Question

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

• A. $[Cr(CN{)}_6{]}^{3-}$
• B. $[Co(N{H}_3{)}_6{]}^{3+}$
• C. $[Co(Cl{)}_6{]}^{3-}$
• D. $[Cr(H_{2}O{)}_6{]}^{3+}$

Answer 1

Answer: (C)

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

${\left[Co{\left(N{H}_3\right)}_6\right]}^{3+}>{\left[Cr{\left(CN\right)}_6\right]}^{3-}\simeq {\left[Cr{\left(H_{2}O\right)}_6\right]}^{3+}>{\left[Co{\left(Cl\right)}_6\right]}^{3-}$

Because order of field strength is

$C{l}^{-}<H_{2}O<N{H}_3<C{N}^{-}.$

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

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

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

and y = number of electrons in $e_g$

For $C{o}^{3+}{d}^6$ high spin in ${\left[Co{\left(Cl\right)}_6\right]}^{3-}$

${\Delta }_{{}^{\circ }}=4\times \left(-0.4\right)+\left(2\times 0.6\right)=-0.4{\Delta }_{\circ }$

For $C{r}^{3+}{d}^3$ low spin ${\left[Cr{\left(CN\right)}_6\right]}^{3-},{\left[Cr{\left(H_{2}O\right)}_6\right]}^{3+}$

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

For $C{o}^{3+}{d}^6$ low spin in ${\left[Co\left(N{H}_3\right)\right]}^{3+}$

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

$\therefore$ For ${\left[Co{\left(Cl\right)}_6\right]}^{-}$ complex, magnitude of ${\Delta }_{\circ }$ CFSE is minimum.