(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 ]
Explanation
(a)(i) State Pauli's exclusion principle: Pauli's exclusion principle states that two electrons in the same orbita of an atom cannot have same values for all four quantum numbers/no two electrons can have the same four quantum numbers/no two electrons in the same orbital if an atom can have the same spin. (ii) Hund's rule of maximum multiplicity. Hund's rule states that electrons occupy each orbital supply first before pairinc takes place in degenerate orbital/most stable arrangement of electrons in sunshells is the one with the greatest number of parallel spins.
(d) (i) Li\(^+\), Na\(^+\), K\(^+\), size increase down a group as more shells are being added. (ii) F, O\(^{2-}\), N\(^{2-}\), size increase as nuclear charge decrease in the same period/isoelectronic.
(e) Phosphorus (V) oxide = P\(_{4}\)O\(_{10}\)/P\(_{2}\)Q\(_{5}\) = (31 x 4) + (16 x 10) or (31 x 2) + (16 x 5) = 124 + 160 or 62 + 80 = 284 or 142. Therefore % by mass of phosphorus = \(\frac{4P}{P_4Q_{10}}\) x 100% or \(\frac{2P}{P_2Q_{5}}\) x 100% \(\frac{124}{284}\) x 100% or \(\frac{62}{142}\) x 100% = 43.7% or 100 - 43.7 = 56.3% oxygen.
