The sum of the coordination number and the ox | vedclass

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(C) In the complex $[M(en)_2(C_2O_4)]Cl$,let the oxidation state of $M$ be $x$.Ethylenediamine $(en)$ is a neutral bidentate ligand,so its charge is $

Vedclass · Accepted Answer

$9$

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(C) In the complex $[M(en)_2(C_2O_4)]Cl$,let the oxidation state of $M$ be $x$.Ethylenediamine $(en)$ is a neutral bidentate ligand,so its charge is $0$.Oxalate $(C_2O_4^{2-})$ is an anionic bidentate ligand with a charge of $-2$.Chloride $(Cl^-)$ is a counter ion with a charge of $-1$.The total charge of the complex is $0$,so: $x + 2(0) + 1(-2) + 1(-1) = 0$.$x - 2 - 1 = 0$,which gives $x = +3$.The coordination number is calculated as: $2 	imes (	ext{denticity of } en) + 1 	imes (	ext{denticity of } C_2O_4^{2-}) = 2(2) + 1(2) = 4 + 2 = 6$.The sum of the coordination number and the oxidation number is $6 + 3 = 9$.

The sum of the coordination number and the oxidation number of the metal $M$ in the complex $[M(en)_2(C_2O_4)]Cl$ (where $en$ is ethylenediamine) is:

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