiation are used to see inside the body. Unlike X-rays, which have to go through the body, a gamma source can be injected into the body and its radiation produces pictures of specific areas. For example, technetium methylene diphosphonate is a chemical that will build up in bones. This can be made with radioactive technetium-99, which emits gamma radiation. The radiation can be used to expose photographic plates or be picked up by a scanner outside the body giving a very detaile picture of the bone structure. Thallium-201 is a potassium analogue that is taken up by tumours and emits gamma radiation. Again, using photographic plates, radiologists can get very accurate pictures of tumours – because they effectively emit their own radiation. Thallium-201 is also used to evaluate how well regions of the heart are pumping after a coronary artery bypass surgery. The thallium acts as a tracer, giving out gamma radiation. The radiation is captured in special detectors and an image is built up. By using 3 detectors placed around a body, it is possible to create a 3D image. This technique is called Emission Computed Tomography (ECT). In the case of the heart, the method is refined and is known as Single Photon ECT (SPECT). Industrial tracers Gamma sources can be used to detect leaks in pipes and to trace pollution. Scientists can label possible sources of the leak or pollution by introducing a short lived gamma emitter. If they detect gamma radiation later on, they can work out the source of the leak. Similarly if they're not sure where a leak is in a buried pipe, they can add a gamma source to the inflow and search for a build up of radiation. Sterilisation Gamma-emitting radioisotopes, such as cobalt-60, can be used to sterilise pharmaceuticals and cosmetics. Their gamma radiation kills bacteria without needing to heat up these heat-sensitive products. This technique can also be used in hospitals for medical equipment and is useful because instruments can be sterilised without opening the packages. Crack detection Gamma radiation can be used to search for cracks in welds and pipes and give an early warning of failure where they are wearing thin – useful for aeroplane safety checks. This works in a similar way to X-rays – the gamma radiation penetrates the metal a little but will show up as hot spots where there is a fault. However, unlike X-rays, the gamma source can be put in awkward places – it doesn't involve carrying around a high energy piece of machinery.

Ionising radiations

2015-07-10T01:04:00.000+06:00

	Ionising radiations

		Radioactive isotopes decay into other elements. When they decay, they give out ionising radiations – so called because they cause ionisation of the medium that they travel through. There are three main types of radiation called alpha (a), beta (b) and gamma (g). They were first identified at the end of the 19th century and were classified according to their penetrating power (alpha is the least penetrative and gamma the most). We now know more about what they are and why they behave as they do.

Picture 2.4 Alpha radiation.

Alpha radiation - a

What is it?	Alpha radiation is a stream of particles. These are the nuclei of helium – two protons and two neutrons. This means that they have a (relatively) large mass. Alpha particles from a given radioactive decay all have the same energy (of the order of a few MeV).

What's its charge?	The atomic number of helium is 2 so an alpha particle carries a double positive charge.

What's its symbol?	or

How fast do they go?	Alpha particles don’t travel very fast (compared with beta particles) because they have such a large mass. They carry energy away from a radioactive decay; this energy is of the order of a few MeV and is given to the kinetic energy of the alpha particle; they have a mass of 6.6 x 10^–27 kg so their speed is about 7 x 10⁶ m.s^-1 – i.e about a fiftieth of the speed of light.

How ionising are they?	Their low speed means that they tend to collide with plenty of other atoms and cause a lot of ionisation on their way. An alpha particle can ionise thousands of air particles before it slows down to thermal speeds.

How far do they go?	All these collisions mean that they lose energy quickly, so they have a short range in air.

What do they go through?	They are easily stopped by anything solid – even a piece of paper will stop alpha radiation.

Effect of magnetic field	Alpha particles are deflected slightly in a magnetic field. Their large mass means that they don't get deflected much. They will travel in the arc of a circle in a uniform field (see page 19).

Effect of electric field	They are deflected slightly in an electric field and will move in a parabolic curve in a uniform field.

Picture 2.5 Beta radiation.

Picture 2.6 Bending beta particles in magnetic field. They must take a curved path to reach the detector.

Beta radiation - b

What is it?	Beta radiation (the sort that you'll come across in school) is a stream of fast moving electrons. Beta plus radiation is a stream of particles called positrons. Beta radiation from a given radioactive decay is given out with a range of energies (see page 16).

What's its charge?	You will usually come across beta minus (b-) radiation, which has a negative charge.

What's its symbol?	or

How fast do they go?	These particles have very little mass (about 7, 000 times lighter than an alpha particle) and travel close to the speed of light (300, 000 km.s^-1).

How ionising are they?	They tend to pass through the air and solid matter without many collisions with other atoms. So beta radiation is only weakly ionising.

How far do they go?	However, it means that it has a long range in air.

What do they go through?	It will pass through paper, aluminium and steel. However, it is stopped by lead or thick pieces of other metals.

Effect of magnetic field	Beta particles are easily deflected in a magnetic field because of their small mass. They will travel in a circular path in a uniform field (see page 19). The direction of their deflection tells us whether they are positively or negatively charged.

Effect of electric field	They are deflected in an electric field and will move in a parabolic curve in a uniform field.

Picture 2.7 Gamma radiation.

Gamma radiation - g

What is it?	Gamma radiation is at the high frequency end of the electromagnetic spectrum. It has a very short wavelength (much less than the radius of an atom) and will pass through atoms with very little chance of being deflected or absorbed. Gamma radiation is often given out with alpha and beta radiation.

What's its charge?	It has no charge.

What's its symbol?

How fast does it go?	As it is part of the electromagnetic spectrum, it travels at the speed of light – 300, 000 km.s^-1.

How ionising is it?	It will tend to pass through matter without causing much ionisation.

How far does it go?	It has an extremely long range in air but gets weaker with distance. Its intensity obeys an inverse square law – getting weaker with the square of distance, i.e. doubling the distance quarters the intensity. Tripling the distance leads to a ninth of the intensity.

What does it go through?	It will get through thin samples of most materials without any noticeable decrease in intensity. However, its intensity is reduced by lead or very thick pieces of other metals. The thicker the sample, the greater the reduction in intensity.

Effect of magnetic field	No effect because it has no charge.

