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 :

 In\[{{[Mn{{({{H}_{2}}O)}_{6}}]}^{2+}},Mn\]is present as\[M{{n}^{2+}}\]or\[Mn\] (II), so its electronic configuration \[=1{{s}^{2}},2{{s}^{2}}2{{p}^{6}},3{{s}^{2}}3{{p}^{6}},3{{d}^{5}}\] In\[{{[Mn{{({{H}_{2}}O)}_{6}}]}^{2+}}\]the co-ordination number of \[Mn\]is six, but in presence of weak ligand field, there will be no pairing of electrons in 3d. So it will form high spin complex due to presence of five unpaired electron. In\[{{[Mn{{({{H}_{2}}O)}_{6}}]}^{2+}}\]