2.2.1 (a,b,c) Orbitals, Shells and Sub-Shells

Syllabus

Learners should be able to demonstrate and apply their knowledge and understanding of:

Energy levels, shells, sub-shells, atomic orbitals, electron configuration

(a) the number of electrons that can fill the first four shells

(b) atomic orbitals, including:

(i) as a region around the nucleus that can hold up to two electrons, with opposite spins

(ii) the shapes of s- and p-orbitals

(iii) numbers of orbitals in s-, p- & d-sub-shells, & the number of electrons that can fill s-, p- & d- sub-shells

(c) filling of orbitals:

(i) for the first three shells and the 4s and 4p orbitals in order of increasing energy

(ii) for orbitals with the same energy, occupation singly before pairing

{Learners are expected to be familiar with the 'electrons in box' representations.}

What does this mean?

What you already know and why it's (almost) all wrong.

At GCSE you learned 2.8.8

Hopefully you also learned to pair electrons, as below.

That model still works for many aspects of A' Level Chemistry - but not all.

Firstly, atoms are not 3 dimensional.

Electrons do not go around the nucleus in circular orbits - your Physics teacher can bore you with an explanation of why this would be impossible.

And we never really know exactly where an electron is, let alone where all the electrons in an atom are at the same time.

Oh, and the third shell can hold more than 8 electrons.

But, other than that, it was all fine, honest.

What you need to learn now.

What you learn at A level isn't entirely true either - but it works for almost everything up to degree level.

Orbital - "an orbital is an area of space about the nucleus that can hold a maximum of two electrons with opposite spins"

Shell - "a number of orbitals with the same principal quantum number(n)"

Sub-shell - "all the orbitals of the same type within one shell."

I hope that's clear.

But what does it mean?

Electrons fill up an atom from the lowest energy upwards.

In practice this means that all the electrons will go into the first shell until its full, then they'll fill the next shell until it is full and so on.

The first shell only has one orbital - which is why it can take a maximum of two electrons.

This first orbital is an s-orbital - it is spherical and is called the 1s orbital.

Every shell will have an s-orbital, getting larger as the principal quantum number rises from 1 to 2 and 3 etc.

But the second shell has both s and p orbitals, an s subshell and a p sub-shell.

The third has s, p and d orbitals (in an s subshell, p subshell and d subshell) - and that's as far as we generally go.

Though there are also f subshells containing f orbitals in later shells, which we don't talk about at A level (like Fight club)

S subshells always have one orbital

P subshells have 3

D subshells have 5

The first 20 elements

Hydrogen

₁H has 1 electron in the 1S orbital.

We would say its electronic configuration is 1S¹.

We could draw it in the GCSE way as shown right.

But more frequently we do the electron-box model shown below.

Helium

₂H has 2 electrons in the 1S orbital.

We would say its electronic configuration is 1S².

We should draw its electron-box with the electrons spinning oppositely.

Lithium

₃Li has 2 electrons in the 1S orbital but this is full.

There are two sub-shells in shell 2 - the s-subshell and the p-subshell.

The extra electron must go into the 2S orbital because it is at a slightly lower energy than the 2P orbitals (this is true of all shells)

We would say its electronic configuration is 1S²2S¹.

Beryllium

₄Be has 2 electrons in the 1S orbital and two in the 2S also, also oppositely spinning.

We would say its electronic configuration is 1S²2S².

Boron

₅B has 2 electrons in the 1S orbital and two in the 2S.

The fifth electron must go into one of the three 2P orbitals

We would say its electronic configuration is 1S²2S²2P¹.

Carbon

₆C has 2 electrons in the 1S orbital and two in the 2S, and two in the 2P

Making its electronic configuration is 1S²2S²2P².

Only model (c) would be correct.

(a) is incorrect because you can't start spin-pairing electrons until there's no alternative because they repel.

(b) is also incorrect because having neighbouring electrons in parallel spins is the lower energy arrangement.

Following this the arrangements would be:

N = 1S²2S²2P3

O = 1S²2S²2P⁴

F = 1S²2S²2P5

Ne = 1S²2S²2P6

What do P orbitals look like?

P orbitals are not spherical - though no one will ask you to put their shape into words. (Propeller shape? Dumb-bell shape?)

But they do have a direction - along the x, y or z axis

And the three directions are at 90^o to each other.

They are orthogonal.

Element 20 and beyond.

The gap between energy levels n=1 and n=2 is bigger than the gap between n=2 and n=3, which is bigger than the gap between n=3 and n= 4 and so on.

In this diagram we see an electron in n=2 falling back to n=1

This emits light of a particular frequency.

Which is how we can know anything about the differences in energy.

eV is a measure of energy (Physics, ignore it!).

Notice that the energy levels are getting closer as they get bigger.

Eventually there is no difference between levels and the electrons are no longer in the atom.

In reality, every energy level except n=1 is split into sub-shells

By the time we get to n=3 and n=4 a problem occurs.

The 3d subshell is at a very slightly higher energy level than the 4s subshell.

