JEE MAIN - Chemistry (2006 - No. 39)
The “spin-only” magnetic moment [in units of Bohr magneton, (µB)] of Ni2+ in aqueous solution would
be : (Atomic number of Ni = 28)
2.84
4.90
0
1.73
Explication
The number of unpaired electrons in $$N{i^{2 + }}\left( {aq} \right) = 2$$
Water is weak ligand hence no pairing will take place spin magnetic moment
$$\eqalign{ & = \sqrt {n\left( {n + 2} \right)} = \sqrt {2\left( {2 + 2} \right)} \cr & = \sqrt 8 = 2.82 \cr} $$
Water is weak ligand hence no pairing will take place spin magnetic moment
$$\eqalign{ & = \sqrt {n\left( {n + 2} \right)} = \sqrt {2\left( {2 + 2} \right)} \cr & = \sqrt 8 = 2.82 \cr} $$