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Atomic Accountancy Worksheet 2 Answers to Critical Thinking, Exercises of Chemistry

CHEM1611 Chemistry A Pharmacy University of Sydney, first year chemistry worksheet with answers to all critical questions about atomic orbitals and electronic configuration of atoms.

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CHEM1611 Worksheet 2: Atomic Accountancy
Model 1: Atomic Orbitals
Throughout history, the model of the atom and how/where the electrons exist and move has changed as our
scientific knowledge has increased. The current model describes the motions of electrons using atomic orbitals.
Orbitals gives us information about the probability of an electron being in a particular place around the nucleus.
Orbitals have different shapes and sizes, depending on the energy of the electron.
To understand how orbitals work, we must first consider quantum numbers. Each electron has a set of four
quantum numbers, each describing a certain property. There are four quantum numbers:
n: the principal quantum number identifies the size and energy of the orbital. Its value can be any non-
zero integer: 1, 2, 3, 4... with the size of the orbital increasing with n.
l: the angular momentum quantum number identifies the shape of the orbital. For each value of n, it has
values from 0 to (n-1). For historic reasons, a code is used for the l value:
o l = 0 means an s orbital
o l = 1 means a p orbital
o l = 2 means a d orbital
o l = 3 means an f orbital
ml: the magnetic orbital quantum number identifies the subshell and the orientation of the orbital For
each value of l, it has values from l…0…-l
ms: the spin quantum number which describes the spin of the electron. It has values of +½ or -½ which
are sometimes called ‘spin up’ and ‘spin down’ respectively. Each orbital can be used for a maximum of
2 electrons – one ‘spin up’ and one ‘spin down’.
This all sounds very complicated! However, as you will see below, these quantum numbers lead to the Periodic
Table and hence to all of chemistry (and so all biochemistry, biology, medicine etc.). The shapes and sizes of
all of the orbitals with n = 1, 2 and 3 are shown below.
Critical thinking questions
1. What are the characteristic shapes of s, p, and d orbitals?
2. Which quantum number identifies the shape of an orbital?
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CHEM1611 Worksheet 2 : Atomic Accountancy

Model 1 : Atomic Orbitals

Throughout history, the model of the atom and how/where the electrons exist and move has changed as our scientific knowledge has increased. The current model describes the motions of electrons using atomic orbitals. Orbitals gives us information about the probability of an electron being in a particular place around the nucleus. Orbitals have different shapes and sizes, depending on the energy of the electron.

To understand how orbitals work, we must first consider quantum numbers. Each electron has a set of four quantum numbers, each describing a certain property. There are four quantum numbers:

  • n : the principal quantum number identifies the size and energy of the orbital. Its value can be any non- zero integer: 1, 2, 3, 4... with the size of the orbital increasing with n.
  • l : the angular momentum quantum number identifies the shape of the orbital. For each value of n , it has values from 0 to ( n - 1). For historic reasons, a code is used for the l value: o l = 0 means an s orbital o l = 1 means a p orbital o l = 2 means a d orbital o l = 3 means an f orbital
  • ml : the magnetic orbital quantum number identifies the subshell and the orientation of the orbital For each value of l , it has values from l …0…- l
  • m s: the spin quantum number which describes the spin of the electron. It has values of +½ or - ½ which are sometimes called ‘spin up’ and ‘spin down’ respectively. Each orbital can be used for a maximum of 2 electrons – one ‘spin up’ and one ‘spin down’.

This all sounds very complicated! However, as you will see below, these quantum numbers lead to the Periodic Table and hence to all of chemistry (and so all biochemistry, biology, medicine etc.). The shapes and sizes of all of the orbitals with n = 1, 2 and 3 are shown below.