Effect of electric field	No effect because it has no charge.

Isotopes and stability

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	Isotopes and stability

		The nucleus of an atom contains protons and neutrons and it accounts for nearly all the mass of an atom. We can denote the number of nucleons in the nucleus using: the atomic number: this is number of protons in the nucleus of an atom. It has the symbol Z. It is fixed for a given element. the atomic mass number: this is the number of protons added to the number of neutrons and is given the symbol A.

Picture 2.1 Showing the atomic number and mass number of an atom. Notice that they come before the symbol. A is at the top and Z is at the bottom.

	For example . . .

		Nitrogen has the chemical symbol N. It has an atomic number of 7 (Z=7) because it has 7 protons in its nucleus. We write this in front of the chemical symbol to the bottom-left. Some nitrogen atoms have 15 nucleons in the nucleus and therefore have an atomic mass number of 15. Again, the atomic mass number, A, comes before the chemical symbol and is positioned top-left. How does this tell us the number of neutrons? A is the number of neutrons plus protons in the nucleus. However, we already know that there are 7 protons. Therefore, there must be 8 neutrons in the nucleus to add up to 15.

name	symbol	A	Z	neutrons
nitrogen-12		12	7	5
nitrogen-13		13	7	6
nitrogen-14		14	7	7
nitrogen-15		15	7	8

Table 3. Some of the isotopes of nitrogen. The atomic number is always 7. However, the atomic mass number, A, changes.

	Number of neutrons

		However, not all nitrogen nuclei have 8 neutrons in a nucleus. They can have 5, 6, 7, 8, 9, 10 or 11 neutrons. We call these different isotopes of nitrogen. There are always 7 protons because the nucleus is always a nitrogen nucleus (Z=7). Since different isotopes of an element have different numbers of neutrons (but always the same number of protons) they have different mass numbers. The isotopes of nitrogen have mass numbers ranging from 12 to 18. We call them nitrogen-12, nitrogen-13, nitrogen-14 and so on. Some isotopes of nitrogen are unstable. They give out radiation to become more stable; we say they are radioactive. Nitrogen-14 and nitrogen-15 are both stable isotopes of nitrogen. However, the other 5 isotopes are all unstable. Nitrogen-12 and nitrogen-13 will decay by beta plus emission and nitrogen-16, nitrogen-17 and nitrogen-18 decay by beta minus emission.

	Stable isotopes

		Every element has a number of different isotopes. Some of these are radioactive and some are stable. All the elements up to Z = 82 have at least one stable isotope. We can plot a graph of stable isotopes. The atomic number, Z, is on the x-axis and the number of neutrons (N) is on the y-axis. The graph is a curve – see picture 2.2.

Picture 2.2 Graph of stable isotopes.

		For light elements (the bottom left of the graph), the stable isotopes are the ones with the same numbers of protons and neutrons and hence the ‘line of stability’ follows the straight line of N = Z. For example, the stable isotope carbon-12 has 6 protons and 6 neutrons. If we got a straight line all the way, then this would tell us that the stable isotopes have the same number of neutrons as protons. However, this is not the case. The line curves upwards. Stable isotopes of the heavier elements (top right of the graph) have more neutrons than protons. For example, Gold-197 is stable. It contains 79 protons and 118 neutrons. The neutrons in a nucleus can be thought of as acting as a kind of glue to hold the nucleus together. The positively charged protons are in a very confined space but would rather not be, due to the fact that they repel each other. However, protons and neutrons are all attracted to each other as a result of another force - the strong nuclear force (see below). The neutrons don't contribute any repulsive effects because they are neutral. So having more neutrons around can help to hold the nucleus together. Notice that no amount of neutrons can hold a nucleus together once it has more that 82 protons – the line stops at Z = 82! All of the elements with an atomic number greater than 82 have only unstable isotopes.

Picture 2.3 Protons are held together by the strong nuclear force.

	Forces in nucleus

		We now know that the nucleus of helium is 1.9 x 10^–15 m across. It contains two protons. They are both positively charged and will repel each other. However, they stay bound in a helium nucleus. Therefore there must be another force that holds them together. This is the strong nuclear force. It is an attractive force that only has an effect over a very short range in nucleus (about 10^–15 m - the size of the nucleus). The strong nuclear force binds protons and neutrons together to make the nucleus. Neutrons and protons are made from quarks (see page 17).The strong nuclear force is actually a force between quarks and is carried by particles called gluons. Protons and neutrons are made of quarks and they feel the strong nuclear force as well. Electrons do not feel the strong nuclear force. Fundamental particles that don't feel the strong nuclear force are all in the family of leptons.

Inside the nucleus

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1. The atom

Picture 1.6 An atom has a nucleus with electrons outside it.

	Inside the nucleus

		The atom is made from a central nucleus with electrons around the outside. The electrons have a negative charge and very little mass. Most of the mass of the atom is concentrated in the nucleus, which has a positive charge. The positive charge in the nucleus is balanced by the negative electrons, making an atom neutral. But what's inside the nucleus? And what are the fundamental particles that make up an atom?

	proton	neutron	electron

m	1	1	1 1800
q	+ 1	0	- 1

Table 1. The relative mass(m) and the relative charge (q) of the three main sub–atomic particles.

	What's in the nucleus?

		There are two types of particle in the nucleus of an atom: the proton and the neutron. The proton has a positive charge and a large mass (1800 times more than an electron – see table 1). The number of protons in a nucleus is the same for all the atoms of a particular element. It corresponds to the atomic number of that element. For example, if a nucleus contains three protons, then it is part of an atom of lithium (atomic number 3). The neutron has a similar mass to the proton but has no charge. The neutrons help keep the nucleus together.

Picture 1.7 Lithium has 3 positve protons in its nucleus.

name	symbol	protons	Z
hydrogen	H	1	1
helium	He	2	2
lithium	Li	3	3
carbon	C	6	6

Table 2. Some atoms and their atomic numbers.

What do protons do?

