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

24

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27

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30

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33

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Solution

The correct option is D $33$
Effective atomic number, $E A N$ = number of electrons in Cr+2×[number of ligands –charge]

$= 24 + 2 \times 6 - 3 = 33$ .

Hence option D is correct.

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