Question
Which of the following statements is correct for $${\left[ {Mn{{\left( {CN} \right)}_6}} \right]^{3 - }}$$ according to valence bond theory?
A.
$${d^2}s{p^3},$$ inner orbital complex, paramagnetic, 2.82 $$B.M.$$
B.
$${d^2}s{p^3},$$ inner orbital complex, diamagnetic, zero magnetic moment.
C.
$${d^2}s{p^3},$$ outer orbital complex, paramagnetic, 3.87 $$B.M.$$
D.
$$ds{p^2},$$ outer orbital complex, diamagnetic, zero magnetic moment.
Answer :
$${d^2}s{p^3},$$ inner orbital complex, paramagnetic, 2.82 $$B.M.$$
Solution :
In $${\left[ {Mn{{\left( {CN} \right)}_6}} \right]^{3 - }},$$ oxidation state of $$Mn = + 3,M{n^{3 + }} = 3{d^4}$$
It has two unpaired electrons.
$$\eqalign{ & \mu = \sqrt {n\left( {n + 2} \right)} \cr & \,\,\,\,\, = \sqrt {2\left( {2 + 2} \right)} \cr & \,\,\,\,\, = \sqrt 8 \cr & \,\,\,\,\, = 2.82\,B.M. \cr} $$
In $${\left[ {Mn{{\left( {CN} \right)}_6}} \right]^{3 - }},$$ oxidation state of $$Mn = + 3,M{n^{3 + }} = 3{d^4}$$
It has two unpaired electrons.
$$\eqalign{ & \mu = \sqrt {n\left( {n + 2} \right)} \cr & \,\,\,\,\, = \sqrt {2\left( {2 + 2} \right)} \cr & \,\,\,\,\, = \sqrt 8 \cr & \,\,\,\,\, = 2.82\,B.M. \cr} $$