All atoms of a given element have the same number of protons in the nucleus. This number is the atomic number of the element and we give it the symbol Z. For example, lithium atoms contain 3 protons. Therefore, the atomic number of lithium is 3. Any atom that has 3 protons must be a lithium atom.
We can show this information about an atom using its symbol and atomic number:

(The symbol for lithium is Li - note it starts with a capital letter although its name, lithium, starts with a small letter.)

The atomic number is written to the left bottom of the symbol.

	Neutrons

		The elements in the periodic table are arranged in increasing atomic mass. Their atomic numbers correspond to their position in the sequence (as well as to the number of protons in their nuclei). However, the atomic mass is not proportional to the atomic number (see page 6). This tells us that there must be something else in the nucleus as well as protons. For small atomic numbers, the relative atomic mass is approximately double the atomic number. For higher atomic numbers, it is more like 2.5 times the atomic number. i.e. they are not proportional. We can see this clearly by looking at hydrogen and helium.

Picture 1.8 The atomic mass is not proportional to the number of protons. Therefore, there must be something else in the nucleus too.

		Helium has atomic number 2 and a relative atomic mass of about 4. Hydrogen has atomic mass and relative atomic mass of 1. If protons were the only particles in the nucleus, then a helium nucleus would have two protons and therefore only twice the mass of hydrogen. However, it actually has four times the mass of hydrogen. This suggests that there is something else in the nucleus as well as protons. This new particle – the neutron - would have no charge. So how could it be detected?

	Finding the neutron

		The difficulty with finding the neutron is that it has no effect on charged particles (because it is neutral). This means that it will not knock electrons out of atoms making ions. So it will not leave any visible trace in a cloud chamber or register in a Geiger counter because both of these rely on ionisation to register the presence of particles. However, the neutron does feel the strong nuclear force – the short-range force that holds the nucleus together. So if it passes close enough to a proton it will make it move.

Picturing sub-atomic particles
In diagrams, all these particles are made to look a bit like small coloured snooker balls. This is because we have to draw them somehow. However, it is important to understand that they are not like tiny coloured snooker balls – solid with well-defined edges.

•	They do not have any colour because they are smaller than the wavelength of visible light. So light waves pass straight over them without changing direction.
•	We cannot talk about their being solid because a solid is a collection of atoms. These particles are smaller than atoms.
•	They do not have well-defined edges but can influence other particles over a long distance.

		In 1932, James Chadwick, who had worked with Rutherford, detected neutrons and measured their mass in an invisible game of billiards. He fired the neutrons at a block of paraffin wax, which has a high concentration of hydrogen and is therefore rich in protons. Some of the neutrons collided with protons in the wax and knocked them out. Chadwick could then detect these protons and measure their energy. Using his knowledge of energy and momentum, he was able to work out the mass of the neutrons from the range of energies of the protons that they knocked out. He found that its mass was slightly more than that of a proton. Chadwick, like Rutherford, used an ingenious method to probe into what cannot be seen.

Atom

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One hundred years ago, amidst glowing glass tubes and the hum of electricity, the British physicist J.J. Thomson was venturing into the interior of the atom. At the Cavendish Laboratory at Cambridge University, Thomson was experimenting with currents of electricity inside empty glass tubes. He was investigating a long-standing puzzle known as "cathode rays." His experiments prompted him to make a bold proposal: these mysterious rays are streams of particles much smaller than atoms, they are in fact minuscule pieces of atoms. He called these particles "corpuscles," and suggested that they might make up all of the matter in atoms. It was startling to imagine a particle residing inside the atom--most people thought that the atom was indivisible, the most fundamental unit of matter.
	*A simple cathode ray tube.*
*Thomson in his office.*		homson's speculation was not unambiguously supported by his experiments. It took more experimental work by Thomson and others to sort out the confusion. The atom is now known to contain other particles as well. Yet Thomson's bold suggestion that cathode rays were material constituents of atoms turned out to be correct. The rays are made up of electrons: very small, negatively charged particles that are indeed fundamental parts of every atom.
"Could anything at first sight seem more impractical than a body which is so small that its mass is an insignificant fraction of the mass of an atom of hydrogen?" -- J.J. Thomson.		odern ideas and technologies based on the electron, leading to television and the computer and much else, evolved through many difficult steps. Thomson's careful experiments and adventurous hypotheses were followed by crucial experimental and theoretical work by many others in the United Kingdom, Germany, France and elsewhere. These physicists opened for us a new perspective--a view from inside the atom.

Calculations and titration

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Calculations and titration

It is important to know the correct quantities of reactants to use in a synthesis, and how much product has been produced. It is also important to know how pure the product is - this can be tested using the method of titration.

HOW TO CARRY OUT TIRATION?

Calculating relative formula masses

Relative atomic mass

The relative atomic mass of an element shows its mass compared with the mass of atoms of other elements. The relative atomic mass of carbon is 12, while the relative atomic mass of magnesium is 24. This means that each magnesium atom is twice the mass of a carbon atom.
The relative atomic mass of each element can be found in the Periodic Table.

Relative formula mass

Relative atomic masses can be used to find the relative formula mass of a compound.
Here are two examples:

Question: Find the relative formula mass of sulfuric acid H₂SO₄.

Answer>>>: Sulfuric acid has the formula H₂SO₄. The relative atomic masses of the elements in sulfuric acid are H = 1, O = 16 and S = 32. Adding together the relative atomic masses of each atom in the formula of sulfuric acid gives the relative formula mass of sulfuric acid: H₂SO₄ relative formula mass = (2 x 1) + 32 + (4 x 16) = 98

Question: Find the relative formula mass of sodium hydroxide NaOH.

Answer>>>

Sodium hydroxide has the formula NaOH. The relative atomic masses of the elements in sodium hydroxide are H = 1, Na = 23 and O = 16. Adding together the relative atomic masses of each atom in the formula of sodium hydroxide gives the relative formula mass: NaOH relative formula mass = 23 + 16 + 1 = 40

Percentage yield

The yield from a chemical reaction is the mass of product made. A balanced symbol equation shows the number of atoms and molecules in the reactants and products. The reactants are on the left of the arrow and the products are on the right of the arrow. A balanced equation can tell us the theoretical yield for a reaction, by using the relative formula masses of reactants and products involved. For example the formula for the neutralisation of sulfuric acid with sodium hydroxide is: sulfuric acid + sodium hydroxide → sodium sulfate + water H₂SO₄ + 2NaOH → Na₂SO₄ + 2H₂O By using the relative formula masses it can be calculated that 6.9g of sulfuric acid theoretically produces 10g of sodium sulfate. If this neutralisation reaction is carried out, the actual yield is found by measuring the mass of product - sodium sulfate - that has been made. The actual yield of sodium sulfate produced will be less than the theoretical yield calculated using the equation.

