The coordination number and the oxidation sta | vedclass

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(A) The complex is $[M(en)_{2}(C_{2}O_{4})]NO_{2}$.In this complex,the counter ion is $NO_{2}^{-}$,which has a charge of $-1$. Therefore,the complex i

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$6$ and $3$

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(A) The complex is $[M(en)_{2}(C_{2}O_{4})]NO_{2}$.In this complex,the counter ion is $NO_{2}^{-}$,which has a charge of $-1$. Therefore,the complex ion $[M(en)_{2}(C_{2}O_{4})]$ must have a charge of $+1$.Let the oxidation state of $M$ be $x$.$en$ (ethane-$1,2$-diamine) is a neutral bidentate ligand (charge $= 0$).$C_{2}O_{4}^{2-}$ (oxalate) is a bidentate ligand with a charge of $-2$.Setting up the equation for the charge of the complex ion:$x + (2 	imes 0) + (1 	imes -2) = +1$$x - 2 = +1$$x = +3$Thus,the oxidation state of $M$ is $+3$.Coordination number calculation:$en$ is bidentate ($2$ donor sites each,$2 	imes 2 = 4$ sites).$C_{2}O_{4}^{2-}$ is bidentate ($2$ donor sites).Total coordination number $= 4 + 2 = 6$.Therefore,the coordination number is $6$ and the oxidation state is $3$.

The coordination number and the oxidation state of the element $M$ in the complex $[M(en)_{2}(C_{2}O_{4})]NO_{2}$ [where $en$ is ethane-$1,2$-diamine] are respectively:

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