Question

The crystal field stabilization energy (CFSE) $$\text{[Fe(}{\text{H}}_2\text{O}{\text{)}}_6]\text{C}{\text{l}}_2$$ and $${\text{K}}_2\text{[NiC}{\text{l}}_4]$$, respectively, are:
a.) $$\text{-0}\text{.4}{\mathrm{\Delta }}_{\text{0}}\text{ and -0}\text{.8}{\mathrm{\Delta }}_{\text{t}}$$
b.) $$\text{-0}\text{.4}{\mathrm{\Delta }}_{\text{0}}\text{ and -1}\text{.2}{\mathrm{\Delta }}_{\text{t}}$$
c.) $$\text{-2}\text{.4}{\mathrm{\Delta }}_{\text{0}}\text{ and -1}\text{.2}{\mathrm{\Delta }}_{\text{t}}$$
d.) $$\text{-0}\text{.6}{\mathrm{\Delta }}_{\text{0}}\text{ and -0}\text{.8}{\mathrm{\Delta }}_{\text{t}}$$

Answer 1

Hint: Crystal field theory states that there is a difference of $$\text{10}{\text{D}}_{\text{q}}$$ between $${\text{e}}_{\text{g}}$$ and $${\text{t}}_{\text{2g}}$$ energy levels. When an electron goes into $${\text{t}}_{\text{2g}}$$ which is a lower energy level, it stabilizes the system by an amount of $$\text{-4}{\text{D}}_{\text{q}}$$ and when an electron goes into $${\text{e}}_{\text{g}}$$ level, it destabilizes the system by $$\text{+6}{\text{D}}_{\text{q}}$$.

Complete answer:
The stability which results due to placing of a transition metal ion into a field that is caused by the set of ligands which surround it, is called Crystal field stabilization energy. In the complex, $$[\text{ Fe(}{\text{H}}_2\text{O}{\text{)}}_6]\text{C}{\text{l}}_2$$, Fe forms a +2 ion i.e. $${[\text{ Fe(}{\text{H}}_2\text{O}{\text{)}}_6]}^{+2}$$.
The electronic configuration of Iron in +2 becomes $$[\text{ Ar }]\text{ 3}{\text{d}}^3$$.
is a weak field ligand and hence pairing of electrons does not occur. Electrons in $${\text{t}}_{\text{2g}}$$ are given as $${\text{t}}_{\text{2g}}^{\text{2, 1, 1}}$$.

Formula for calculating crystal field stabilization energy is $$\mathrm{\Delta }\text{ = no}\text{. of }{\text{e}}^{-}\text{ in }{\text{t}}_{\text{2g}}\times \text{ (-0}\text{.4) + no}\text{. of }{\text{e}}^{-}\text{ in }{\text{e}}_g\times \text{ (0}\text{.6)}$$

Thus, for this complex - $$\mathrm{\Delta }\text{ = 4}{\mathrm{\Delta }}_0\times \text{ (-0}\text{.4) + 2}{\mathrm{\Delta }}_0\times \text{ (0}\text{.6)}$$
                                              $$\mathrm{\Delta }\text{ = - 0}\text{.4}{\mathrm{\Delta }}_0$$
In the complex, $${\text{K}}_2[\text{ NiC}{\text{l}}_4]$$ charge on Nickel is +2 as it forms $${[\text{ NiC}{\text{l}}_4]}^{-2}$$.

    The electronic configuration of Nickel in +2 is $$[\text{ Ar }]\text{ 3}{\text{d}}^8$$.

 is also a weak field ligand and hence pairing does not occur and the complex formed has tetrahedral geometry. Electrons in $${\text{e}}_{\text{g}}$$ are $${\text{e}}_{\text{g}}^{\text{2, 2}}$$.
Putting these values in the formula we get, $$\mathrm{\Delta }\text{ = -0}\text{.6 }\times \text{ 4}{\mathrm{\Delta }}_{\text{t}}\text{ + 0}\text{.4 }\times \text{ 4}{\mathrm{\Delta }}_{\text{t}}$$
                                                                                  $$\text{ = -2}\text{.4}{\mathrm{\Delta }}_{\text{t}}\text{ + 1}\text{.6}{\mathrm{\Delta }}_{\text{t}}$$
                                                                             $$\mathrm{\Delta }\text{ = -0}\text{.8}{\mathrm{\Delta }}_{\text{t}}$$

So, the correct answer is “Option A”.

Note: Complexes which have a higher number of unpaired electrons are called as high spin complexes and the ones which have low number of unpaired electrons are called as low spin complexes. Most of the time, high spin complexes have weak field ligands and hence their splitting energy has lower value. Conversely, low spin complexes have strong field ligands and hence have higher value of splitting energy.