EAN of $C r$ in $[C r (N H_{3})_{6}]$ $C l_{3}$ is:

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Solution

The correct option is B 33
$C r$ is in +3 oxidation state.
EAN = Atomic number - oxidation state + 2 $\times$ Number of ligands
= 24 - 3 + 2 $\times$ 6 = 33

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EAN of Cr in $[C r (N H_{3})_{6}] C l_{3}$ is

C r

[C r (N H_{3})_{6}]

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The effective atomic number $(E A N)$ of $_{24} C r$ in $[C r (N H_{3})_{6}] C l_{3}$ is :

[C r (N H_{3})_{6}] C l_{3}

[C r (N H_{3})_{6}] C l_{3}

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