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 $M{n}^{2+}$ or $Mn$ (II), so its electronic configuration
$=1s^2,2s^{2}2p^6,3s^{2}3p^6,3d^5$

$In{\left[Mn{\left(H_{2}O\right)}_6\right]}^{2+}$ the co-ordination number of $Mn$ is six, but in presence of weak ligand field, there will be no pairing of electrons in $3d$. So it will form high spin complex due to presence of five unpaired electron. $In{\left[Mn{\left(H_{2}O\right)}_6\right]}^2+$