Q23.93 - Chemistry: Molecular Nature Of Matter And Change

Question

How many d electrons (n of dⁿ) are in the central metal ion in 

(a) $\left[\mathrm{Ru}{\left({\mathrm{NH}}_3\right)}_5\mathrm{Cl}\right]{\mathbf{SO}}_{\mathbf{4}}\mathbf{;}$(b) ${\mathbf{Na}}_{\mathbf{2}}\left[\mathrm{Os}{\left(\mathrm{CN}\right)}_6\right]$; (c) $\left[\mathrm{Co}{\left({\mathrm{NH}}_3\right)}_4{\mathrm{CO}}_{3}I\right]$?

Step-by-Step Solution

Answer

1. 5 d electrons
2. 4 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: Explanation

1. Electron configuration of Ru:$1{\mathrm{s}}^{2}2{\mathrm{s}}^{2}2{\mathrm{p}}^{6}3{\mathrm{s}}^{2}3{\mathrm{p}}^{6}3{\mathrm{d}}^{10}4{\mathrm{s}}^{2}4{\mathrm{p}}^{6}4{\mathrm{d}}^{7}5{\mathrm{s}}^1$ In $\left[\mathrm{Ru}{\left({\mathrm{NH}}_3\right)}_5\mathrm{Cl}\right]{\mathrm{SO}}_4$, the oxidation state of Ru is $+3\left({\mathrm{Ru}}^{3+}\right)$, so we have to remove 1 electron from the 5s subshell and another 2 electrons from the 4d subshell, which leaves $\left({\mathrm{Ru}}^{3+}\right)$ with 5, 4d electrons.

 

1. The electron configuration of Os:$1{\mathrm{s}}^{2}2{\mathrm{s}}^{2}2{\mathrm{p}}^{6}3{\mathrm{s}}^{2}3{\mathrm{p}}^{6}3{\mathrm{d}}^{10}4{\mathrm{s}}^{2}4{\mathrm{p}}^{6}4{\mathrm{d}}^{10}5{\mathrm{s}}^{2}5{\mathrm{p}}^{6}4{\mathrm{f}}^{14}5{\mathrm{d}}^{6}6{\mathrm{s}}^2$  In Naz[Os(CN)  ], the oxidation state of Os is $+4\left({\mathrm{Os}}^{4+}\right)$, so we have to remove 2 electrons from the 6s subshell and another 2 electrons from the 5d subshell, which leaves ${\mathrm{Os}}^{4+}$ with 4, 5d electrons.

 

1. The electron configuration of $\mathrm{Co}:1 {\mathrm{s}}^{2}2 {\mathrm{s}}^{2}2{\mathrm{p}}^{6}3 {\mathrm{s}}^{2}3{\mathrm{p}}^{6}3 {\mathrm{d}}^{7}4 {\mathrm{s}}^2\ln \left[\mathrm{Co}{\left({\mathrm{NH}}_3\right)}_4{\mathrm{CO}}_3\mathrm{l}\right]$, the oxidation state of Co is $+3\left({\mathrm{Co}}^{3+}\right)$, so we have to remove 2 electrons from the 4 s subshell and another 1 electron from the 3d subshell, which leaves ${\mathrm{Co}}^{3+}$ with 6, 3d electrons.