Question

Which of the following statement is correct about $\lbrack Co(H_{2}O)_{6}\rbrack^{2 +}$complex ?

• A. Electronic configurations $= 3d^{7} \rightarrow t_{2}^{5}e_{g}^{2},$ No, of unpaired electrons = $3,\mu = 3.87B.M.$
• B. Electronic configurations $= 3d^{6} \rightarrow t_{2}^{4}e_{g}^{2},$ No. of unpaired electrons = $2,\mu = 2.87B.M.$
• C. Electronic configurations $= 3d^{7} \rightarrow t_{2}^{4}e_{g}^{1},$ No. of unpaired electrons = $1,\mu = 2.87B.M.$
• D. Electronic configurations $= 3d^{7} \rightarrow t_{2}^{3}e_{g}^{4},$ No. of unpaired electrons $= 3,\mu = 3.87B.M.$

Answer 1

Answer: (A)

Electronic configurations $= 3d^{7} \rightarrow t_{2}^{5}e_{g}^{2},$

No, of unpaired electrons = $3,\mu = 3.87B.M.$

Answer 2

In $\lbrack Co(H_{2}O)_{6}\rbrack^{2 +}$Oxidations state of Co = +2, $Co^{2 +} = 3d^{7}$

$\mu = \sqrt{n(n + 2)} = \sqrt{3(3 + 2)} = \sqrt{15} = 3.87B.M.$