Q23.91P - Chemistry: Molecular Nature Of Matter And Change

Question

Give the number of d electrons (n of dⁿ) for the central metal ion in 

(a) $\left[\mathrm{Cr}{\left(H_{2}O\right)}_6\right]{\left({\mathrm{ClO}}_3\right)}_{\mathbf{2}}$; (b) ${\left[\mathrm{Mn}{\left(\mathrm{CN}\right)}_6\right]}^{\mathbf{2}\mathbf{-}}$; (c) $\left[\mathrm{Ru}\left(\mathrm{NO}\right){\left(\mathrm{en}\right)}_2\mathrm{Cl}\right]\mathbf{Br}$.

Step-by-Step Solution

Answer

1. 4 d electrons
2. 3 d electrons
3. 5 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: Explanation

1. In Electron configuration of $\mathrm{Cr}:1 {\mathrm{s}}^{2}2 {\mathrm{s}}^{2}2{\mathrm{p}}^{6}3 {\mathrm{s}}^{2}3{\mathrm{p}}^{6}3 {\mathrm{d}}^{5}4 {\mathrm{s}}^1\mathrm{lnCr}{\left({\mathrm{H}}_2\mathrm{O}\right)}_{62}$, the oxidation state of Cr is $+2\left({\mathrm{Cr}}^{2+}\right)$, so we have to remove 1 electron from the 4 s subshell and 1 electron from the 3d subshell, which leaves ${\mathrm{Cr}}^{2+}$ with 4, 3d electrons.
2. In Electron configuration of Mn: $1{\mathrm{s}}^{2}2{\mathrm{s}}^{2}2{\mathrm{p}}^{6}3{\mathrm{s}}^{2}3{\mathrm{p}}^{6}3{\mathrm{d}}^{5}4{\mathrm{s}}^2\ln {\left[\mathrm{Mn}(\mathrm{CN}{)}_6\right]}^{2-}$, the oxidation state of Mn is $+4\left({\mathrm{Mn}}^{4+}\right)$, so we have to remove 2 electrons from the 4s subshell and another 2 electrons from the 3d subshell, which leaves ${\mathrm{Mn}}^{4+}$ with 3d electrons.
3. In Electron configuration of Ru:$1 {\mathrm{s}}^{2}2 {\mathrm{s}}^{2}2{\mathrm{p}}^{6}3 {\mathrm{s}}^{2}3{\mathrm{p}}^{6}3 {\mathrm{d}}^{10}4 {\mathrm{s}}^{2}4{\mathrm{p}}^{6}4 {\mathrm{d}}^{7}5 {\mathrm{s}}^1\ln \left[\mathrm{Ru}\left(\mathrm{NO}\right){\left(\mathrm{en}\right)}_2\mathrm{Cl}\right]\mathrm{Br}$ the oxidation state of Ru is  $+3\left({\mathrm{Ru}}^{3+}\right)$, so we have to remove 1 electron from the 5 s subshell and another 2 electrons from the 4 d subshell, which leaves $\left({\mathrm{Ru}}^{3+}\right)$ with 5, 4d electrons.