JEE Main 2025 — Coordination Compounds Question with Solution

${\Delta }_0$ value of the octahedral complexes can be calculated using the formula

CASE $=\left(-0,4{\Delta }_0\right)\times$ number of electrons in $f_{2g}$ orbitals + $\left(+0.6{\Delta }_0\right)\times$ number of electrons in eg orbitals.

The ligand CN${}^{-}$ is a strong field ligand and all the complexes form low-spin complex (pairing occurs).

I) ${\left[M_N(\mathrm{CN}{)}_6\right]}^{3-}$

Oxidation state of $M_n:+3$

$Mn$ configuration $\to d^5{s}^2$

${\mathrm{Mn}}^{+3}$ configuration $\to {\mathrm{d}}^4$

$$\begin{array}{r}x+6(-1)=-3\\ x=-3+b=+3\end{array}$$

$$\begin{array}{rl}CFSE& =-0.4{\Delta }_0\times 4+\left(0.6{\Delta }_0\right)\times 0\\ & =-1.6{\Delta }_0\end{array}$$

II) ${\left[\mathrm{Co}(\mathrm{CN}{)}_6\right]}^{4-}$

Oxidation state of $C_0:+2$

$C_0$ configuration $\to d^7{s}^2$

${\mathrm{Co}}^{+2}$ configuration $\to d^7$

$$\begin{array}{rl}\text{ CFSE }& =-0.4{\Delta }_0\times 6+\left(0.6{\Delta }_0\right)\times 1\\ & =-2.4{\Delta }_0+0.6{\Delta }_0=-1.8{\Delta }_0\end{array}$$

$$\begin{array}{rl}x+6(-1)& =-4\\ x& =-4+6=+2\end{array}$$

$$\begin{array}{rl}\text{ CFSE }& =-0.4{\Delta }_0\times 6+\left(0.6{\Delta }_{10}\right)\times 1\\ & =-2.4{\Delta }_0+0.6{\Delta }_0=-1.8{\Delta }_0\end{array}$$

III) ${\left[\mathrm{Fe}(\mathrm{CN}{)}_6\right]}^{4-}$

Oxidation state of $\mathrm{Fe}:+2$

Fe configuration $\to d^6{s}^2$

${\mathrm{Fe}}^{+2}$ configuration $\to {\mathrm{d}}^6$

$$\begin{array}{r}x+(-1)\times 6=-4\\ x=-4+6\\ =+2\end{array}$$

$$\begin{array}{rl}CFSE& =-0.4{\Delta }_0\times 6+0.6{\Delta }_0\times 0\\ & =-2.4{\Delta }_0\end{array}$$

IV) ${\left[\mathrm{Fe}(\mathrm{CN}{)}_6\right]}^{3-}$

Oxidation state of $\mathrm{Fe}:+3$

$Fe$ configuration $\to d^6{s}^2$

${\mathrm{Fe}}^{+3}$ configuration $\to {\mathrm{d}}^5$

$$\begin{array}{c}x+6(-1)=-3\\ x=-3+6\\ =+3\end{array}$$

I) ${\left[{\mathrm{Mn}}_n(\mathrm{CN}{)}_6\right]}^{3-}\to -1.6{\Delta }_0$

II) ${\left[\mathrm{Co}(\mathrm{CN}{)}_6\right]}^{4-}\to -1.8{\Delta }_0$

III) ${\left[\mathrm{Fe}(\mathrm{CN}{)}_6\right]}^{4-}\to -2.4{\Delta }_0$

IV) $[\mathrm{Fe}{\left((\mathrm{N}{)}_6\right]}^{3-}\to -2.0{\Delta }_0$

Stability of complexes based on ${\Delta }_0$ value, greater the CFSE value, more stable the complex.

The more negative the CFSE value, the move stable the complex.

So, complex with CFSE $-2.4\Delta$ is most stable and complex with CFSE $-1.6{\Delta }_0$, is least stable. Stability order:

$$\begin{array}{rl}& \begin{array}{rl}& -1.6{\Delta }_0<-1.8{\Delta }_0<-2.0{\Delta }_0<-2.4{\Delta }_0\\ & {\left[Mn(\mathrm{CN}{)}_6\right]}^{3-}<{\left[\mathrm{C}(\mathrm{CN}{)}_6\right]}^{4-}<{\left[\mathrm{Fe}(\mathrm{CN}{)}_6\right]}^{3-}<{\left[\mathrm{Fe}(\mathrm{CN}{)}_6\right]}^{4-}\\ & I<II<IV<III\end{array}\\ & \text{ Answer: Option (1) }\end{array}$$