question_answer

Considering \[{{H}_{2}}O\]as weak field ligand, the number of unpaired electrons in\[{{[Mn{{({{H}_{2}}O)}_{6}}]}^{2+}}\]will be (At. no. of \[Mn=25\]):                                                                                                                                                    

A) three

B)                five

C) two               

D) four

Correct Answer: B

Solution :

\[C{{N}^{-}}\] is strong field ligand because it is an example of pseudohalide. Pseudohalideions are stronger coordinating ligands and they have the ability to form c bond and n bond. In\[\left[ \text{Mn}{{\left( {{\text{H}}_{\text{2}}}\text{O} \right)}_{\text{6}}} \right]{{\text{ }}^{\text{2}+}},\text{ Mn}\] is present as \[M{{n}^{2+}}\]or \[\text{Mn}\] (II), so its electronic configuration \[=1{{s}^{2}},\text{ }2{{s}^{2}}2{{p}^{6}},\text{ }3{{s}^{2}}3{{p}^{6}},\text{ }3{{d}^{5}}\] In\[{{\left[ \text{Mn}{{\left( {{\text{H}}_{\text{2}}}\text{O} \right)}_{\text{6}}} \right]}^{\text{2}+}}\]the co-ordination number of Mn is six, but in presence of weak ligand field, there will beno pairing of electrons in 3d. So it will form high spin complex due to presence of five unpaired electron. In \[{{[Mn{{({{H}_{2}}O)}_{6}}]}^{2+}}\]