Q23.90P - Chemistry: Molecular Nature Of Matter And Change

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)  ${\mathbf{\left[}{\mathbf{RhCl}}_{\mathbf{6}}\mathbf{\right]}}^{\mathbf{3}\mathbf{-}}$.

Step-by-Step Solution

Answer

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

Step 1: Definition of the electronic configuration

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

Step 2: 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 $0 3d$ electrons.
2. Electron configuration on $\mathrm{Au}:1 {\mathrm{s}}^{2}2 {\mathrm{s}}^{2}2{\mathrm{p}}^{6}3 {\mathrm{s}}^{2}3{\mathrm{p}}^{6}3 {\mathrm{d}}^{1}04 {\mathrm{s}}^{2}4{\mathrm{p}}^{6}4 {\mathrm{d}}^{10}5 {\mathrm{s}}^{2}5{\mathrm{p}}^{6}4{\mathrm{f}}^{14}5 {\mathrm{d}}^{10}6 {\mathrm{s}}^1\mathrm{InK}[\mathrm{AuCl}4$ 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.