Magnetic moment and unpaired electrons

Started by Pranjal Singh, August 31, 2020, 08:54:21 AM

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Pranjal Singh

Here it is
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uma

Magnetic moment=[n(n+2)]1/2=1.73
n is the number of unpaired electrons

Its value is given so we can calculate n value ( you get a quadratic equation n2 +2n -2.9929=0)
n = 1
It means vanadium has only one unpaired electron
In VCl2 V is with +2 Oxdation Number
V 2+  electronic configurations is 
1s22s22p63s23p6 3d3
All three d electrons are unpaired due to Hund's rule

In VCl3 V is with +3 Oxdation Number
V 3+  electronic configurations is 
1s22s22p63s23p6 3d2
All two d electrons are unpaired due to Hund's rule

In VCl4 V is with +4 Oxdation Number
V 4+  electronic configurations is 
1s22s22p63s23p6 3d1
one d electrons is unpaired and this should be the answer.

In VCl5 V is with +5 Oxdation Number
V 5+  electronic configurations is 
1s22s22p63s23p6
No unpaired electron and it should be diamagentic.

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