The coordination number and the oxidation sta | vedclass

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(D) In the given complex,there are two $en$ ligands and one $C_2O_4$ ligand. Both are bidentate ligands.Coordination number = $(2 	imes 2) + (1 	ime

Vedclass · Accepted Answer

$6$ and $3$

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(D) In the given complex,there are two $en$ ligands and one $C_2O_4$ ligand. Both are bidentate ligands.Coordination number = $(2 	imes 2) + (1 	imes 2) = 4 + 2 = 6$.Let the oxidation state of $E$ be $x$.The complex is $[E(en)_2(C_2O_4)]NO_2$. Since $NO_2$ is a counter ion with a charge of $-1$,the complex cation $[E(en)_2(C_2O_4)]^+$ has a charge of $+1$.$x + 2(0) + 1(-2) = +1$$x - 2 = +1$$x = +3$.Therefore,the coordination number is $6$ and the oxidation state is $+3$. Thus,option $(d)$ is correct.

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

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