Using crystal field theory, draw energy level diagram, write elec… - Vidyadip

Question

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:

$[\text{CoF}_6]^{3-}, [\text{Co(H}_2\text{O})_6]^{2+}, [\text{Co(CN)}_6]^{3-}$
$[\text{FeF}_6]^{3-}, [\text{Fe(H}_2\text{O})_6]^{2+}, [\text{Fe(CN})_6]^{4-}$

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Answer

$\text{Co}^{2+}=3\text{d}^7$
Number of unpaired electrons = 4
Magnetic moment
$=\sqrt{\text{n}(\text{n}+2)}=\sqrt{4(4+2)}=4.9\text{B.M.}$
$[\text{Co(H}_2\text{O})_6]^{2+}:$

$\text{Co}^{2+}=3\text{d}^7$
Number of unpaired electrons = 3
Magnetic moment = $\sqrt{3(3+2)}=3.87\text{B.M.}$
$[\text{Co(CN)}_6]^{3-}:$

$\text{Co}^{3+}=3\text{d}^6$
Number of unpaired electrons = 0
Diamagnetic.

$\text{FeF}^{3-}_6:$

$\text{Fe}^{3+}=3\text{d}^5$
Number of unpaired electrons = 5
Magnetic moment = $\sqrt{5(5+2})=5.92\text{B.M.}$
$[\text{Fe(H}_2\text{O})_6]^{2+}:$

$\text{Fe}^{2+}=3\text{d}^6$
Number of unpaired electrons = 4
Magnetic moment = $\sqrt{4(4+2})=4.9\text{B.M.}$
$[\text{Fe(CN)}_6]^{4-}:$

$\text{Fe}^{2+}=3\text{d}^6$
Diamagnetic.

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