Question

Which of these ions cannot form both high-spin and low-spin octahedral complexes:

(a)${\mathbf{Ti}}^{\mathbf{3}\mathbf{+}}$; (b)${\mathbf{Co}}^{\mathbf{2}\mathbf{+}}\mathbf{;}$(c)${\mathbf{Fe}}^{\mathbf{2}\mathbf{+}}$; (d)${\mathbf{Cu}}^{\mathbf{2}\mathbf{+}}\mathbf{?}$

Short answer

${\mathrm{Ti}}^{3+}$and ${\mathrm{Cu}}^{2+}$cannot form both high-spin and low-spin complexes.

Step-by-step solution

Formation of octahedral complex

High-spin and low spin octahedral complexes are possible to form only when the number of d electrons is 4, 5, 6 or 7. The metal ions with 1, 2 or 3delectrons always form high-spin complexes because of no need for pairing up electrons. The metal ions with 8 or 9 d electrons will always form high-spin complexes because low energy d orbitals are filled with 6 electrons and higher energy d orbitals will have one or two unpaired electrons$\left(d^8\text{or}{d}^9\right)$ .

Electronic configuration

a) Electron configuration of ${\mathrm{Ti}}^{3+}:1{\mathrm{s}}^{2}2{\mathrm{s}}^{2}2{\mathrm{p}}^{6}3{\mathrm{s}}^{2}3{\mathrm{p}}^{6}3{\mathrm{d}}^1$; so, the ${\mathrm{Ti}}^{3+}$metal ion can only form only high-spin octahedral complexes, because it has 1d electron.

b) Electron configuration of${\mathrm{Co}}^{2+}:1{\mathrm{s}}^{2}2{\mathrm{s}}^{2}2{\mathrm{p}}^{6}3{\mathrm{s}}^{2}3{\mathrm{p}}^{6}3{\mathrm{d}}^7$; so, the${\mathrm{Co}}^{2+}$metal ion can form both high-spin and low-spin octahedral complexes, because it has 7 d electrons.

c) Electron configuration for${\mathrm{Fe}}^{2+}:1{\mathrm{s}}^{2}2{\mathrm{s}}^{2}2{\mathrm{p}}^{6}3{\mathrm{s}}^{2}3{\mathrm{p}}^{6}3{\mathrm{d}}^6;$so, the${\mathrm{Fe}}^{2+}$metal ion can form both high-spin and low-spin octahedral complexes, because it has 6 d electrons.

d) Electron configuration of ${\mathrm{Cu}}^{2+}:1{\mathrm{s}}^{2}2{\mathrm{s}}^{2}2{\mathrm{p}}^{6}3{\mathrm{s}}^{2}3{\mathrm{p}}^{6}3{\mathrm{d}}^9$; so, the ${\mathrm{Cu}}^{2+}$metal ion can form only high-spin octahedral complexes, because it has 9d electrons.