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

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, $${\mu _s} = \sqrt {n(n + 2)} BM$$

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

For [Cr(NH₃)₄Cl₂]Cl,

For [Ti(H₂O)₅Cl](NO₃)₂,

The complex is paramagnetic due to presence of unpaired electron.

Since the groups between the complex ion and ions outside can be exchanged, it shows ionisation isomerism.

For [Pt(en)(NH₃)Cl]NO₃,

The complex is square planar and all electrons are paired. So it is diamagnetic. Since the groups between the complex ion and ions outside can be exchanged, it shows ionisation isomerism.

For [Co(NH₃)₄(NO₃)₂]NO₃, NH₃ is a strong ligand and causes pairing