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JEE Advanced · Chemistry · 23. Coordination Compounds

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

  1. A \(\left[\mathrm{Cr}\left(\mathrm{H}_2 \mathrm{O}\right)_4\left(\mathrm{O}_2 \mathrm{~N}\right)\right] \mathrm{Cl}_2\)
  2. B \(\left[\mathrm{Cr}\left(\mathrm{H}_2 \mathrm{O}\right)_4 \mathrm{Cl}_2\right]\left(\mathrm{NO}_2\right)\)
  3. C \(\left[\mathrm{Cr}\left(\mathrm{H}_2 \mathrm{O}\right)_4 \mathrm{Cl}(\mathrm{ONO})\right] \mathrm{Cl}\)
  4. D \(\left[\mathrm{Cr}\left(\mathrm{H}_2 \mathrm{O}\right)_4 \mathrm{Cl}_2\left(\mathrm{NO}_2\right)\right] \cdot \mathrm{H}_2 \mathrm{O}\)
Verified Solution

Answer & Solution

Correct Answer

(B) \(\left[\mathrm{Cr}\left(\mathrm{H}_2 \mathrm{O}\right)_4 \mathrm{Cl}_2\right]\left(\mathrm{NO}_2\right)\)

Step-by-step Solution

Detailed explanation

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
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