Nuclear Physics and RadioactivityHard
Question
A magnetic moment of 1.73 BM will be shown by one among the following :-
Options
A.[CoCl6]4-
B.[Cu(NH3)4]2+
C.[Ni(CN)4]2-
D.TiCl4
Solution
Magnetic moment 1.73 BM
n = no. of unpaired e-
μ = 1.73 =
n = 1
• [CoCl6]4- ⇒ Co+2 ; d7
Cl- (weak field ligand) t2g5eg2 unpaired e-= 3
• [Cu(NH3)4]2+ ⇒ Cu+2 ; d9
NH3 Strong field ligand, hybridisation dsp2
• one e- of 3d jumps into 4p subshell.
unpaired e- = 1
• [Ni(CN)4]2- ⇒ Ni+2 ; d8 unpaired e- = 0
CN- = Strong field ligand dsp2
• TiCl4 ⇒ Ti+4 ; do unpaired e- = zero.
n = no. of unpaired e-
μ = 1.73 =
n = 1
• [CoCl6]4- ⇒ Co+2 ; d7
Cl- (weak field ligand) t2g5eg2 unpaired e-= 3
• [Cu(NH3)4]2+ ⇒ Cu+2 ; d9
NH3 Strong field ligand, hybridisation dsp2
• one e- of 3d jumps into 4p subshell.
unpaired e- = 1
• [Ni(CN)4]2- ⇒ Ni+2 ; d8 unpaired e- = 0
CN- = Strong field ligand dsp2
• TiCl4 ⇒ Ti+4 ; do unpaired e- = zero.
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