(a) State (i) Pauli's Excusion principle;
(ii) Hund's rule of maximum multiplicity.
(b)(i) Write the electronic configuration of each of the following ions of copper: I. Cu\(_+\) II. Cu\(_{(2+)}\) [\(_{29}Cu\)]
(ii) Give the number of unpaired-electrons in each of the ions in (b)(i) above.
(iii) State the type of reaction represented by the following equation:
2Cu\(^+_{(aq)}\) \(\to\) Cu\(^{2+}_{(aq)}\) + Cu\(_{(s)}\)
(iv) Write the formula of one compound of Cu+.
(c)(i) Name the type of radiation that will I. penetrate lead block; II. be stopped by thin paper
(ii) Give the charge on each of the radiations mentioned in I(c)(i) above.
(iii) What term is used to describe each of the following nuclear processes?
I. Combination of two lighter nuclei to form a heavy nucleus II. Splitting of a heavy nucleus into two or more lighter nuclei
III. Time required for one-half of the atoms of a radioactive substance to decay.
(d) Arrange the following ions in order of increasing size. Give a reason for your answer in each case. I. Li\(^{+}\), K\(^{+}\), Na\(^{+}\); II. O\(^{2-}\), F\(^{-}\), N\(^{3-}\)
(e) Determine the percentage composition of phosphorus and oxygen in phosphorus (V) oxide [ P = 31, O = 16 ]
(a) Statements of the rules
- (i) Pauli's Exclusion Principle: No two electrons in the same atom can have the same set of four quantum numbers. In effect, an orbital can hold a maximum of two electrons, and these two must have opposite (paired) spins.
- (ii) Hund's Rule of Maximum Multiplicity: When electrons are being placed in orbitals of equal energy (degenerate orbitals such as the three p or five d orbitals), each orbital is singly occupied first, all with parallel spins, before any orbital receives a second electron. This gives the maximum number of unpaired electrons and the lowest energy arrangement.
(b) Copper ions (Cu has atomic number 29; atom = \([Ar]3d^{10}4s^{1}\)). Electrons are lost from the 4s orbital first, then from 3d.
- (i) I. Cu\(^{+}\): \(1s^{2}2s^{2}2p^{6}3s^{2}3p^{6}3d^{10}\) i.e. \([Ar]3d^{10}\)
- I. Cu\(^{2+}\): \(1s^{2}2s^{2}2p^{6}3s^{2}3p^{6}3d^{9}\) i.e. \([Ar]3d^{9}\)
- (ii) Unpaired electrons: Cu\(^{+}\) (\(3d^{10}\)) has 0 unpaired electrons (all paired); Cu\(^{2+}\) (\(3d^{9}\)) has 1 unpaired electron.
- (iii) The reaction \(2Cu^{+} \to Cu^{2+} + Cu\) is a disproportionation reaction (a redox reaction in which the same species, Cu\(^{+}\), is simultaneously oxidized to Cu\(^{2+}\) and reduced to Cu).
- (iv) A compound of Cu\(^{+}\): copper(I) oxide, Cu\(_2\)O (or copper(I) chloride, CuCl).
(c) Radioactivity
- (i) I. Radiation that penetrates a lead block: gamma (\(\gamma\)) radiation. II. Radiation stopped by thin paper: alpha (\(\alpha\)) radiation.
- (ii) Charges: gamma radiation carries no charge (0); alpha radiation carries a charge of +2 (it is a helium nucleus, \(^{4}_{2}He^{2+}\)).
- (iii) I. Nuclear fusion. II. Nuclear fission. III. Half-life.
(d) Order of increasing ionic size
- I. Li\(^{+}\) < Na\(^{+}\) < K\(^{+}\). Reason: these are ions of the same group with the same charge; going down the group the number of occupied electron shells increases, so the ionic radius increases.
