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\[{{[Mn{{({{H}_{2}}O)}_{6}}]}^{2+}}\]You need to login to perform this action.
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