Question

Which of these ions cannot form both high- spin and low-spin octahedral complexes: (a) ${\mathbf{Mn}}^{\mathbf{3}\mathbf{+}}\mathbf{;}$(b)${\mathbf{Nb}}^{\mathbf{3}\mathbf{+}}\mathbf{;}$(c)^{${\mathbf{Ru}}^{\mathbf{3}\mathbf{+}}\mathbf{;}$}(d)role="math" localid="1663323480799" ${\mathbf{Ni}}^{\mathbf{2}\mathbf{+}}$

Short answer

Nb3+ and Ni2+ cannot form both high-spin and low-spin complexes.

Step-by-step solution

Unpaired electrons 

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 3 d electrons 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 ${\mathrm{d}}^8$or ${\mathrm{d}}^9.$

Electronic configuration

1. Electron configuration of $M{n}^{3+}:1s^{2}2s^{2}2p^{6}3s^{2}3p^{6}3d^4;$so, the $M{n}^{3+}$metal ion can form both high-spin and low-spin octahedral complexes, because it has 4d electrons.

1. Electron configurationof $N{b}^{3+}:1s^{2}2s^{2}2p^{6}3s^{2}3p^{6}3d^{10}4s^{2}4p^{6}4d^2$so, the ${\mathrm{Nb}}^{3+}$metal ion can form only high-spin octahedral complexes, because it has 2d electrons.

1. Electron configuration of $R{u}^{3+}:1s^{2}2s^{2}2p^{6}3s^{2}3p^{6}3d^{10}4s^{2}4p^{6}4d^5$so, the $R{u}^{3+}$metal ion can form both high-spin and low-spin octahedral complexes, because it has 5d electrons.

1. Electron configuration of ${\mathrm{Ni}}^{2+}:1{\mathrm{s}}^{2}2{\mathrm{s}}^{2}2{\mathrm{p}}^{6}3{\mathrm{s}}^{2}3{\mathrm{p}}^{6}3{\mathrm{d}}^8$so, the ${\mathrm{Ni}}^{2+}$metal ion can form only high-spin octahedral complexes, because it has 8 d electrons.