Critical thinking questions

  1. What are the characteristic shapes of s , p , and d orbitals?
  2. Which quantum number identifies the shape of an orbital?
  1. For each value of n = 1, 2, and 3, what are the possible values for l , and what labels correspond to these orbitals? ( Hint : re-read the second bullet point in Model 1.)

n Possible l values Orbital labels

1

2

3

  1. For each value of l = 0, 1, 2, what are the possible values for ml? For each value of l , how many values can ml have? ( Hint : re-read the third bullet point in Model 1.). Complete the first 3 columns in the table below.
  2. Complete the 4th^ column in the table below. ( Hint : re-read the fourth bullet point in Model 1.)

l Possible ml values Number of values of ml Number of orbitals Maximum number of electrons 0 1 2

Model 2 : Electronic Configurations of Atoms

The electronic configuration of an atom provides information on which orbitals its electrons are in. The Periodic Table below shows how the orbitals are occupied by the valence electrons. The number in front of the orbital ( 1 s , 2 s , 2 p , etc) gives the shell (the n quantum number). The letter ( s , p , d , etc) gives the subshell (the l quantum number). The superscript (1 s^1 , 1 s^2 , 2 p^3 etc) tells us how many electrons are in the sub-shell.

An element has the electronic configuration of the previous element plus an extra electron in the orbital shown on the figure. For example, H is 1 s^1 and He is 1 s^2. Li has the electron configuration of He plus one more electron so is 1 s^2 2 s^1. The valence electrons of an atom are those in the outermost shell of the atom; core electrons are those in the completely filled inner shells.

Critical thinking questions

  1. What are the electronic configurations the following elements?

(a) Li (b) Be (c) B (d) C

(e) N (f) O (g) F (h) Ne

The figure on the right shows an easy way to remember the order in which atomic orbitals are filled. The arrows are followed starting from the top and remembering the ideas that you have discovered above:

  • each s orbital can accommodate a maximum of 2 electrons
  • each set of p orbitals can accommodate a maximum of 6 electrons
  • each set of d orbitals can accommodate a maximum of 10 electrons
  • each set of f electrons can accommodate a maximum of 14 electrons
  1. What is the electron configuration of Fe? Give your answer using both long hand and the [X] shorthand.
  2. What is the electron configuration of As?

Model 3 : Electronic Configurations of Ions

The electronic configurations of cations and anions can be worked out from those of the neutral atoms by removing or adding electrons respectively:

  • O has the electron configuration [He] 2s^2 2p^4. O^2 -^ has the electron configuration [He] 2s^2 2p^6 : the two electrons were added to the 2 p orbitals.
  • Mg has the electron configuration [Ne] 3s^2. Mg2+^ has the electron configuration [Ne] : the two electrons are removed from the 2 s orbital.

Critical thinking questions

  1. Draw the box representation for the valence electrons of O, O^2 - , Mg and Mg2+. What do you notice?
  2. Write down the electronic configurations of the following ions.

(a) S^2 -^ (b) Al3+^ (c) H+^ (d) H-

Extension questions

  1. Discuss within your group, the best position in the periodic table for H and He considering their chemical nature and their electronic configuration.
  2. The first ionisation energy (I.E.) of the first 18 elements is plotted below. Identify the general trends that

exist moving across a period and down a group of the periodic table and discuss in your groups the reasons why IE(B) < IE(Be) and IE(O) < IE(N).

1s

2s 2p

3s 3p 3d

4s 4p 4d 4f

5s 5p 5d 5f

0

500

1000

1500

2000

2500

H He Li Be B C N O F Ne Na Mg Al Si P Si Cl Ar

I.E. (kJ/mol)

CHEM1611 Worksheet 2 – Answers to Critical Thinking Questions

The worksheets are available in the tutorials and form an integral part of the learning outcomes and experience for this unit.

Model 1: Atomic Orbitals

  1. s orbitals consist of a single lobe. p orbitals consist of two large lobes. d orbitals consist of four lobes.

(The shape of 3 d z 2 is a little different to that of the other 3 d orbitals. It consists of two large lobes with a ring around the centre.)

