Question

Low spin complex of $d^6$ cation in an octahedral field will have which of the following energy?

• A. $\frac{-12}{5}{\Delta }_0+P$
• B. $\frac{12.5}{5}{\Delta }_0+3P$
• C. $\frac{-12}{5}{\Delta }_0+3P$
• D. $\frac{-2}{5}{\Delta }_0+2P$

Answer 1

The correct option is C $\frac{-12}{5}{\Delta }_0+3P$

Electronic configuration of Low spin complex of $d^6$ cation

Low spin complexes are the inner orbital complexes which are formed by inner $(n-1)d$ orbitals. Low spin complexes are generally formed in the presence of strong field ligands.

For low spin complex ${\Delta }_0>$ pairing energy. This implies that pairing takes place. The splitting of octahedral complex takes place into $t_{2}g$ and $e_g.$

Electronic configuration of $d^6=t_2{g}^6{e}_g^0$

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Calculating total energy

Total energy:

$=[-\frac{2}{5}\times n(t_{2}g)+\frac{3}{5}\times n(e_g\left)\right]{\Delta }_o+P$

${\Delta }_o=$ Splitting energy

$n\left(t_{2}g\right)=$ Number of electrons in $t_{2}g$ orbital

$n\left(e_g\right)=$ Number of electrons in $e_g$ orbital

$P=$ Pairing energy

Total energy:

$-\frac{2}{5}\times 6\times {\Delta }_0+\frac{3}{5}\times 0\times {\Delta }_0+3$

Where,

$n\left(t_{2}g\right)=6$

$n\left(e_g\right)=0$

Pairing energy $=3P$ (as 3 electrons got paired)

Total energy $=-\frac{12}{5}{\Delta }_0+3P$

So, option (A) is the correct answer