Question

What is the number of unpaired electron in $\left[Fe\right(H_{2}O{)}_6{]}^{2+}$ if is known to be a high spin complex ?

• A. 0
• B. 2
• C. 3
• D. 4

Answer 1

The correct option is D 4

The configuration of $F{e}^{2+}\text{is}3d^6$.
Water is a weak ligand, and hence the pairing energy is higher than the crystal field splitting energy $({\Delta }_o<P)$. So, of the six electrons only two electrons are paired and the four electrons occupy the other four available d-orbitals forming a high spin complex. the configuration will be $t_{2g}^4{e}_g^2$. Thus the number of unpaired electron is four. Hence option (d) is the correct answer.