IB DP Chemistry Structure 1.2 The nuclear atom HL Paper 2- Exam Style Questions - New Syllabus
▶️ Answer/Explanation
Markscheme (with detailed working)
(a)
Relative atomic mass from weighted average: \[ A_r(\mathrm{Ni}) = 58(0.68) + 60(0.26) + 61(0.01) + 62(0.04) + 64(0.01) = 39.44 + 15.60 + 0.61 + 2.48 + 0.64 = \boxed{58.77}. \] [1]
Relative atomic mass from weighted average: \[ A_r(\mathrm{Ni}) = 58(0.68) + 60(0.26) + 61(0.01) + 62(0.04) + 64(0.01) = 39.44 + 15.60 + 0.61 + 2.48 + 0.64 = \boxed{58.77}. \] [1]
(b)
(i) Nickel has \(Z=28\). If it has 26 electrons, charge is \(2+\). Nuclear symbol: \(\boxed{{}^{58}_{28}\mathrm{Ni}^{2+}}\). [1]
(ii) Electron configuration of \(\mathrm{Ni^{2+}}\): \([\mathrm{Ar}]\,3d^8\) (4s empty). Orbital diagram shows eight \(3d\) electrons with two unpaired: 4s: [ ] 3d: [↑↓][↑↓][↑↓][↑][↑].
[1]
(iii) \(\boxed{\text{Paramagnetic}}\) — because there are unpaired electrons in \(3d\). [1]
(i) Nickel has \(Z=28\). If it has 26 electrons, charge is \(2+\). Nuclear symbol: \(\boxed{{}^{58}_{28}\mathrm{Ni}^{2+}}\). [1]
(ii) Electron configuration of \(\mathrm{Ni^{2+}}\): \([\mathrm{Ar}]\,3d^8\) (4s empty). Orbital diagram shows eight \(3d\) electrons with two unpaired: 4s: [ ] 3d: [↑↓][↑↓][↑↓][↑][↑].
[1](iii) \(\boxed{\text{Paramagnetic}}\) — because there are unpaired electrons in \(3d\). [1]
(c)
(i) From the spectrochemical series, \(\mathrm{NH_3}\) is a stronger-field ligand than \(\mathrm{H_2O}\) (\(\cdots < \mathrm{H_2O} < \mathrm{NH_3} \cdots\)), so \(\Delta_\mathrm{o}\) (d-orbital splitting) is larger in \([\mathrm{Ni(NH_3)_6}]^{2+}\). A larger \(\Delta_\mathrm{o}\) means absorption of higher-energy (shorter-wavelength) light. Blue–violet is observed when yellow–orange light (\(\sim 585\text{–}647\ \mathrm{nm}\)) is absorbed, hence \([\mathrm{Ni(NH_3)_6}]^{2+}\) appears blue–violet. [2]
(ii) \(\mathrm{NH_3}\) acts as a \(\boxed{\text{Lewis base}}\) — it donates a lone pair to \(\mathrm{Ni^{2+}}\) to form a coordinate bond. [1]
(iii) In water, \(\mathrm{NH_3}\) acts as a \(\boxed{\text{Brønsted–Lowry base}}\) (proton acceptor): \[ \mathrm{NH_3(aq) + H_2O(l) \rightleftharpoons NH_4^+(aq) + OH^-(aq)}. \] (Alternative: \(\mathrm{NH_4OH(aq)}\) formulation also acceptable.) [2]
(i) From the spectrochemical series, \(\mathrm{NH_3}\) is a stronger-field ligand than \(\mathrm{H_2O}\) (\(\cdots < \mathrm{H_2O} < \mathrm{NH_3} \cdots\)), so \(\Delta_\mathrm{o}\) (d-orbital splitting) is larger in \([\mathrm{Ni(NH_3)_6}]^{2+}\). A larger \(\Delta_\mathrm{o}\) means absorption of higher-energy (shorter-wavelength) light. Blue–violet is observed when yellow–orange light (\(\sim 585\text{–}647\ \mathrm{nm}\)) is absorbed, hence \([\mathrm{Ni(NH_3)_6}]^{2+}\) appears blue–violet. [2]
(ii) \(\mathrm{NH_3}\) acts as a \(\boxed{\text{Lewis base}}\) — it donates a lone pair to \(\mathrm{Ni^{2+}}\) to form a coordinate bond. [1]
(iii) In water, \(\mathrm{NH_3}\) acts as a \(\boxed{\text{Brønsted–Lowry base}}\) (proton acceptor): \[ \mathrm{NH_3(aq) + H_2O(l) \rightleftharpoons NH_4^+(aq) + OH^-(aq)}. \] (Alternative: \(\mathrm{NH_4OH(aq)}\) formulation also acceptable.) [2]
Total Marks: 9