Magnetic moment of $$C{e^{3 + }}$$  ion on the basis of 'spin-only' formula will be ______ B.M.

Solution :
The outer electronic configuration of $$C{e^{3 + }}$$  is $$4{f^1}.$$
$$\eqalign{ & {\text{Hence, }}\mu = \sqrt {n\left( {n + 2} \right)} \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = \sqrt {1\left( {1 + 2} \right)} \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = 1.732\,B.M. \cr} $$