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$EAN$ of the element are equal in

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$[Ni {(en)}_{2}]^{2 +}, [Fe {(H_{2} O)}_{6}]^{2 +}$

$[CO {(CN)}_{6}]^{3 -}, [Fe {(CN)}_{6}]^{4 -}$

${[CO (EDTA)]}^{2 -}, [Sc {(H_{2} O)}_{6}]^{3 +}$

$[Ni {(CO)}_{4}], [Fe {(CN)}_{6}]^{4 -}$

answer is D.

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Detailed Solution

$[Ni {(CO)}_{4}]; E . A . N of Ni = 28 + 0 + 4 \times 2 = 36 (kr) [Fe {(CN)}_{6}]^{4 -}; E . A . N of E = 26 - 2 + 6 \times 2 = 36 (kr)$

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