Hard MCQ +4 / -1 PYQ · JEE Mains 2022

Octahedral complexes of copper(II) undergo structural distortion (Jahn-Teller). Which one of the given copper (II) complexes will show the maximum structural distortion? (en - ethylenediamine; $\mathrm{H}_{2} \mathrm{~N}_{-} \mathrm{CH}_{2}-\mathrm{CH}_{2}-\mathrm{NH}_{2}$)

  1. A $$\left[\mathrm{Cu}\left(\mathrm{H}_{2} \mathrm{O}\right)_{6}\right] \mathrm{SO}_{4}$$ Correct answer
  2. B $$\left[\mathrm{Cu}(\mathrm{en})\left(\mathrm{H}_{2} \mathrm{O}\right)_{4}\right] \mathrm{SO}_{4}$$
  3. C cis-$\left[\mathrm{Cu}\left(\mathrm{en})_{2} \mathrm{Cl}_{2}\right]\right.$
  4. D trans-$\left[\mathrm{Cu}\left(\mathrm{en})_{2} \mathrm{Cl}_{2}\right]\right.$

Solution

<p>According to John teller any nonlinear molecular system in a degenerate electronic state will be unstable and will undergo some kind of distortion which will lower its symmetry and energy and split the degenerate state.</p> <p>In case of octahedral $d^{9}$ configuration, the last electron may occupy either $d^{2}$ or $d_{x^{2} y^{2}}$ orbitals of $e_{g}$ set.</p> <p>If it occupies $\mathrm{dz}^{2}$ orbital most of the electron density will be concentrated between the metal and the two ligands on the $\mathrm{z}$ axis. Thus there will be greater electrostatic repulsion associated with these ligands than with the other four on xy plane.</p> <p>The Jahn Teller effect is mostly observed in octahedral environments. The considerable distortions are usually observed in high spin $\mathrm{d}^{4}$, low spin $\mathrm{d}^{7}$ and $\mathrm{d}^{9}$ configuration.</p>

About this question

Subject: Chemistry · Chapter: Coordination Compounds · Topic: Ligands and Coordination Number

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