Q3 / 84 · JEE Advanced 2010 (Paper 1) · Chemistry

The ionization isomer of $\left[\mathrm{Cr}\left(\mathrm{H}_2 \mathrm{O}\right)_4 \mathrm{Cl}\left(\mathrm{NO}_2\right)\right] \mathrm{Cl}$ is

  • A
    $\left[\mathrm{Cr}\left(\mathrm{H}_2 \mathrm{O}\right)_4\left(\mathrm{O}_2 \mathrm{~N}\right)\right] \mathrm{Cl}_2$
  • Correct answer
    B
    $\left[\mathrm{Cr}\left(\mathrm{H}_2 \mathrm{O}\right)_4 \mathrm{Cl}_2\right]\left(\mathrm{NO}_2\right)$
  • C
    $\left[\mathrm{Cr}\left(\mathrm{H}_2 \mathrm{O}\right)_4 \mathrm{Cl}(\mathrm{ONO})\right] \mathrm{Cl}$
  • D
    $\left[\mathrm{Cr}\left(\mathrm{H}_2 \mathrm{O}\right)_4 \mathrm{Cl}_2\left(\mathrm{NO}_2\right)\right] \cdot \mathrm{H}_2 \mathrm{O}$

Solution

The ionization isomer of $\left[\mathrm{Cr}\left(\mathrm{H}_2 \mathrm{O}\right)_4 \mathrm{Cl}\left(\mathrm{NO}_2\right)\right] \mathrm{Cl}$ is $\left[\mathrm{Cr}\left(\mathrm{H}_2 \mathrm{O}\right)_4 \mathrm{Cl}_2\right] \mathrm{NO}_2$ because of exchanging of ligand and counter ions Coordination chemistry Straight conceptual II

Hence, the correct answer is option: 'B'.

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