Theoretical vs actual yield

The difference between the theoretical yield and the actual yield for any reaction may have one or more of a number of causes:

Some of the reactants may remain unreacted when the reaction is complete.
Some of the product may be lost when liquids or solids are transferred from one container to another.
Some of the reactants may form other products.

If you have both the theoretical yield and actual yield, you can work out the percentage yield for the synthesis:

percentage yield = (actual yield ÷ theoretical yield) × 100

Question: In the neutralisation of sulfuric acid with sodium hydroxide, the theoretical yield from 6.9g of sulfuric acid is 10g. In a synthesis, the actual yield is 7.2g. What is the percentage yield for this synthesis?

Answer>>>

percentage yield = (actual yield ÷ theoretical yield) × 100 percentage yield = (7.2 ÷ 10) x 100 = 72%

Carrying out a titration

Measuring purity

For some synthesised chemical compounds - pharmaceuticals, for example - it is very important that they have a high level of purity. Just a tiny amount of an impurity in a drug could cause a great deal of harm to a patient. Samples of chemicals that are synthesised must be checked for purity. This is often done by carrying out a titration. A titration is used to measure the volume of one solution that exactly reacts with another solution.

Steps of titration

A titration is carried out using a number of steps:

If the sample is a solid, it is weighed using an accurate balance, and then dissolved to make up a known volume of solution (usually 100cm³).
A pipette is used to measure accurately a volume of this solution - for example, 10cm³. A safety pipette filler is used to draw solution into the pipette. This is emptied into a conical flask.
A few drops of an indicator may be added to the conical flask. This will show a change of colour when the titration is complete.
A second chemical is placed in a burette. This other solution is of a chemical that will react with the synthesised chemical sample in the conical flask. Often the solution in the burette is an acid or alkali, and it must be of a precise, known concentration.
The solution from the burette is run into the conical flask. The solution is added one drop at a time, with swirling to mix the solutions as the end-point is approached. Eventually, a colour change shows that the correct amount has been added to react completely with the synthesised chemical in the sample.
The volume of solution added from the burette is noted. The titration results can then be used to calculate the amount of the synthesised chemical in the sample, and therefore find its purity.

Apparatus required for titration (from left to right): burette, conical flask, safety pipette filler and pipette

Calculations from equations - Higher

A balanced symbol equation can be used as the starting point to calculate the quantities of reactants used, and the quantities of products made, in a reaction.

Step 1: write down the balanced symbol equation

Sulfuric acid is neutralised by sodium hydroxide, to make sodium sulfate and water, according to the following equation:
H₂SO₄ + 2NaOH → Na₂SO₄ + 2H₂O

Step 2: work out the relative formula mass

The relative formula masses of the reactants and products can be worked out using relative atomic masses obtained from the periodic table.
H₂SO₄ relative formula mass = (2 x 1) + 32 + (4 x 16) = 98
NaOH relative formula mass = 23 + 16 + 1 = 40
Na₂SO₄ relative formula mass = (2 x 23) + 32 + (4 x 16) = 142
H₂O relative formula mass = (2 x 1) + 16 = 18

Step 3: work out the masses of reactants and products

The masses of reactants and products can now be worked out from the equation:
H₂SO₄ + 2NaOH → Na₂SO₄ + 2H₂O
98 + (2x40) = 142 + (2x18)
98 + 80 = 142 + 36
These numbers can now be applied to any unit of mass - grams, kilograms, tonnes, and so on. Now try the questions below.

Question......: What mass of sulfuric acid reacts with sodium hydroxide to produce 10g of sodium sulfate?

Question.......: What mass of sodium sulfate is made when 5.0 tonnes of sodium hydroxide is reacted with sulfuric acid?

Gas volume in calculations

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The molar gas volume in calculations, moles, gas volumes and Avogadro's Law

Avogadro's Law states that equal volumes of gases under the same conditions of temperature and pressure contain the same number of molecules.
- So the volumes have equal moles of separate particles (molecules or individual atoms) in them.
- Therefore one mole of any gas (formula mass in g), at the same temperature and pressure occupies the same volume .
- This is 24dm³ (24 litres) or 24000 cm³, at room temperature of 25^oC/298K and normal pressure of 101.3 kPa/1 atmosphere (such conditions are often referred to as RTP).
- The molar volume for s.t.p is 22.4 dm³ (22.4 litres) at 0^oC and 1atmosphere pressure.
- Historically, s.t.p unfortunately stands for standard temperature and pressure, but these days 25^oC/298K is usually considered the standard temperature (RTP).
Some handy relationships for substance Z below:
moles Z = mass of Z gas (g) / atomic or formula mass of gas Z (g/mol)
- mass of Z in g = moles of Z x atomic or formula mass of Z
- atomic or formula mass of Z = mass of Z / moles of Z
- 1 mole = formula mass of Z in g.
gas volume of Z = moles of Z x volume of 1 mole
- rearranging this equation gives ...
- moles of Z = gas volume of Z / volume of 1 mole
- moles = V(dm³) / 24 (at RTP)
- The latter form of the equation can be used to calculate molecular mass from experimental data because
  - moles = mass / molecular mass = gas volume / volume of 1 mole
  - mass / molecular mass = gas volume / volume of 1 mole
  - molecular mass = mass x volume of 1 mole/volume
  - therefore at RTP: M_r = mass(g) x 24 / V(dm³)
  - so, if you know the mass of a gas and its volume, you can work out moles of gas and then work out molecular mass.
  - This has been done experimentally in the past, but these days, molecular mass is readily done very accurately in a mass spectrometer.
Note (i): In the following examples, assume you are dealing with room temperature and pressure i.e. 25^oC and 1 atmosphere pressure so the molar volume is 24dm³ or 24000cm³.
Note (ii):
- Apart from solving the problems using the mole concept (method (a) below, and reading any equations involved in a 'molar way' ...
- It is also possible to solve them without using the mole concept (method (b) below). You still use the molar volume itself, but you think of it as the volume occupied by the formula mass of the gas in g and never think about moles!