  1. The orbital quantum number, l.
  2. See table below.

n Possible l values Orbital labels

1 0 only 1 s

2 0 and 1 2 s and 2 p

3 0, 1 and 2 3 s , 3 p and 3 d

  1. See table below.
  2. See table below.

l Possible ml values Number of values of ml Number of orbitals Maximum number of electrons

0 0 1 1 1 × 2 = 2

1 1, 0 and -1 3 3 3 × 2 = 6

2 2, 1, 0, -1 and -2 5 5 5 × 2 = 10

Model 2: Electronic Configurations of Atoms

  1. (a) 1 s^2 2 s^1 or [He] 2 s^1 (b) 1 s^2 2 s^2 or [He] 2 s^2

(c) 1 s^2 2 s^2 2 p^1 or [He] 2 s^2 2 p^1 (d) 1 s^2 2 s^2 2 p^2 or [He] 2 s^2 2 p^2 (e) 1 s^2 2 s^2 2 p^3 or [He] 2 s^2 2 p^3 (f) 1 s^2 2 s^2 2 p^4 or [He] 2 s^2 2 p^4 (g) 1 s^2 2 s^2 2 p^5 or [He] 2 s^2 2 p^5 (h) 1 s^2 2 s^2 2 p^6 or [He] 2 s^2 2 p^6

  1. 1 s^2 2 s^2 2 p^6 3 s^1 or [Ne] 3 s^1. Both Na and Li have 1 electron in an s orbital outside a noble gas

configuration.

  1. 2 s is lower in energy than 2 p. (2 s is filled before 2 p is.).
  2. The arrows refer to m s - the spin quantum number. According to the Pauli Principle, a maximum of one

‘up’ and one ‘down’ spin (arrow) can occupy each orbital (box).

  1. Hund’s Rule.

Extension questions

  1. As H is 1 s^1 and He is 1 s^2 , a Periodic Table based purely on electronic configurations would have the

elements in the s -block, above Li and Mg respectively. The Periodic Table, however, was worked out from experimental observations of chemical and physical properties. He is an unreactive gas and for this reason, it is placed in the Nobel Gases. This can perhaps be justified from its electronic configuration as it has the maximum number of electrons possible for an element with n = 1 just as Ne has the maximum number of electrons possible for an element with n = 2. H is more awkward! Chemically it behaves a little like the Group 1 elements – its commonest oxidation number is +1, just like the Group 1 elements. However, compounds containing it with this oxidation number are very different to those of Group 1 elements: LiOH, NaOH and KOH contain M+^ and OH-^ ions and are strongly basic. HOH is a covalent molecule and is not basic. In its elemental form, hydrogen exists as a gas made up of H 2 molecules whereas the Group 1 elements exist as metallic solids. (Note however that at very high pressure such as in the cores of planets, hydrogen probably is also metallic). In forming the diatomic H 2 molecule, hydrogen acts more like the halogens (F 2 , Cl 2 , Br 2 and I 2 ). When it reacts with metals, it forms hydrides such as LiH which contain H-^ ions. This is akin to the halogens which also forms anions in compounds like LiF and NaCl. There is no correct answer!

  1. There are a number of trends:

(a) Ionisation energies decrease down each group as the electron to be removed is in an orbital with a higher n quantum number and orbits further from the nucleus. As a result, the attraction to the nucleus is reduces. (b) Ionisation energies generally increase across a period as the n quantum number does not change but the positive charge of the nucleus increases. As a result, the attraction to the nucleus increases. The combination of (a) and (b) means that the first element in each row (H, Li, Na) have the lowest ionisation energy in that row whilst the last element in each row (He, Ne and Ar) have the highest ionisation energy in that row. (c) Trend (b) is broken between the between Be and B and between Mg and Al. This is due to the change in the orbital of the electron which is being removed. When Be (2 s^2 ) is ionised, an s - electron is removed. When B (2 s^22 p^1 ) is ionised, a p electron is removed. As 2 p -orbitals have higher energies than 2 s orbitals, it is easier to ionise B than Be despite the nuclear charge being higher. (d) Trend (b) is also broken between N and O and between P and S. This is due to the change in the spin of the electron being ionised. In N (2 s^22 p^3 ), the p -electron being ionised has the same spin as the other p -electrons. In O, (2 s^22 p^4 ), the p -electron being ionised has the opposite spin as the other p -electrons. (Draw the electronic configurations using the box notation to see this.) Electrons with the opposite spin repel each other more than electrons with the same spin. It is then easier to remove an electron from O.