Q23.92 - Chemistry: Molecular Nature Of Matter And Change

Question

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

 (a) $\mathbf{Ca}\left[{\mathrm{IrF}}_6\right]$; (b) ${\left[{\mathrm{HgI}}_4\right]}^{\mathbf{2}\mathbf{-}}$; (c) $\mathbf{\left[}\mathbf{Co}\mathbf{\right(}\mathbf{EDTA}\mathbf{)}{\mathbf{]}}^{\mathbf{2}\mathbf{-}}$?

Step-by-Step Solution

Answer

1. 5 d electrons
2. 10 d electrons
3. 7 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. The electron configuration of $\mathrm{Ir}:1{\mathrm{s}}^{2}2{\mathrm{s}}^{2}2{\mathrm{p}}^{6}3{\mathrm{s}}^{2}3{\mathrm{p}}^{6}3{\mathrm{d}}^{10}4{\mathrm{s}}^{2}4{\mathrm{p}}^{6}4{\mathrm{d}}^{10}5{\mathrm{s}}^{2}5{\mathrm{p}}^{6}4{\mathrm{f}}^{14}5{\mathrm{d}}^{7}6{\mathrm{s}}^2$ In Ca[IrF6], the oxidation state of Ir is $+4\left({\mathrm{Ir}}^{4+}\right)$, so we have to remove 2 electrons from the 6 s subshell and another 2 electrons from the 5d subshell, which leaves ${\mathrm{Ir}}^{4+}$ with 5, 5d electrons.
2. The electron configuration of Hg: $1{\mathrm{s}}^{2}2{\mathrm{s}}^{2}2{\mathrm{p}}^{6}3{\mathrm{s}}^{2}3{\mathrm{p}}^{6}3{\mathrm{d}}^{10}4{\mathrm{s}}^{2}4{\mathrm{p}}^{6}4{\mathrm{d}}^{10}5{\mathrm{s}}^{2}5{\mathrm{p}}^{6}4{\mathrm{f}}^{14}5{\mathrm{d}}^{10}6{\mathrm{s}}^2$ In ${\left[{\mathrm{H}}_4\right]}^{2-}$, the oxidation state of Hg is $+2\left({\mathrm{Hg}}^{2+}\right)$, so we have to remove 2 electrons from the 6 s subshell, which leaves ${\mathrm{Hg}}^{2+}$ with 10, 5d electrons.
3. The electron configuration of Co:$1{\mathrm{s}}^{2}2{\mathrm{s}}^{2}2{\mathrm{p}}^{6}3{\mathrm{s}}^{2}3{\mathrm{p}}^{6}3{\mathrm{d}}^{7}4{\mathrm{s}}^2$ In $\left[\mathrm{Co}(\mathrm{EDTA}){]}^{2-}\right.$ the oxidation state of Co is $+2\left({\mathrm{Co}}^{2+}\right)$, so we have to remove 2 electrons from the 4s subshell, which leaves ${\mathrm{Co}}^{2+}$ with 7, 3d electrons.