The sum of the coordination number and the ox | vedclass

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(D) Let the oxidation state of metal $M$ be $x$.The complex is $[M(en)_2(C_2O_4)]Cl$. The charge on $en$ is $0$,the charge on $C_2O_4^{2-}$ is $-2$,an

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$9$

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(D) Let the oxidation state of metal $M$ be $x$.The complex is $[M(en)_2(C_2O_4)]Cl$. The charge on $en$ is $0$,the charge on $C_2O_4^{2-}$ is $-2$,and the charge on the complex ion is $+1$ (since $Cl$ is $-1$).$x + 2(0) + 1(-2) = +1$$x - 2 = 1$$x = +3$Thus,the oxidation number of $M$ is $3$.Now,for the coordination number:$en$ (ethylenediamine) is a bidentate ligand,and $C_2O_4^{2-}$ (oxalate) is also a bidentate ligand.Coordination Number $(C.N.) = (2 	imes 2) + (1 	imes 2) = 4 + 2 = 6$.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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