- II. F\(^{-}\) < O\(^{2-}\) < N\(^{3-}\). Reason: these are isoelectronic ions (each has 10 electrons). As the nuclear charge (number of protons) decreases from F (9) to O (8) to N (7), the pull on the fixed number of electrons weakens, so the electron cloud expands and the ionic radius increases.
(e) Percentage composition of P\(_2\)O\(_5\) (phosphorus(V) oxide; P = 31, O = 16)
Molar mass \(= (2\times31) + (5\times16) = 62 + 80 = 142\ \text{g mol}^{-1}\).
\[\%\,P = \frac{62}{142}\times100 = 43.66\%\]
\[\%\,O = \frac{80}{142}\times100 = 56.34\%\]
So P\(_2\)O\(_5\) is about 43.7% phosphorus and 56.3% oxygen.
(a) Statements of the rules
- (i) Pauli's Exclusion Principle: No two electrons in the same atom can have the same set of four quantum numbers. In effect, an orbital can hold a maximum of two electrons, and these two must have opposite (paired) spins.
- (ii) Hund's Rule of Maximum Multiplicity: When electrons are being placed in orbitals of equal energy (degenerate orbitals such as the three p or five d orbitals), each orbital is singly occupied first, all with parallel spins, before any orbital receives a second electron. This gives the maximum number of unpaired electrons and the lowest energy arrangement.
(b) Copper ions (Cu has atomic number 29; atom = \([Ar]3d^{10}4s^{1}\)). Electrons are lost from the 4s orbital first, then from 3d.
- (i) I. Cu\(^{+}\): \(1s^{2}2s^{2}2p^{6}3s^{2}3p^{6}3d^{10}\) i.e. \([Ar]3d^{10}\)
- I. Cu\(^{2+}\): \(1s^{2}2s^{2}2p^{6}3s^{2}3p^{6}3d^{9}\) i.e. \([Ar]3d^{9}\)
- (ii) Unpaired electrons: Cu\(^{+}\) (\(3d^{10}\)) has 0 unpaired electrons (all paired); Cu\(^{2+}\) (\(3d^{9}\)) has 1 unpaired electron.
- (iii) The reaction \(2Cu^{+} \to Cu^{2+} + Cu\) is a disproportionation reaction (a redox reaction in which the same species, Cu\(^{+}\), is simultaneously oxidized to Cu\(^{2+}\) and reduced to Cu).
- (iv) A compound of Cu\(^{+}\): copper(I) oxide, Cu\(_2\)O (or copper(I) chloride, CuCl).
(c) Radioactivity
- (i) I. Radiation that penetrates a lead block: gamma (\(\gamma\)) radiation. II. Radiation stopped by thin paper: alpha (\(\alpha\)) radiation.
- (ii) Charges: gamma radiation carries no charge (0); alpha radiation carries a charge of +2 (it is a helium nucleus, \(^{4}_{2}He^{2+}\)).
- (iii) I. Nuclear fusion. II. Nuclear fission. III. Half-life.
(d) Order of increasing ionic size
- I. Li\(^{+}\) < Na\(^{+}\) < K\(^{+}\). Reason: these are ions of the same group with the same charge; going down the group the number of occupied electron shells increases, so the ionic radius increases.
- II. F\(^{-}\) < O\(^{2-}\) < N\(^{3-}\). Reason: these are isoelectronic ions (each has 10 electrons). As the nuclear charge (number of protons) decreases from F (9) to O (8) to N (7), the pull on the fixed number of electrons weakens, so the electron cloud expands and the ionic radius increases.
(e) Percentage composition of P\(_2\)O\(_5\) (phosphorus(V) oxide; P = 31, O = 16)
Molar mass \(= (2\times31) + (5\times16) = 62 + 80 = 142\ \text{g mol}^{-1}\).
\[\%\,P = \frac{62}{142}\times100 = 43.66\%\]
\[\%\,O = \frac{80}{142}\times100 = 56.34\%\]
So P\(_2\)O\(_5\) is about 43.7% phosphorus and 56.3% oxygen.
