
Jahn-Teller effect is not observed in high spin complexes of:
$\left( A \right)$ ${d^4}$
$\left( B \right)$ ${d^9}$
$\left( C \right)$ ${d^7}$
$\left( D \right)$ ${d^8}$
Answer
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Hint: In octahedral, the Jahn Teller effect is due to the extent of interactions of ligands with d orbitals of metals. A nonlinear molecular system undergoes distortion. Due to this distortion, the symmetry and the energy of the system is reduced.
Complete step by step answer:
Jahn Teller effect is not observed in high spin complexes of ${d^8}$ because symmetry filling of electrons in d orbitals there is $8{e^ - }$ due to weak ligands first $3{e^ - }$filling in ${t_{2g}}$ and $2{e^ - }$in ${e_g}$ and remaining ${e^ - }$ get paired in ${t_{2g}}$ due to symmetry filling of electron there is no distortion in $d$ orbital therefore, Jahn Teller effect is not observed.
In ${d^4}$ , there is $4{e^ - }$ due to weak ligand first filling of $3{e^ - }$ in ${t_{2g}}$ then remaining one ${e^ - }$ in ${e_g}$ because of asymmetry filling of electron distortion present in ${e_g}$ therefore, Jahn teller effect is observed.
In ${d^9}$ , there is $9{e^ - }$ due to weak ligand first filling of $3{e^ - }$ in ${t_{2g}}$ and $2{e^ - }$ in ${e_g}$ orbital the remaining $4{e^ - }$ first paired ${t_{2g}}$ then remaining $1{e^ - }$ paired only one orbital of in ${e_g}$ because of asymmetry filling of electron distortion present in ${e_g}$ therefore, Jahn teller effect is observed.
In ${d^7}$ , there is $7{e^ - }$ due to weak ligand first filling of $3{e^ - }$ in ${t_{2g}}$ and $2{e^ - }$ in ${e_g}$ orbital the remaining $2{e^ - }$ first paired ${t_{2g}}$ but one orbital is unpaired in ${t_{2g}}$ because of asymmetry filling of electron distortion present in ${e_g}$ therefore, Jahn teller effect is observed.
As we discussed above, the Jahn Teller effect is not observed in the high spin complex of the ${d^8}$ configuration.
So, the correct option is $\left( D \right)$ .
Note:
Jahn Teller effect is an important mechanism of spontaneous symmetry breaking in the molecular and solid state system which has far reaching consequences in different fields, and is responsible for a variety of phenomena in spectroscopy, material science.
Complete step by step answer:
Jahn Teller effect is not observed in high spin complexes of ${d^8}$ because symmetry filling of electrons in d orbitals there is $8{e^ - }$ due to weak ligands first $3{e^ - }$filling in ${t_{2g}}$ and $2{e^ - }$in ${e_g}$ and remaining ${e^ - }$ get paired in ${t_{2g}}$ due to symmetry filling of electron there is no distortion in $d$ orbital therefore, Jahn Teller effect is not observed.
In ${d^4}$ , there is $4{e^ - }$ due to weak ligand first filling of $3{e^ - }$ in ${t_{2g}}$ then remaining one ${e^ - }$ in ${e_g}$ because of asymmetry filling of electron distortion present in ${e_g}$ therefore, Jahn teller effect is observed.
In ${d^9}$ , there is $9{e^ - }$ due to weak ligand first filling of $3{e^ - }$ in ${t_{2g}}$ and $2{e^ - }$ in ${e_g}$ orbital the remaining $4{e^ - }$ first paired ${t_{2g}}$ then remaining $1{e^ - }$ paired only one orbital of in ${e_g}$ because of asymmetry filling of electron distortion present in ${e_g}$ therefore, Jahn teller effect is observed.
In ${d^7}$ , there is $7{e^ - }$ due to weak ligand first filling of $3{e^ - }$ in ${t_{2g}}$ and $2{e^ - }$ in ${e_g}$ orbital the remaining $2{e^ - }$ first paired ${t_{2g}}$ but one orbital is unpaired in ${t_{2g}}$ because of asymmetry filling of electron distortion present in ${e_g}$ therefore, Jahn teller effect is observed.
As we discussed above, the Jahn Teller effect is not observed in the high spin complex of the ${d^8}$ configuration.
So, the correct option is $\left( D \right)$ .
Note:
Jahn Teller effect is an important mechanism of spontaneous symmetry breaking in the molecular and solid state system which has far reaching consequences in different fields, and is responsible for a variety of phenomena in spectroscopy, material science.
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