Question

What is the magnetic moment (spin only) and hybridisation of the brown ring complex $\left[Fe\right(H_{2}O{)}_{5}NO]S{O}_4$?

• A. $\sqrt{3}BM,s{p}^3{d}^2$
• B. $\sqrt{3}BM,d^{2}s{p}^3$
• C. $\sqrt{15}BM,s{p}^3{d}^2$
• D. $\sqrt{15}BM,d^{2}s{p}^3$

Answer 1

The correct option is C $\sqrt{15}BM,s{p}^3{d}^2$

In $\left[Fe\right(H_{2}O{)}_{5}NO]S{O}_4$, the $F{e}^{+}\left(d^7\right)$ ion undergoes $s{p}^3{d}^2$ hybridization to form outer orbital or high-spin (spin-free) complex with 3 unpaired electrons.

The spin only magnetic moment is given by the formula ${\mu }_{eff}=\sqrt{n(n+2)}B.M.$.

For three unpaired electrons, the expression becomes ${\mu }_{eff}=\sqrt{3(3+2)}B.M.=\sqrt{15}B.M.$.

Option C is correct.