QJEEAdv21Chf0e7: JEE Advanced 2021 Chemistry - Class 12 JEEAdv21Chf0e7 Solution: The calculated spin only magnetic moments of [ Cr(NH 3 ) 6 ] 3+ …

${[Cr{(N{H_3})_6}]^{3 + }}$

Atomic number of Cr = 24

Electron configuration of Cr = 1s² 2s² 2p⁶ 3s² 3p⁶ 3d⁵ 4s¹

Electronic configuration of Cr³⁺ = 1s² 2s² 2p⁶ 3s² 3p⁶ 3d³ 4s⁰

NH₃ is a weak field ligand. The splitting of d-orbital occur.


Number of unpaired electrons, n = 3

Magnetic moment (spin only), ${\mu _s} = \sqrt {n(n + 2)} BM$

$= \sqrt {3(3 + 2)} = \sqrt {3 \times 5} = \sqrt {15} = 3.87BM$

${[Cu{F_6}]^{3 - }}$ Atomic number of Cu = 29

Electronic configuration of Cu = 1s² 2s² 2p⁶ 3s² 3p⁶ 3d¹⁰ 4s¹

Electronic configuration of Cu²⁺ = 1s² 2s² 2p⁶ 3s² 3p⁶ 3d⁸ 4s⁰

F ^$-$ is a weak field ligand, so splitting of d-orbitals occur as follows:


Number of unpaired electrons, n = 2

Spin only magnetic moment, ${\mu _s} = \sqrt {n(n + 2)} BM$

$= \sqrt {2(2 + 2)} = \sqrt {2 \times 4} = 2\sqrt 2 = 2.84BM$