Question

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

Answer 1

Answer: (B)

$In\left[ Mn \left( H _{2} O \right)_{6}\right]^{2+}, Mn$ is present as $Mn ^{2+}$ or $Mn$ (II), so its electronic configuration
$=1 s^{2}, 2 s^{2} 2 p^{6}, 3 s^{2} 3 p^{6}, 3 d^{5}$

$In\left[ Mn \left( H _{2} O \right)_{6}\right]^{2+}$ the co-ordination number of $Mn$ is six, but in presence of weak ligand field, there will be no pairing of electrons in $3 d$. So it will form high spin complex due to presence of five unpaired electron. $In\left[ Mn \left( H _{2} O \right)_{6}\right]^{2}+$