The effective atomic number of Cr(atomic number $= 24$ ) in $[C r (N H_{3})_{6}] C l_{3}$ is?

35

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27

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33

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36

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Solution

The correct option is D $33$
$E A N = Z - X + Y$

$E A N$ effective atomic number
$Z =$ atomic number of the metal
$X =$ number of electrons lost during the formation of the metal ion from its atom
$Y =$ number of electrons donated by the ligands.
$E A N = 24 - 3 + 12 = 33$
The effective atomic number of Cr(atomic number $= 24$ ) in $[C r (N H_{3})_{6}] C l_{3}$ is $33$ .

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

C r

= 24

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

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

K_{4} F e (C N)_{6}

C r (C O)_{6}

F e (Z = 26)

C r (Z = 24)

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