The coordination number and the oxidation state of the element E in the complex $[E (e n)_{2} (C_{2} O_{4})] N O_{2}$ (where (en) is ethylenediamine) are, respectively :

6 and 3

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6 and 2

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4 and 2

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4 and 3

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Solution

The correct option is A 6 and 3
The co-ordination number is six as three bi-dentate ligands are present, two en and one $C_{2} O_{4}^{2 -}$ . The oxidation state of element E is $+ 3$ . The oxidation states of en, $C_{2} O_{4}^{2 -}$ and $N O_{2}^{-}$ are 0,-2 and -1 respectively. Let E be the oxidation state of metal E.
$E + 2 (0) + (- 2) - 1 = 0$
$E - 3 = 0$
$E = + 3$ .

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