Chapter 23: Q23.92 (page 1061)

How many d electrons (n of dⁿ) are in the central metal ion in
(a) $Ca [{IrF}_{6}]$ ; (b) ${[{HgI}_{4}]}^{2 -}$ ; (c) $[Co (EDTA)]^{2 -}$ ?

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

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

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