Using crystal field theory, draw energy level diagram. Write electronic configuration of the central metal atom/ion and determine the magnetic moment value in the following:

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

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

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Solution

l)

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

$\Rightarrow$ The electronic configuration of $C o^{3 +}$ is $[A r] 3 d^{6}$ i.e. $t_{2 g}^{4} e_{g}^{2}$

Since $F^{-}$ is a weak field ligand, $Δ_{0} < P$ . Therefore, a high-spin complex is formed.

Number of unpaired electron (n) = 4. So, it is paramagnetic in nature.

Magnetic moment $μ = \sqrt{n (n + 2)}$

= $\sqrt{4 (4 + 2)}$

= $\sqrt{24}$

= $4.9 B M$

$[C o (H_{2} O)_{6}]^{2 +}$

$\Rightarrow$ The electronic configuration of $C o^{2 +}$ is $[A r] 3 d^{7}$ i.e. $t_{2 g}^{5} e_{g}^{2}$

Since $H_{2} O$ is a weak field ligand, $Δ_{0} < P$ . Therefore, a high-spin complex is formed.

Number of unpaired electron (n) = 3. So, it is paramagnetic in nature.

Magnetic moment $μ = \sqrt{n (n + 2)}$

= $\sqrt{3 (3 + 2)}$

= $\sqrt{15}$

= $3.87 B M$

$[C o (C N)_{6}]^{3 -}$

$\Rightarrow$ The electronic configuration of $C o^{3 +}$ is $[A r] 3 d^{6}$ i.e. $t_{2 g}^{6} e_{g}^{0}$

Since $C N^{-}$ is a strong field ligand, $Δ_{0} > P$ . Therefore, a low-spin complex is formed.

The number of unpaired electron (n) = 0
So, it is diamagnetic in nature.

Magnetic moment $μ = \sqrt{n (n + 2)}$

= $\sqrt{0 (0 + 2)}$

= $\sqrt{0}$

= $0 B M$

ll)

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

$\Rightarrow$ The electronic configuration of $F e^{+ 3}$ is $[A r] 3 d^{5}$ i.e. $t_{2 g}^{3} e_{g}^{2}$

Since $F^{-}$ is a weak field ligand, $Δ_{0} < P$ . Therefore, a high-spin complex is formed.

The number of unpaired electron (n) = 5
So,it is paramagnetic in nature.

Magnetic moment $μ = \sqrt{n (n + 2)}$

= $\sqrt{5 (5 + 2)}$

= $\sqrt{35}$

= $5.92 B M$

$[F e (H_{2} O)_{6}]^{2 +}$

$\Rightarrow$ The electronic configuration of $F e^{+ 2}$ is $[A r] 3 d^{6}$ i.e. $t_{2 g}^{4} e_{g}^{2}$

Since $H_{2} O$ is a weak field ligand, $Δ_{0} < P$ . Therefore, a high-spin complex is formed.

The number of unpaired electron (n) = 4
So,it is paramagnetic in nature.

Magnetic moment $μ = \sqrt{n (n + 2)}$

= $\sqrt{4 (4 + 2)}$

= $\sqrt{24}$

= $4.9 B M$

$[F e (C N)_{6}]^{4 -}$

$\Rightarrow$ The electronic configuration of $F e^{+ 2}$ is $[A r] 3 d^{6}$ i.e. $t_{2 g}^{6} e_{g}^{0}$

Since $C N^{-}$ is a weak field ligand, $Δ_{0} \geq P$ . Therefore, a low-spin complex is formed.

The number of unpaired electron (n) = 0
So,it is diamagnetic in nature.

Magnetic moment $μ = \sqrt{n (n + 2)}$

= $\sqrt{0 (0 + 2)}$

= $\sqrt{0}$

= $0 B M$

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Similar questions

Give the electronic configuration of the following complexes on the basis of crystal field splitting theory. $[C o F_{6}]^{3 -}, [F e (C N)_{6}]^{4 -}$ and $[C u (N H_{3})_{6}]^{2 +} .$

Discuss the nature of bonding in the following coordination entities on the basis of valence bond theory:
$[F e (C N_{6})]^{4 -}$

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

$[C o (C_{2} O_{4})_{3}]^{3 -}$

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

The colour of the coordination compounds depends on the crystal field splitting. What will be the correct order of absorption of wavelength of light in the visible region, for the complexes, $[C o (N H_{3})_{6}]^{3 +}, [C o (C N)_{6}]^{3 -}, [C o (H_{2} O)_{6}]^{3 +}$
$[C o (N H_{3})_{6}]^{3 +}, [C o (C N)_{6}]^{3 -}, [C o (H_{2} O)_{6}]^{3 +}$
(a) $[C o (C N)_{6}]^{3 -} > [C o (N H_{3})_{6}]^{3 +} > [C o (H_{2} O)_{6}]^{3 +}$
(b) $[C o (N H_{3})_{6}]^{3 +} > [C o (H_{2} O)_{6}]^{3 +} > [C o (N H_{3})_{6}]^{3 +}$
(c) $[C o (H_{2} O)_{6}]^{3 +} > [C o (N H_{3})_{6}]^{3 +} > [C o (C N)_{6}]^{3 -}$
(d) $[C o (C N)_{6}]^{3 -} > [C o (N H_{3})_{6}]^{3 +} > [C o (H_{2} O)_{6}]^{3 +}$