The sum of the coordination number and the ox | vedclass

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(A) $1$. The coordination number of $M$ is calculated by considering the denticity of the ligands. $en$ (ethylenediamine) is a bidentate ligand and $C

Vedclass · Accepted Answer

$9$

Vedclass · Answer

(A) $1$. The coordination number of $M$ is calculated by considering the denticity of the ligands. $en$ (ethylenediamine) is a bidentate ligand and $C_2O_4^{2-}$ (oxalate) is also a bidentate ligand.$2$. Coordination number $= (2 	imes 2) + (1 	imes 2) = 4 + 2 = 6$.$3$. Let the oxidation number of $M$ be $x$. The charge on $en$ is $0$, the charge on $C_2O_4^{2-}$ is $-2$, and the charge on $Cl^-$ is $-1$.$4$. The total charge of the complex $[M(en)_2C_2O_4]Cl$ is $0$. Thus, $x + 2(0) + 1(-2) - 1 = 0$.$5$. $x - 2 - 1 = 0 \implies x = +3$.$6$. The sum of the coordination number and the oxidation number $= 6 + 3 = 9$.

The sum of the coordination number and the oxidation number of $M$ in $[M(en)_2C_2O_4]Cl$ is

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