Chapter 23: Q23.91P (page 1061)

Give the number of d electrons (n of dⁿ) for the central metal ion in
(a) $[Cr {(H_{2} O)}_{6}] {({ClO}_{3})}_{2}$ ; (b) ${[Mn {(CN)}_{6}]}^{2 -}$ ; (c) $[Ru (NO) {(en)}_{2} Cl] Br$ .

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

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

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