Low spin tetrahedral complexes are not formed because:
A. for tetrahedral complexes, the CFSE is lower than the pairing energy.
B. for tetrahedral complexes, the CFSE is higher than the pairing energy.
C. electrons do not go to ${{{e}}_{{g}}}$ in case of tetrahedral complexes.
D. tetrahedral complexes are formed by weak field ligands only.
Answer
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Hint:This problem can be solved using crystal field theory. There will be a degeneracy, i.e. different energy levels having the same energy, in d or f subshell. The ligands produce electric fields which remove this degeneracy.
Complete step by step answer:
We know that there is a splitting of d orbitals. Due to the repulsion of electrons, the outermost electrons in d orbital will have high energy. This splitting depends upon several factors like:
-Metal ion nature
-Oxidation state of metal
-The way the ligands are arranged around the metal ion
-Ligand nature
-d orbitals are splitted into two ways-one having ${{{d}}_{{{xy}}}},{{{d}}_{{{yz}}}},{{{d}}_{{{xz}}}}$ energy levels and other having ${{{d}}_{{{{x}}^2} - {{{y}}^2}}},{{{d}}_{{{{z}}^2}}}$. The first one forms ${{{t}}_{2{{g}}}}$set and the second one forms ${{{e}}_{{g}}}$ set. In tetrahedral complexes, the last two orbitals have lower energy than the first three orbitals. In such complexes, splitting is less than in octahedral complexes. CFSE of tetrahedral complexes is less than the pairing energy. The electrons are occupied in the higher energy levels. Thus its CFSE will not be enough for pairing of spin to occur. Thus it rarely forms low spin complexes, but forms high spin complexes.
Hence the correct option is A.
Additional information:
Since the ${{{d}}_{{{xy}}}},{{{d}}_{{{yz}}}},{{{d}}_{{{xz}}}}$ orbitals is directed the same as the ligands and ${{{d}}_{{{{x}}^2} - {{{y}}^2}}},{{{d}}_{{{{z}}^2}}}$ lie between the ligands, ${{{d}}_{{{xy}}}},{{{d}}_{{{yz}}}},{{{d}}_{{{xz}}}}$ orbitals have high energy and ${{{d}}_{{{{x}}^2} - {{{y}}^2}}},{{{d}}_{{{{z}}^2}}}$ orbitals have low energy.
Note:
There are four ligands, which are around the central metal atom, in tetrahedral complexes, i.e. the number of ligands in tetrahedral complexes is $\dfrac{2}{3}$ rd of the octahedral complexes. This indicates that the CFSE is lower than that of octahedral complexes. Thus it is very easy to form high spin complexes.
Complete step by step answer:
We know that there is a splitting of d orbitals. Due to the repulsion of electrons, the outermost electrons in d orbital will have high energy. This splitting depends upon several factors like:
-Metal ion nature
-Oxidation state of metal
-The way the ligands are arranged around the metal ion
-Ligand nature
-d orbitals are splitted into two ways-one having ${{{d}}_{{{xy}}}},{{{d}}_{{{yz}}}},{{{d}}_{{{xz}}}}$ energy levels and other having ${{{d}}_{{{{x}}^2} - {{{y}}^2}}},{{{d}}_{{{{z}}^2}}}$. The first one forms ${{{t}}_{2{{g}}}}$set and the second one forms ${{{e}}_{{g}}}$ set. In tetrahedral complexes, the last two orbitals have lower energy than the first three orbitals. In such complexes, splitting is less than in octahedral complexes. CFSE of tetrahedral complexes is less than the pairing energy. The electrons are occupied in the higher energy levels. Thus its CFSE will not be enough for pairing of spin to occur. Thus it rarely forms low spin complexes, but forms high spin complexes.
Hence the correct option is A.
Additional information:
Since the ${{{d}}_{{{xy}}}},{{{d}}_{{{yz}}}},{{{d}}_{{{xz}}}}$ orbitals is directed the same as the ligands and ${{{d}}_{{{{x}}^2} - {{{y}}^2}}},{{{d}}_{{{{z}}^2}}}$ lie between the ligands, ${{{d}}_{{{xy}}}},{{{d}}_{{{yz}}}},{{{d}}_{{{xz}}}}$ orbitals have high energy and ${{{d}}_{{{{x}}^2} - {{{y}}^2}}},{{{d}}_{{{{z}}^2}}}$ orbitals have low energy.
Note:
There are four ligands, which are around the central metal atom, in tetrahedral complexes, i.e. the number of ligands in tetrahedral complexes is $\dfrac{2}{3}$ rd of the octahedral complexes. This indicates that the CFSE is lower than that of octahedral complexes. Thus it is very easy to form high spin complexes.
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