In the complex [E(en)_2(C_2O_4)]NO_2,the oxid | vedclass

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(A) $1$. The complex is $[E(en)_2(C_2O_4)]NO_2$. $2$. Let the oxidation state of $E$ be $x$. $3$. The ligand $(en)$ (ethylenediamine) is neutral $(0)$

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$3$ and $6$

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(A) $1$. The complex is $[E(en)_2(C_2O_4)]NO_2$. $2$. Let the oxidation state of $E$ be $x$. $3$. The ligand $(en)$ (ethylenediamine) is neutral $(0)$,and the oxalate ion $(C_2O_4^{2-})$ has a charge of $-2$. The counter ion $NO_2^-$ has a charge of $-1$. $4$. The total charge of the complex is $0$. $5$. Equation: $x + 2(0) + 1(-2) + 1(-1) = 0 \implies x - 2 - 1 = 0 \implies x = +3$. $6$. The coordination number is calculated by the number of donor atoms attached to the central metal ion. $7$. $(en)$ is a bidentate ligand ($2$ donor atoms each,$2 	imes 2 = 4$) and $(C_2O_4^{2-})$ is a bidentate ligand ($2$ donor atoms). $8$. Total coordination number = $4 + 2 = 6$. $9$. Thus,the oxidation state is $3$ and the coordination number is $6$.

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

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