Methods of measuring how much gas is formed (volume can be compared with theoretical prediction!)
- (a) You can collect the gases in a calibrated gas syringe.
  - You must make sure too much gas isn't produced and too fast!
  - A gas syringe is more accurate than collecting the gas in an inverted measuring cylinder under water shown below, but its still only accurate to the nearest cm³.
  - You can collect any gas by this method.
- (b) The gas is collected in a measuring cylinder filled with water and inverted over a trough of water.
  - You can get a more accurate result by using an inverted burette instead of a measuring cylinder.
  - However, this method is no good if the gas is soluble in water!
  - Burettes are calibrated in 0.10 cm³ intervals. measuring cylinders to the nearest cm³ or worse!
  - In both methods the reaction is carried out in conical flask fitted with a sealing rubber bung, but a tube enabling the gas evolved to be collected in some suitable container.
- (c) A third method is to measure the gas loss by carrying out the reaction in a flask set up on an accurate one-pan electronic balance.
- You need to put a cotton wool plug in the neck of the conical flask in case you lose any of the solution in a spray as the gas bubbles up - effervescence can produce an aerosol.
- This method can be used for any reaction that produces a gas, but the gas is released into the laboratory, ok if its harmless.
- It is potentially the most accurate method, BUT, the mass loss may be quite small especially hydrogen [M_r(H₂) = 2], better for the 'heavier' gas carbon dioxide [M_rCO₂) = 44]
Molar gas volume calculation Example 9.1
- What is the volume of 3.5g of hydrogen? [A_r(H) = 1]
- common thinking: hydrogen exists as H₂ molecules, so M_r(H₂) = 2, so 1 mole or formula mass in g = 2g
- method (a)
  - so moles of hydrogen = 3.5/2 = 1.75 mol H₂
  - so volume H₂ = mol H₂ x molar volume = 1.75 x 24 = 42 dm³ (or 42000 cm³)
  - -
- method (b):
  - 2g occupies 24 dm³, so scaling up for the volume of hydrogen ...
  - 3.5 g will have a volume of 3.5/2 x 24 = 42 dm³ (or 42000 cm³)
  - -
Molar gas volume calculation Example 9.2
- Given the equation
- MgCO_3(s) + H₂SO_4(aq) ==> MgSO_4(aq) + H₂O_(l) +CO_2(g)
- What mass of magnesium carbonate is needed to make 6 dm³ of carbon dioxide? [A_r's: Mg = 24, C = 12, O = 16, H =1 and S = 32]
- method (a):
  - since 1 mole = 24 dm³, 6 dm³ is equal to 6/24 = 0.25 mol of gas
  - From the equation, 1 mole of MgCO₃ produces 1 mole of CO₂, which occupies a volume of 24 dm³.
  - so 0.25 moles of MgCO₃ is need to make 0.25 mol of CO₂
  - formula mass of MgCO₃ = 24 + 12 + 3x16 = 84,
  - so required mass of MgCO₃ = mol x formula mass = 0.25 x 84 = 21g
  - -
- method (b):
  - converting the equation into the required reacting masses ..
  - formula masses: MgCO₃ = 84 (from above), CO₂ = 12 + 2x16 = 44
  - MgCO₃ : CO₂ equation ratio is 1 : 1
  - so 84g of MgCO₃ will form 44g of CO₂
  - 44g of CO₂ will occupy 24dm³
  - so scaling down, 6 dm³ of CO₂ will have a mass of 44 x 8/24 = 11g
  - if 84g MgCO₃ ==> 44g of CO₂, then ...
  - 21g MgCO₃ ==> 11g of CO₂ by solving the ratio, scaling down by factor of 4
  - -
Molar gas volume calculation Example 9.3
- 6g of a hydrocarbon gas had a volume of 4.8 dm³. Calculate its molecular mass.
- method (a):
  - 1 mole = 24 dm³, so moles of gas = 4.8/24 = 0.2 mol
  - molecular mass = mass in g / moles of gas
  - M_r = 6 / 0.2 = 30
  - i.e. if 6g = 0.2 mol, 1 mol must be equal to 30g by scaling up
  - -
- method (b):
  - 6g occupies a volume of 4.8 dm³
  - the formula mass in g occupies 24 dm³
  - so scaling up the 6g in 4.8 dm³
  - there will be 6 x 24/4.8 = 30g in 24 dm³
  - so the molecular or formula mass = 30
  - -
Molar gas volume calculation Example 9.4
- Given the equation ... (and A_r's Ca = 40, H = 1, Cl = 35.5)
- Ca_(s) + 2HCl_(aq) ==> CaCl_2(aq) + H_2(g)
- What volume of hydrogen is formed when ...
  - (i) 3g of calcium is dissolved in excess hydrochloric acid?
  - (ii) 0.25 moles of hydrochloric acid reacts with calcium?
- (i) method (a):
  - 3g Ca = 3/40 = 0.075 mol Ca
  - from 1 : 1 ratio in equation, 1 mol Ca produces 1 mol H₂
  - so 0.075 mol Ca produces 0.075 mol H₂
  - so volume H₂ = 0.075 x 24 = 1.8 dm³ (or 1800 cm³)
  - -
- (i) method (b):
  - from equation 1 Ca ==> 1 H₂ means 40g ==> 2g
  - so scaling down: 3g Ca will produce 2 x 3/40 = 0.15g H₂
  - 2g H₂ has a volume 24 dm³, so scaling down ...
  - 0.15g H₂ has a volume of (0.15/2) x 24 = 1.8 dm³ (or 1800 cm³)
  - -
- (ii) method (a) only:
  - from equation: 2 moles HCl ==> 1 mole H₂ (mole ratio 2:1)
  - so 0.25 mol HCl ==> 0.125 mol H₂, volume 1 mole gas = 24 dm³
  - so volume H₂ = 0.125 x 24 = 3 dm³
  - -

