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Question

(C) The complex is $[E(en)_2(C_2O_4)]NO_2$. $en$ (ethylenediamine) is a bidentate ligand,so $2 	imes en$ provides $2 	imes 2 = 4$ coordination sites

Vedclass · Accepted Answer

$6$ and $3$

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(C) The complex is $[E(en)_2(C_2O_4)]NO_2$. $en$ (ethylenediamine) is a bidentate ligand,so $2 	imes en$ provides $2 	imes 2 = 4$ coordination sites. $C_2O_4^{2-}$ (oxalate) is a bidentate ligand,providing $2$ coordination sites. Total coordination number = $4 + 2 = 6$. Let the oxidation state of $E$ be $x$. The charge on $en$ is $0$,the charge on $C_2O_4$ is $-2$,and the charge on $NO_2$ is $-1$. $x + 2(0) + (-2) + (-1) = 0$ $x - 3 = 0$ $x = +3$. Thus,the coordination number is $6$ and the oxidation state is $3$.

The coordination number and oxidation state of metal $E$ in the complex $[E(en)_2(C_2O_4)]NO_2$ (where $en$ is ethylenediamine) are respectively:

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