3.1.1 (c) First Ionisation Energy

Syllabus

(c) first ionisation energy (removal of 1 mol of electrons from 1 mol of gaseous atoms) and successive ionisation energy, and:

(i) explanation of the trend in first ionisation energies across Periods 2 and 3, and down a group, in terms of attraction, nuclear charge and atomic radius.

(ii) prediction from successive ionisation energies of the number of electrons in each shell of an atom and the group of an element

{Definition required for first ionisation energy only.}

{Explanation to include the small decreases as a result of s- and p-sub-shell energies (e.g. between Be and B) and p-orbital repulsion (e.g. between N and O).}

What does this mean?

What is first ionisation energy?

Some elements naturally form ions, some do not.

Some elements naturally form positive cations, other negative anions.

None of that is really relevant to ionisation energies.

We have to compare like with like so since it is possible to remove electrons from any atom if we give them enough energy our definition is for forming positive ions by removing electrons.

And since we can't really deal with individual atoms we deal with molar quantities.

So learn the definition:

The first ionisation energy of an element is the energy necessary to remove of 1 mole of electrons from 1 mole of gaseous atoms, to create 1 mole of gaseous 1⁺ ions.

Notice that we have to start with gaseous elements to have a useful comparison between say Iron and Neon.

And that they must be as atoms rather than molecules to have a useful comparison between say F₂ and He.

Strictly speaking the atoms should be in their ground state - with their electrons in the lowest possible energy levels - for a fair comparison to be made.

Learn this - examiners love this definition and it will gain more than 1 mark in an exam.

You won't be asked to define 2nd or 3rd ionisation energies.

Which is a shame because the definitions are very similar.

You might well be asked to write an equation for a 2nd or 3rd ionisation energy though.

Video

What do first IEs tell us?

You don't have to reproduce this graph in an exam.

But you may have to predict the position of one or two elements that have been removed.

And you'll certainly have to be able to explain what this graph tells us.

And why it is this shape.

It's difficult at first but it helps to explain a great many other things in Chemistry so it is well worth the effort.

Shells

Ever wondered how we know that there are shells?

Or how we know how many elements are in each shell?

We can't image electrons so we base our theories on evidence like this.

Notice that there are particularly large drops between elements 2 & 3 (He and Li), between 10 & 11 (Ne and Na), and 18 & 19 (Ar and K)

Why should it suddenly become easier to remove electrons at these points?

The assumption is because the second element in each pair has its outermost electron is in a new shell, further away from the nucleus where it is less attracted by the protons and suffers extra shielding from inner shells. This lowering of the attraction outweighs the extra attraction one extra proton in the nucleus.

So Shell 1 contains only two elements (H and He)

Shell 2 contains 8 elements from Li to Ne

Shell 3 is a more complicated due to the 3d electrons being a higher energy than 4s - it should contain 18 elements.

Sub-shells

How do we know that there's any difference between s and p-orbitals?

In all periods there is a general upwards trend to 1st IEs

You should be able to explain why in terms of

increasing nuclear charge as more protons are added
no extra shielding (no extra shells)
decreasing atomic radius - atoms get smaller across a period as the same shell is pulled inwards by the extra protons

So why is there always a drop between the second and third element's 1st IE?

We assume that the third electron must be entering a slightly higher energy orbital from which it is easier to remove.

Spin-pairing

Some of the evidence for spin-pairing can also be seen on the graph.

In both Period 2 and Period 3 there is a drop off in 1st IE from the 5th to the 6th element.

This time it isn't due to a new subshell it is due to moving from half-filled orbitals to doubly filled ones.

Obviously, electrons repel each other.

So this makes removing one slightly easier even though the atomic radius is a little smaller and there is one more proton in the nucleus.

Video

Why are we ignoring the d-block elements?

Because they're all very similar and they don't teach us much.

The pattern of Period 2 and 3 is repeated in other Periods but the d-block elements and f-block elements don't have much to show us.

Successive ionisation energies

You may not need to be able to define 2nd and third ionisation energies but you'll still need to think about them.

If you look any element the successive ionisation energies increase each time but by a similar amount each time when we remove electrons from the outer shell.

You should be able to account for this rise by pointing out that the "effective nuclear charge" is increasing because although there are no extra protons there are fewer electrons to attract.

Also, as the effective nuclear charge increases the ions will get smaller.

And removing a negative electron from a 1+ ion would always be easier than from a 2+ or 3+ ion.

Now look at what happens when you move to taking an electron from an inner shell.

There is a sudden increase.

We can use this to identify what group an atom is from its successive ionisation energies.

An examiner might tell you the graph below is for an element in the first 18.

You should note a big jump between 1st and 2nd IE.

This tells us its a Grp 1 element.

It can't be Lithium because ₃Li doesn't have 11 electrons to remove.

So it must be Sodium.

If the graph looked like this...

Then it's clear we have 7 similar IE's and a big jump to 8.

It's Group 7 element, though we couldn't say for sure if it was F or Cl.

If we want to put all of Chlorine's IE's on the same graph we would have to take Logs of them.

This fits them on the same scale but allows the jump between 7 & 8 (and 15 & 16) to be seen.

PowerPoint

Google Presentation

Exam-style Questions

1. The first ionisation energies of Neon, Sodium and Magnesium are 2080, 494 & 736 kJ mol^-1, respectively.

(a) Explain the meaning of the term first ionisation of an atom.

...................................................................................................……………………………………………………………………

(2)

(b) Write an equation to illustrate the process occurring when the second ionisation energy of magnesium is measured.

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(2)

(c) Explain why the value of the first ionisation energy of Magnesium is higher than that of Sodium.

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(2)

(d) Explain why the value of the first ionisation energy of Neon is higher than that of Sodium.

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(2)

(Total 8 marks)

2. The diagram below shows the values of the first ionisation energies of some of the elements in Period 3.

(a) On the above diagram, use crosses to mark the approximate positions of the values of the first ionisation energies for the elements Na, P and S. Complete the diagram by joining the crosses.

(3)

(b) Explain the general increase in the values of the first ionisation energies of the elements Na–Ar.

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(3)

(c) In terms of the electron sub-levels involved, explain the position of Aluminium and the position of Sulphur in the diagram.

Explanation for Aluminium ..............................................................................................................................................................

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Explanation for Sulphur ...................................................................................................................................................................

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(4)

(Total 10 marks)

Answers

1. (a) Enthalpy change/required when an electron is removed/knocked out/displaced (Ignore ‘minimum’ energy) (1)

From a gaseous atom (1)

(could get this mark from equation)

(b) Mg+_(g) → Mg2+_(g) + e– Equation (1)

Or Mg+_(g) + e– → Mg2+_(g) + 2e– State symbols (Tied to M1) (1)

(c) Increased/stronger nuclear charge or more protons (1)

Smaller atom or electrons enter the same shell or same/similar shielding (1)

(d) Electron removed from a shell of lower energy or smaller atom or e– nearer (1)

nucleus or e– removed from 2p rather than from 3s

Less shielding (1)

(Do not accept ‘e– from inner shell’)

2. (a)

(b) Increased nuclear charge / proton number (1)

NOT increased atomic number

Electrons enter same shell / energy level OR atoms get smaller OR same shielding (1)

Stronger attraction between nucleus and (outer) electrons (1)

(c) Explanation for Aluminium: (third) electron in (3)p sub-shell (1)

Sub-shell further away from nucleus OR of higher energy (1)

OR extra shielding from (3)s

Explanation for Sulphur: Pair of electrons in (3)p orbital (1)

Repulsion between electrons (1) tied to reference to e– pair in M3

Penalise ‘2p’ once only

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