Molar gas volume calculation Example 9.5
- Given the equation ... (and A_r's Mg = 24, H = 1, Cl = 35.5)
- Mg_(s) + 2HCl_(aq) ==> MgCl_2(aq) + H_2(g)
- How much magnesium is needed to make 300 cm³ of hydrogen gas?
- method (a)
  - 300 cm³ = 300/24000 = 0.0125 mol H₂ (since 1 mol of any gas = 24000 cm³)
  - from the equation 1 mole Mg ==> 1 mole H₂
  - so 0.0125 mole Mg needed to make 0.0125 mol H₂
  - so mass of Mg = mole Mg x A_r(Mg)
  - so mass Mg needed = 0.0125 x 24 = 0.3g
  - -
- method (b)
  - reaction ratio in equation is 1 Mg ==> 1 H₂,
  - so reacting mass ratio is 24g Mg ==> 2g H₂,
  - 2g H₂ has a volume of 24000 cm³ (volume of formula mass in g)
  - so scaling down: mass Mg needed = 24 x (300/24000) = 0.3g
  - -
Molar gas volume calculation Example 9.6
- A small teaspoon of sodium hydrogencarbonate (baking soda) weighs 4.2g. Calculate the moles, mass and volume of carbon dioxide formed when it is thermally decomposed in the oven. Assume room temperature for the purpose of the calculation.
  - 2NaHCO_3(s) ==> Na₂CO_3(s) + H₂O_(g) + CO_2(g)
  - Formula mass of NaHCO₃ is 23+1+12+(3x16) = 84 = 84g/mole
  - Formula mass of CO₂ = 12+(2x16) = 44 = 44g/mole (not needed by this method)
    - or a molar gas volume of 24000 cm³ at RTP (definitely needed for this method)
  - In the equation 2 moles of NaHCO₃ give 1 mole of CO₂ (2:1 mole ratio in equation)
  - Moles NaHCO₃ = 4.2/84 = 0.05 moles ==> 0.05/2 = 0.025 mol CO₂ on decomposition.
  - Mass = moles x formula mass, so mass CO₂ = 0.025 x 44 = 1.1g CO₂
  - Volume = moles x molar volume = 0.025 x 24000 = 600 cm³ of CO₂
  - -
Molar gas volume calculation Example 9.7
- What volume of carbon dioxide is formed at RTP when 5g of carbon is burned?
  - C(s) + O₂(g) ==> CO₂(g)
  - 1 mole carbon gives 1 mole of carbon dioxide, atomic mass of carbon = 12
  - moles = mass / atomic mass, moles carbon = moles carbon dioxide = 5/12 = 0.417 mol
  - 1 mole of gas at RTP occupies 24 dm³
  - so 0.417 mol occupies a volume of 0.417 x 24 = 10.0 dm³
  - -
Molar gas volume calculation Example 9.8
- What volume of carbon dioxide gas is formed at RTP if 1Kg of propane gas fuel is burned?
  - C₃H₈(g) + 5O₂(g) ==> 3CO₂(g) + 4H₂O(l)
  - 1 mole of propane gas gives 3 moles of carbon dioxide gas on complete combustion
  - 1 kg = 1000g, atomic masses: C = 12, H =1
  - M_r(propane) = (3 x 12) + (8 x 1) = 44
  - moles = mass in g / molecular mass, therefore moles propane = 1000/44 = 22.73 mol
  - from equation molar ratio: moles carbon dioxide = 3 x moles of propane
  - mol propane = 3 x 22.73 = 68.18 mol
  - 1 mole of gas at RTP occupies a volume of 24 dm³
  - so 68.18 mol of gas occupies a volume of 68.18 x 24 = 1636 dm³

Calculation for Solution

2015-07-10T00:01:00.001+06:00

Calculations involving density, percent concentration, molality and molarity
Problems #1 - 10

Problem #1: Phosphoric acid is usually obtained as an 87.0% phosphoric acid solution. If it is 13.0 M, what is the density of this solution? What is its molality?
Solution for density:
1) Determine the moles of H₃PO₄ in 100.0 grams of 87.0% solution:

87.0 g of the 100.0 g is H₃PO₄ moles H₃PO₄ = 87.0 g / 97.9937 g/mol = 0.8878 mol

2) Calculate the volume of 13.0 M solution which contains 0.8878 mol of H₃PO₄:

13.0 mol/L = 0.8878 mol / x x = 0.0683 L = 68.3 mL

3) Determine the density of the solution:

100.0 g / 68.3 mL = 1.464 g/mL = 1.46 g/mL (to three sig fig)

Solution for molality:
1) Let us assume 100.0 grams of solution. Therefore:

87.0 g is H₃PO₄
13.0 g is H₂O

2) Calculate the molality:

moles H₃PO₄ = 87.0 g / 97.9937 g/mol = 0.8878 mol
kg of water = 0.0130 kg molality = 0.8878 mol / 0.0130 kg = 68.3 molal

Problem #2: Concentrated hydrochloric acid is usually available at a concentration of 37.7% by mass. What is its molar concentration? (The density of the solution is 1.19 g/mL.)
Solution Path #1:
1) Determine moles of HCl in 100.0 g of 37.7% solution:

37.7 g of this solution is HCl 37.7 g / 36.4609 g/mol = 1.03398435 mol

2) Determine volume of 100.0 g of solution:

density = mass / volume 1.19 g/mL = 100.0 g / x
x = 84.0336 mL

3) Determine molarity:

1.03398435 mol / 0.0840336 L = 12.3 M

Solution Path #2:
1) Assume 1.00 L of solution. Use density to get mass:

1.19 g/mL = x / 1000 mL x = 1190 g

2) (a) Use percent mass to get mass of HCl, then (b) convert to moles:

1190 g times 0.377 = 448.63 g 448.63 g / 36.4609 g/mol = 12.3 mol

3) Determine molarity:

