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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 :(i) $[ CoF_{6}]^{3-}, [Co(H_{2}O)_{6}]^{2+}, [Co(CN)_{6}]^{3-}$

(ii) $[ FeF_{6}]^{3-}, [Fe(H_{2}O)_{6}]^{2+}, [Fe(CN)_{6}]^{4-}$

Answers (1)

(i)

$[ CoF_{6}]^{3-}$ :

$\text {Co}^{3+}=\text {3d}^{6}$

Number of unpaired electrons =4

Magnetic moment

$=\sqrt{n(n+2)}=\sqrt{4(4+2)}=4.9 B.M$

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

$\text {Co}^{2+}=\text {3d}^{7}$

Number of unpaired electrons =3

Magnetic moment = $\sqrt{3(3+2)}=3.87 B.M$

$[Co(CN)_{6}]^{3-}$

$\text {Co}^{3+}=\text {3d}^{6}$

Number of unpaired electrons =0

Diamagnetic .

(ii)

$[ FeF_{6}]^{3-}$

$Fe^{3+}=3d^{5}$

Number of unpaired electrons =5

Magnetic moment = $\sqrt{5(5+2)}=5.92\; B.M$

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

$Fe^{2+}=3d^{6}$

Number of unpaired electrons =4

Magnetic moment = $\sqrt{4(4+2)}=4.9 \; B.M$

$[Fe(CN)_{6}]^{4-} :$

$Fe^{2+}=3d^{6}$

Diamagnetic.

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