The coordination number and the oxidation sta | vedclass

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(D) In the complex $[E(en)_2(C_2O_4)]NO_2$,$en$ (ethylenediamine) and $C_2O_4^{2-}$ (oxalate) are both bidentate ligands.Coordination number = $(2 	i

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$6$ and $3$

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(D) In the complex $[E(en)_2(C_2O_4)]NO_2$,$en$ (ethylenediamine) and $C_2O_4^{2-}$ (oxalate) are both bidentate ligands.Coordination number = $(2 	imes 2) + (1 	imes 2) = 6$.Let the oxidation state of $E$ be $x$.The complex ion is $[E(en)_2(C_2O_4)]^+$,as the counter ion $NO_2^-$ has a $-1$ charge.$x + 2(0) + 1(-2) = +1$.$x - 2 = +1 \Rightarrow x = +3$.Thus,the coordination number is $6$ and the oxidation state is $+3$.

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 ethylenediamine) are,respectively,

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