What are the coordination number and oxidatio | vedclass

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(A) $1$. The coordination number is determined by the number of donor atoms attached to the central metal ion. The ligand $en$ (ethylenediamine) is a

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$6$ and $3$

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(A) $1$. The coordination number is determined by the number of donor atoms attached to the central metal ion. The ligand $en$ (ethylenediamine) is a bidentate ligand,and $C_2O_4^{2-}$ (oxalate) is also a bidentate ligand.$2$. Coordination number = $(2 	imes 2) + (1 	imes 2) = 4 + 2 = 6$.$3$. To find the oxidation state of $E$,let it be $x$. The charge on $en$ is $0$,the charge on $C_2O_4$ is $-2$,and the charge on the counter ion $NO_2$ is $-1$.$4$. The sum of charges in the complex is $0$: $x + 2(0) + 1(-2) + 1(-1) = 0$.$5$. $x - 2 - 1 = 0 \implies x = +3$.

What are the coordination number and oxidation state of $E$ in the complex $[E(en)_2(C_2O_4)]NO_2$?

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