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

4 and 2

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4 and 3

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6 and 3

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6 and 2

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Solution

The correct option is C 6 and 3
As, $6$ co-ordinate bonds are present, the cordination number is $^{'} 6^{'}$
$[E^{n} (e n)_{2}^{0} (C_{2} O_{4})^{2 -}]^{+ 1} N O_{2}^{-}$
$⟹ n - 2 = + 1$
$⟹ n = 3$
Therefore oxidation state $= 3$

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