In the coordination complex [E(en)_2(C_2O_4)] | vedclass

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

Vedclass · Accepted Answer

$6$ and $3$

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(B) In the complex $[E(en)_2(C_2O_4)]NO_2$,both $en$ (ethylenediamine) and $C_2O_4^{2-}$ (oxalate) are bidentate ligands.Coordination Number $= (2 	imes 2) + (1 	imes 2) = 4 + 2 = 6$.To find the oxidation state $(x)$ of $E$:$x + 2(0) + (-2) + (-1) = 0$$x - 3 = 0$$x = +3$.Therefore,the coordination number is $6$ and the oxidation state is $3$.

In the coordination complex $[E(en)_2(C_2O_4)]NO_2$,the coordination number and oxidation state of $E$ are respectively:

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