JEE Chemistry 2025 Question with Solution

Given: Four octahedral iron complexes are given.

Find: Which complex has CFSE = $0$.

CFSE depends on the oxidation state of iron, its $d$-electron configuration, and whether the complex is high spin or low spin.

From the solution:

1. $[Fe(NH_3)_6]Br_2$

• Oxidation state of $Fe$ is $+2$
• Configuration is $d^6$

2. $[Fe(en)_3]Cl_3$

• Oxidation state of $Fe$ is $+3$
• Configuration is $d^5$

3. $K_4[Fe(CN)_6]$

• Oxidation state of $Fe$ is $+2$
• Configuration is $d^6$

4. $K_3[Fe(SCN)_6]$

• Oxidation state of $Fe$ is $+3$
• Configuration is $d^5$

For an octahedral complex, CFSE becomes zero for a high-spin $d^5$ configuration because the electrons are distributed as:

Hence the stabilization and destabilization contributions cancel.

As stated in the solution, in $[Fe(SCN)_6]^{3-}$, iron is $Fe^{3+}$, so it is $d^5$. Since $SCN^-$ acts here to give a high-spin arrangement, the distribution is:

Therefore, CFSE is zero.

Thus, the complex $K_3[Fe(SCN)_6]$ has CFSE = $0$. The correct option is D.