So when electrons are filling up an atom they will fill the 4s before the 3d.

Although there are two exceptions to even this rule, it does at least explain something about the shape of the periodic table.

Notice the 3d orbitals appear after the 4s orbitals.

The Exceptions

Because everything up to element 18 (Neon) is entirely predictable we often use a shortened form of electronic configuration for larger elements where [Ne] represents 1s² 2s² 2p⁶ 3s² 3p⁶

Notice Chromium is [Ne] 4s¹ 3d⁵rather than [Ne] 4s² 3d⁴- which is what we would expect.

You won't be asked why but it is clear that there must be some energetic advantage to having six half-filled orbitals of around the same energy instead of one filled orbital, one empty one and four half-filled ones.

Similarly, notice Copper is [Ne] 4s¹ 3d¹⁰rather than [Ne] 4s² 3d⁹

There must be some energetic advantage to having all five 3d orbitals filled and the 4s half-filled ones rather than one filled 4s, and one half-filled 3d alongside four filled ones.

S,P,D

It is important that you know that electrons fill sub-shells (generally) in this order.

You should also know that there's only ever one s-orbital in each shell.

That there are three p-orbitals in every shell after n=1

That there are five d-orbitals in every shell after n=2

So the shell maximums are really 2.8.18 - but we don't notice because of the 3d/4s problem.

If you needed to work out how many electrons are in subsequent shells you can use the formula:

Total electrons per shell = 2n²

{where n= the principal quantum number}

So,

Shell 1 ===> 2(1)² = 2

Shell 2 ===> 2(2)² = 8

Shell 3 ===> 2(3)² = 18

Shell 4 ===> 2(4)² = 32

Shell 5 ===> 2(5)² = 50

You don't really need to know what d orbitals look like but you know you want to;

And you absolutely don't need to know what f orbitals look like:

Notice that there are 7 - containing a maximum of 14 electrons.

Which is why there are 14 Lanthanides and 14 Actinide elements in the f block.

Videos

Exam-style Questions

1. (a) One isotope of Sodium has a relative mass of 23.

(i) Define, in terms of the fundamental particles present, the meaning of the term isotopes.

.......................................................................................................................................................................

(ii) Explain why isotopes of the same element have the same chemical properties.

.......................................................................................................................................................................

....................................................................................................................................................................... (3)

(b) Give the electronic configuration, showing all sub-levels, for a Sodium atom.

.......................................................................................................................................................................(1)

(c) Explain why Chromium is placed in the d block in the Periodic Table.

.......................................................................................................................................................................

.......................................................................................................................................................................(1)

(d) An atom has half as many protons as an atom of 28Si and also has six fewer neutrons than an atom of 28Si. Give the symbol, including the mass number and the atomic number, of this atom.

.......................................................................................................................................................................(2)

2. (a) In terms of sub-levels, give the complete electronic configuration of the Nitrogen atom,

N, and of the Nitride ion, N^3-.

N....................................................................................................................................................................

N^3- .................................................................................................................................................................

(2)

(b) Complete the electronic configurations for the metals Sodium and Iron.

Electronic configuration of Sodium 1s² .......................................................................................................................................................................

Electronic configuration of Iron 1 s² .......................................................................................................................................................................

(2)

(c) Complete the following electronic configurations. [Ar] represents the electronic configuration of the Argon atom.

A Copper atom [Ar] .......................................................................................................................................................................

A Copper(II) ion [Ar] .......................................................................................................................................................................

(2)

(d) Give the electronic configuration of the F ^- ion in terms of levels and sub-levels.

.......................................................................................................................................................................(1)

(e) Complete the following to show the electronic configuration of silicon.

1s² 2s² .......................................................................................................................................................................

(1)

(Total 8 marks)

Answers

1(a) (i) Atoms with the same number of protons / proton number (1)

NOT same atomic number

with different numbers of neutrons (1)

NOT different mass number / fewer neutrons

(ii) Chemical properties depend on the number or amount of (outer) electrons (1)

OR, isotopes have the same electron configuration / same number of e–

(b) 1s2 2s2 2p6 3s1 (1) (1)

accept subscripted figures

(c) Highest energy e– / outer e–s / last e– in (3)d sub-shell (1)

OR d sub-shell being filled / is incomplete

OR highest energy sub-shell is (3)d

NOT transition element / e– configuration ends at 3d

(d) N correct symbol (1)

Mass number = 15 AND atomic number = 7 (1)

2. (a)

1s2 2s2 2p3 (1)

1s2 2s2 2p6 (1)

(b)

Na 1s2 2s2 2p6 3s1 (1)

Fe 1s2 2s2 2p6 3s2 3p6 3d6 4s2 (1)

accept 3d and 4s transposed

(c)

3d10 4s1 or reverse order (1) 3d9 (1) (allow full configuration if correct; if wrong penalise once)

(d) 1s2 2s2 2p6 (1)

(e) (1s22s2) 2p63s23p2 (1)

Back to 2.2.1?

Click here