Ncert Exercises & Solutions

Q. 46 Using crystal field theory, draw energy level diagram, wrile electronic

configuration of the central metal atom/ion and determine the magnetic

moment value in the following.

(a) ${[C o F_{6}]}^{3 -}, {[C o {(H_{2} 0)}_{6}]}^{2 +}, {[C o {(C N)}_{6}]}^{3 -}$

(b) $F e {F_{6}}^{3 -} {[F e {(H_{2} 0)}_{6}]}^{2 +}, {[F e {(C N)}_{6}]}^{4 -}$

${[C o F_{6}]}^{3 -}$

F is a weak field ligand.

Configuration of Co $^{3 +} = 3 d^{6} (o r t_{2 g}^{4} e_{g}^{2}$

Number of unpaired electrons (n) =4

Magnetic moment ( $μ$ )= $\sqrt{n (n + 2)} = \sqrt{4 (4 + 2)} = \sqrt{24} = 4.9 B M$
${[C o (H_{2} O_{6}]}^{2 +}$

$H_{2} O$ is a weak field ligand.

Configuration of Co $^{2 +} = 3 d^{7} (o r t_{2 g}^{5} e_{g}^{2})$
$N u m b e r o f u n p a i r e d e l e c t o n s (n) = 3$
$μ = \sqrt{3} (3 + 2) = \sqrt{15} = 3.87 B M$
${[C o {(C N)}_{6}]}^{3 -} e . e ., C o^{3 +}$
$∴ C N i s s t r o n g f i e l d l i g a n d .$

There is no unpaired electon, so it is diamagnetic

$μ = 0$

(b) ${[F e F_{6}]}^{3 -}$

$F e^{3 +} = 3 d^{5} (o r t_{2 g}^{3} e_{2}^{g})$
$N u m b e r o f u n p a i r e e d e l e c t o n s, n = 5$
$μ = \sqrt{5 (5 + 2)}$
$= \sqrt{35} = 5.92 B M$
${[F e (H_{2} O_{6}]}^{2 +}$

$F e^{3 +} = 3 d^{5} (o r t_{2 g}^{3} e_{2}^{g})$
$N u m b e r o f u n p a i r e e d e l e c t o n s, n = 4$
$μ = \sqrt{4 (4 + 2)}$
$= \sqrt{24} =$
${[F e {(C N)}_{6}]}^{4 -}$

Since , CN is a strong filed ligand, all the electrons get paired.

$F e^{2 +} = 3 d^{6} (o r t_{2 g}^{6} e_{2}^{0})$

Because there is no unpaired electron, so it is diamagnetic in nature.

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