1.

The "spin-only" magnetic moment [in units of Bohr magneton, (mu_(B))] of Ni^(2+) in aqueous solution would be (At. NO. Ni=28)

Answer»

6
1.73
2.84
4.9

Solution :The number of UNPAIRED electrons in `Ni^(2+)(aq)=2` Water is weak LIGAND hence no pairing will take place spin magnetic moment
`=sqrt(N(n+2))=sqrt(2(2+2))=sqrt(8)=2.82`


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