Question

Assertion: Low spin tetrahedral complexes are rarely observed
Reason: Crystal field splitting energy is less than pairing energy, for tetrahedral complexes.
A.Both Assertion and Reason are correct and Reason is the correct explanation for Assertion.
B.Both Assertion and Reason are correct but Reason is not the correct explanation for Assertion
C.Assertion is correct but Reason is incorrect
D.Assertion is incorrect but reason is correct

Answer 1

We can say that the energy penalty for placing two electrons in the same orbital, ensuing from the electrostatic repulsion between electrons is called pairing energy (P). We can define Crystal Field Stabilization Energy as the energy difference of the electron configuration in the ligand field to the energy of the electronic configuration in the isotropic field.

Complete step by step answer:
We can write the expression for the crystal field stabilization energy as,
$CFSE\left( \Delta \right)E = {E_{ligand field - }}{E_{isotropic field}}$
We have to know that Crystal Field Stabilization Energy is dependent on factors like number of d-electrons, ligand character, geometry, and spin pairing energy.
We employ the term spin pairing energy P, if the electrons are paired within a single orbital.
We can give the spectrochemical series as,
\begin{array}{*{20}{c}} {{I^ - } < B{r^ - } < {S^{2 - }} < SC{N^ - } < C{l^ - } < {N_3}^ - < {F^ - } < NC{O^ - } < O{H^ - } < {C_2}{O_4}^ - < {O^{2 - }} < {H_2}O < NC{S^ - }} \\\ { < C{H_3}CN < gh < py < N{H_3} < bipy < NO_2^ - < PP{h_3} < C{N^ - } < CO} \end{array}
If ${{{\Delta }}_{{o}}}{\text{ > P}}$ , the complex will be low spin and if ${{{\Delta }}_{{o}}}{\text{ < P}}$ , the complex will be high spin.
A tetrahedral compound would have $s{p^3}$ hybridization. This conveys that the last d-orbital is not vacant as if it was then rather than $s{p^3},ds{p^2}$ will have been followed and compound will have been square planar geometry instead of tetrahedral.
Now the formation of the low spin complexes occurs when a bond is formed between strong field ligands with the metal or ion. The strong field ligand leads to pairing of electrons and it creates empty d-orbital and so tetrahedral is not formed.
Also, the d-orbital splitting is less when compared to octahedral.
Hence, to force pairing the splitting energy of the orbital is not sufficient.
Due to which, low spin configurations are rarely noticed in tetrahedral complexes.
We can say both Assertion and Reason are correct and the correct explanation for Assertion is the reason.
Therefore, the option (A) is correct.

Note:
We have to remember that the binding of a strong field ligand results in a more energy difference between the higher energy level and lower energy level orbitals.
Examples: $C{N^-}$ (cyanide ligands), $N{O_2}^-$ (nitro ligand) and ${\text{CO}}$ (carbonyl ligands)
The binding of a weak field ligand results in lower energy difference between the higher energy and lower energy level orbitals because the low difference between the two orbital levels leads to repulsions between electrons in those energy levels, the higher energy orbitals could be quickly be filled with electrons when compared to that in low energy orbitals.
Examples: ${I^-}$ (iodide ligand), $B{r^-}$ (bromide ligand), etc..