Question

The complex ion having minimum magnitude of $\Delta _0 (CFSE) $is

• A. $[Cr(CN)_6]^{3-}$
• B. $[Co(NH_3)_6]^{3+}$
• C. $[Co(Cl)_6]^{3-}$
• D. $[Cr(H_2O)_6]^{3+}$

Answer 1

Answer: (C)

Trick CFSE has lower value, when the complex contains weak field ligand. Thus, CFSE decreases in order

$\left[Co\left(NH_{3}\right)_{6}\right]^{3+} >\left[ Cr\left(CN\right)_{6}\right]^{3-} \simeq\left[Cr\left(H_{2}O\right)_{6}\right]^{3+} >\left[Co\left(Cl\right)_{6}\right]^{3-}$

Because order of field strength is

$Cl^{-} < H_{2}O < NH_{3} < CN^{-}.$

CFSE $\left(\Delta_{{\circ}}\right)$ is calculated by using the formula

$\quad CFSE = \left(- 0.4x + 0.6y\right)\Delta_{{\circ}}$

where, $\quad$ x = number of electrons in $t_{2g}$

and$\quad\quad$ y = number of electrons in $e_{g}$

For $Co^{3+} d^{6}$ high spin in $\left[Co\left(Cl\right)_{6}\right]^{3-}$

$\quad\quad\quad\quad\Delta_{^{\circ}} = 4\times\left(-0.4\right)+\left(2\times0.6\right)=-0.4\Delta_{{\circ}}$

For $Cr^{3+} d^{3}$ low spin $\left[Cr\left(CN\right)_{6}\right]^{3-},\left[Cr\left(H_{2}O\right)_{6}\right]^{3+}$

$\quad \quad \quad \quad \Delta _{{\circ }} = 3\times \left(-0.4\right)=-1.2\Delta _{{\circ }}$

For $Co^{3+} d^{6}$ low spin in $\left[Co\left(NH_{3}\right)\right]^{3+}$

$\quad \quad \quad \quad \Delta _{{\circ }} = 6\times \left(-0.4\right)+\left(0\times 0.6\right)=-2.4\Delta _{^{\circ }}$

$\therefore $ For $\left[Co\left(Cl\right)_{6}\right]^{-}$ complex, magnitude of $\Delta_{{\circ}}$ CFSE is minimum.