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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.
Typology: Exercises
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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:
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.
n Possible l values Orbital labels
1
2
3
l Possible ml values Number of values of ml Number of orbitals Maximum number of electrons 0 1 2
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.
(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:
The electronic configurations of cations and anions can be worked out from those of the neutral atoms by removing or adding electrons respectively:
(a) S^2 -^ (b) Al3+^ (c) H+^ (d) H-
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).
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H He Li Be B C N O F Ne Na Mg Al Si P Si Cl Ar
I.E. (kJ/mol)
The worksheets are available in the tutorials and form an integral part of the learning outcomes and experience for this unit.
(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.)
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
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
(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
configuration.
‘up’ and one ‘down’ spin (arrow) can occupy each orbital (box).
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!
(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.