12.3 mol / 1.00 L = 12.3 M

Problem #3: I have a bottle of NH₃. Its strength is 32.0% and its density is 0.89 g/mL. How do I figure out the molarity?
Solution:
1) Assume 100.0 g of solution is present. Calculate moles of ammonia present:

32.0 grams of NH₃ are present. 32.0 g / 17.0307 g/mol = 1.879 mol

2) Determine volume of 100.0 g of solution:

100.0 g / 0.87 g/mL = 112.36 mL

3) Calculate molarity:

1.879 mol / 0.11236 L = 16.7 M

Problem #4: An aqueous solution of hydrofluoric acid is 30.0% HF, by mass, and has a density of 1.101 g cm^-3. What are the molality and molarity of HF in this solution?
Solution for molality:
1) Let us assume 100.0 grams of solution. Therefore:

30.0 g is HF
70.0 g is H₂O

2) Calculate the molality:

moles HF = 30.0 g / 20.0059 g/mol = 1.49956 mol
kg of water = 0.0700 kg molality = 1.49956 mol / 0.0700 kg = 21.4 molal

Solution for molarity:
1) Determine moles of HF in 100.0 g of 30.0% solution:

30.0 g of this solution is HF 30.0 g / 20.0059 g/mol = 1.49956 mol

2) Determine volume of 100.0 g of solution:

density = mass / volume 1.101 g/mL = 100.0 g / x
x = 90.8265 mL

3) Determine molarity:

1.49956 mol / 0.0908265 L = 16.5 M

Problem #5: Concentrated nitric acid is a solution that is 70.4% HNO₃ by mass. The density of this acid is 1.42 g/mL. What is the molarity and the molality of the acid?
Solution for molarity:
1) Determine moles of HNO₃ in 100.0 g of 70.4% solution:

70.4 g of this solution is HNO₃ 70.4 g / 63.0119 g/mol = 1.11725 mol

2) Determine volume of 100.0 g of solution:

density = mass / volume 1.42 g/mL = 100.0 g / x
x = 70.422535 mL

3) Determine molarity:

1.11725 mol / 0.070422535 L = 15.86 M = 15.9 M (to three sf)

Solution for molality:
1) Let us assume 100.0 grams of solution. Therefore:

70.4 g is HNO₃
29.6 g is H₂O

2) Calculate the molality:

moles HNO₃ = 70.4 g / 63.0119 g/mol = 1.11725 mol
kg of water = 0.0296 kg molality = 1.11725 mol / 0.0296 kg = 37.7 molal

Special bonus part to Problem #5: In the above solution, what is the mole fraction of HNO₃?
Solution:
1) Let us assume 100.0 grams of solution. Therefore:

70.4 g is HNO₃
29.6 g is H₂O

2) Determine the moles of each substance:

HNO₃ = 1.11725 mol
H₂O = 29.6 g / 18.0152 g/mol = 1.643057 mol

3) Determine the mole fraction of HNO₃:

1.11725 mol / (1.11725 mol + 1.643057 mol) = 0.405

Problem #6: The density of toluene (C₇H₈) is 0.867 g/mL, and the density of thiophene (C₄H₄S) is 1.065 g/mL. A solution is made by dissolving 9.660 g of thiophene in 260.0 mL of toluene.

a) Calculate the molality of thiophene in the solution.
b) Assuming that the volumes of the solute and solvent are additive, determine the molarity of thiophene in the solution.

Solution to part a:
1) Determine the moles of thiophene:

9.660 g / 84.142 g/mol = 0.1148 mol

2) Determine mass of toluene in 260.0 mL

260.0 mL x 0.867 g/mL = 225.42 g

3) Calculate the molality:

0.1148 mol / 0.22542 kg = 0.509 molal

Solution to part b:
1) Determine volume of thiophene:

9.660 g ÷ 1.065 g/mL = 9.07 mL

2) Determine total volume:

260.0 mL + 9.07 mL = 269.07 mL

3) Calculate molarity:

0.1148 mol / 0.26907 L = 0.427 M

Problem #7: An aqueous acetic acid solution is simultaneously 6.0835 molar and 8.9660 molal. Compute the density of this solution.
Solution:
1) Use molality to get moles:

8.9660 molal = x moles / 1.00 kg solvent x = 8.9660 moles

2) Compute the mass of the above solution:

1000 g solvent + (8.9660 moles x 60.05 g/mol) = 1538.41 g

3) Use molarity to get the volume of solution containing 8.9660 moles

6.0835 mol/L = 8.9660 moles / x x = 1.47382 L

4) Compute the density:

1538.41 g / 1473.82 mL = 1.0438 g/mL

Problem #8: What is the density (in g/mL) of a 3.60 M aqueous sulfuric acid solution that is 29.0% H₂SO₄ by mass?
Solution:
1) Assume 100.0 g of solution is present.
2) Detemine the volume of solution that weighs 100.0 g:

29.0 g of the 100.0 g is H₂SO₄.
use MV = grams / molar mass to determine the volume.
(3.60 mol/L) (x) = 29.0 g / 98.1 g/mol
x = 0.0821 L

3) Determine the density:

100.0 g / 82.1 mL = 1.22 g/mL

Problem #9: The density of an aqueous solution of nitric acid is 1.430 g/mL. If this solution contained 36.00% nitric acid by mass, how many mL of the solution would be needed to supply 150.20 grams of nitric acid?
Solution path #1:
1) 1.000 mL of solution weight 1.430 g, of which 36.00% is HNO₃
2) 1.430 x 0.3600 tells you the grams of HNO₃ in each mL of solution.
3) 150.20 g divided by the grams of HNO₃ per one mL of solution.
Solution path #2:
1) Determine how many mL of the solution you need:

150.20 g nitric acid x (1 g solution/0.3600 g nitric acid) = 417.2 g solution.

2) Determine the volume of solution that weighs 417.2 g:

417.2 g solution x (1 mL/1.430 g solution)

Problem #10: A bottle of commercial sulfuric acid (density 1.787 g/cc) is labeled as 86% by weight. What is the molarity of acid?
Solution:
1) Determine moles of H₂SO₄ in 100.0 g of 86% solution:

86 g of this solution is H₂SO₄ 86 g / 98.08 g/mol = 0.876835237 mol

2) Determine volume of 100.0 g of solution:

density = mass / volume 1.787 g/mL = 100.0 g / x
x = 55.96 mL

3) Determine molarity:

0.876835237 mol / 0.05596 L = 15.67 M = 15.7 M (to three sf)

Note: Try this problem with 96.0% and a density of 1.84 g/mL. The answer is 18.0 M.

