Question

Give the number of d electrons ( n of ${\mathit{d}}^{\mathbf{n}}$) for the central metal ion in (a) ${\mathbf{\left[}{\mathbf{TiCl}}_{\mathbf{6}}\mathbf{\right]}}^{\mathbf{2}\mathbf{-}}$; (b) $\mathbf{K}\mathbf{\left[}{\mathbf{AuCl}}_{\mathbf{4}}\mathbf{\right]}$; (c) ${\mathbf{\left[}{\mathbf{RhCl}}_{\mathbf{6}}\mathbf{\right]}}^{\mathbf{3}\mathbf{-}}$.

Short answer

1. 0 d electrons
2. 8 d electrons
3. 6 d electrons

Step-by-step solution

Definition of the electronic configuration

The electron configuration of an element describes how electrons are distributed in its atomic orbitals.

Electronic configuration of a given ion

1. Electron configuration of Ti: $1{\mathrm{s}}^{2}2{\mathrm{s}}^{2}2{\mathrm{p}}^{6}3{\mathrm{s}}^{2}3{\mathrm{p}}^{6}3{\mathrm{d}}^{2}4{\mathrm{s}}^2\ln {\left[{\mathrm{TCl}}_6\right]}^{2-}$the oxidation state of Ti is $+4\left({\mathrm{Ti}}^{4+}\right)$, so it means we have to remove 2 electrons from the 4 s subshell and 2 electrons from the 3d subshell, which leaves ${\mathrm{Ti}}^{4+}$with $03d$electrons.
2. Electron configuration on localid="1663328440485" $\mathrm{Au}:1{\mathrm{s}}^{2}2{\mathrm{s}}^{2}2{\mathrm{p}}^{6}3{\mathrm{s}}^{2}3{\mathrm{p}}^{6}3{\mathrm{d}}^{1}04{\mathrm{s}}^{2}4{\mathrm{p}}^{6}4{\mathrm{d}}^{10}5{\mathrm{s}}^{2}5{\mathrm{p}}^{6}4{\mathrm{f}}^{14}5{\mathrm{d}}^{10}6{\mathrm{s}}^1\mathrm{InK}[\mathrm{AuCl}4$the oxidation state of Au is$+3\left({\mathrm{Au}}^{3+}\right)$, so it means we have to remove 1 electron from the 6s subshell and 2 electrons from the 5d subshell, which leaves${\mathrm{Au}}^{3+}$with 8, 5d electrons.
3. Electron configuration of $\mathrm{Rh}:1s^{2}2s^{2}2p^{6}3s^{2}3p^{6}3d^{10}4s^{2}4p^{6}4d^{8}5s^1\ln {\left[{\mathrm{RhCl}}_6\right]}^{3-}$the oxidation state of Rh is$+3\left({\text{Rh}}^{3+}\right)$, so we have to remove 1 electron from the 5 s subshell and 2 electrons from the 4d subshell, which leaves${\mathrm{Rh}}^{3+}$with 6, 4d electrons.