Question

$\left[Fe\right(H_{2}O{)}_6{]}^{2+}$ has a crystal field splitting energy value of $10,400c{m}^{-1}$ and a pairing energy value of $17,600c{m}^{-1}$. Then, the compound is :

• A. a low spin complex
• B. paramagnetic in nature
• C. diamagnetic in nature
• D. none of these

Answer 1

The correct option is B paramagnetic in nature

In given complex, the oxidation state of Fe is +2, so, its configuration is $d^6$.

As CFSE is less than pairing energy according to given values :-

$CFSE=10,400c{m}^{-1}$

$P.E=17,600c{m}^{-1}$

It means weak field or high spin complex.
Thus, there is presence of unpaired electrons and hence it is paramagnetic in nature.
Its configuration is $t_{2g}^4{e}_g^2$

Hence, correct answer is option B.