Methods of balancing Eqn

2015-07-09T23:57:00.004+06:00

Methods of balancing chemical equations

1.: Balancing by inspection: Consider the example:

We notice that there are 2 nitrogens on both the left and right sides, so nitrogen is already balanced. There are 3 oxygens on the left but only 2 on the right, so oxygen is not balanced. Similarly, there are 4 hydrogens on the left and only 2 on the right, so hydrogen is not balanced. The balancing of oxygen and hydrogen can be handled together by making the coefficient of HO equal to 2. Then, the balanced equation reads

Now there are equal quantities of all elements on both sides of the reaction, and mass conservation is satisfied. Balancing by inspection in this way is quick and useful when you can see the solution easily.
2.: Algebraic method: Consider the reaction:

This equation could be balanced by inspection. However, we will use it to illustrate another approach - the algebraic approach. To balance a reaction algebraically, we start by putting unknown coefficients in front of each molecular species in the equation:

Then we write down the balance conditions for each element in terms of the unknowns. In this case, there are are four unknowns, , , and . From carbon balance, we have the condition:

The hydrogen balance condition is:

and the oxygen balance condition gives:

This constitutes a set of only three equations for the four unknowns. The fourth condition comes from recognizing that chemical equations specify relative amounts of reactants and products. Thus, one of the coefficients is arbitrary, and all the others can be expressed relative to it. Thus, we may take one of the coefficients to be 1, arbitrarily, which then leaves only three unknowns and three conditions to determine them. The choice of which coefficient to set to 1 is completely arbitrary and may be taken to be the one that is most convenient for solving the problem. In this case, the choice allows us to arrive at a solution almost immediately, for then it follows that and . From the oxygen balance condition, can be determined to be so . Thus, the balanced equation could be written as

This is a perfectly fine way of writing the equation, however, customarily, it is preferable to express all the coefficients as integers, in accordance with the law of combining volumes and Avogadro's hypothesis. Thus, we can multiply through the equation by 2 on both sides (it can be treated as an algebraic equation in this way) and obtain the final result:

As a second example, consider the reaction:

Again, we put unknown coefficients in front of each molecular species:

Writing down the balance conditions on each element gives:
Sodium balance:

Chlorine balance:

Sulfur balance:

Oxygen balance:

Hydrogen balance:

Setting arbitrarily, gives the immediate solution:

and or . In order to clear the fraction, we multiply all the coefficients by 2 and write down the balanced equation:

Element	Predicted Electron Configuration	Actual Electron Configuration
copper, Cu	[Ar] 3d⁹ 4s²	[Ar] 3d¹⁰4s¹
silver, Ag	[Kr] 4d⁹ 5s²	[Kr] 4d¹⁰ 5s¹
gold, Au	[Xe] 4f¹⁴ 5d⁹ 6s²	[Xe] 4f¹⁴ 5d¹⁰ 6s¹
palladium, Pd	[Kr] 4d⁸ 5s²	[Kr] 4d¹⁰
chromium, Cr	[Ar] 3d⁴ 4s²	[Ar] 3d⁵ 4s¹
molybdenum, Mo	[Kr] 4d⁴ 5s²	[Kr] 4d⁵ 5s¹

Made Easy Chemistry

Why are copper and chromium exceptions to electron configuration?

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Skills to develop

Energy Levels in Many-Electron Atoms

The Pauli Exclusion Principle

Hund's RULE

The Aufbau Procedure

Block of elements by highest occupied atomic orbitals

Special Electronic Configurations

Confidence Building Questions

Electron Configuration Anomalies

Filling Electron Shells

Filling Electron Shells

Aufbau principle

Hund's Rule

Pauli Exclusion principle

Orbital Order

How to Write the Electron Configuration of an Isolated Atom

Rule of Stability

Magnetism

Order of Filling of Electron States

Shapes of Atomic Orbitals

s Sublevels are Spherically Shaped

When the Azimuthal Quantum Number is 1, then ml Can Only Be -1, 0 or +1

The p Orbitals are Often Described as Dumb-bell Shaped

The d Orbitals and f Orbitals are Not Easily Visualized

Quantum Numbers and Atomic Orbitals

Quantum Numbers and Atomic Orbitals

Table of Allowed Quantum Numbers

Writing Electron Configurations

Properties of Monatomic Ions

Atomic Sub Shells

Atomic Sub Shells

Atomic Orbitals

Filling the orbitals

Difference between shells, subshells and orbitals

Nucleons Calculation

Nucleons

Example:

Ionization Energy

Ionization Energy Definition

Ionization Energy Trend

First Ionization Energy

Ionization Energy Periodic Table

Second Ionization Energy

Ionization Energy Chart

Factors Affecting Ionization Potential

1. Charge in the nucleus

2. Atomic radius

3. Completely and partially filled in shells

Successive Ionization Energies

Factors Affecting Ionization Potential

Factors Affecting Ionization Potential

1. Charge in the nucleus

2. Atomic radius

3. Completely and partially filled in shells

Electron Affinity

Ionization Energy

ELECTRONIC STRUCTURES

Inside the Atom

Using radioactivity safely

Using radioactivity safely

Ionising radiations

Isotopes and stability

Inside the nucleus

Atom

Calculations and titration

HOW TO CARRY OUT TIRATION?

Calculating relative formula masses

Relative atomic mass

Relative formula mass

Percentage yield

Theoretical vs actual yield

Carrying out a titration

Measuring purity

Steps of titration

Calculations from equations - Higher

Step 1: write down the balanced symbol equation

Electronic Configuration of Elements

Electronic Configuration of Elements

When the Azimuthal Quantum Number is 1, then m_l Can Only Be -1, 0 or +1

Calculations involving density, percent concentration, molality and molarity
Problems #1 - 10