1.

Calculated spin only magnetic moment of Cr^(x+) is 4.9 BM. Find the 'x' value.

Answer»

Solution :Magnetic moment `mu= 4.9 BM = sqrt(n(n+2))`
Number of unpaired ELECTRONS `= n = 4`
with FOUR unpaired electrons, the configuration is `3d^(4)4s^(0)`.
`Cr^(2+)` ion posses 4 unpaired electrons, THUS .X. value is 2.


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