Question

\(\left[\mathrm{Cr}(\mathrm{CN})_{6}{ }^{4-}\right]\) is less paramagnetic than \(\left[\mathrm{Cr}\left(\mathrm{H}_{2} \mathrm{O}\right)_{6}{ }^{2+}\right]\). Explain.

Short answer

Question: Determine if the given statement, "\(\left[\mathrm{Cr}(\mathrm{CN})_{6}{ }^{4-}\right]\) is less paramagnetic than \(\left[\mathrm{Cr}\left(\mathrm{H}_{2} \mathrm{O}\right)_{6}{ }^{2+}\right]\)" is correct or incorrect. Answer: The given statement is incorrect, as both complexes have the same paramagnetic properties with four unpaired electrons each.

Step-by-step solution

Determine the oxidation state of Chromium in each complex

To find the oxidation state of Chromium, we need to consider that the sum of oxidation states of all species in the complex should be equal to the overall charge of the complex. For the first complex, \(\left[\mathrm{Cr}(\mathrm{CN})_{6}{ }^{4-}\right]\): Let the oxidation state of Cr be x. Since Cyanide (CN) has a charge of -1 and there are six Cyanide ions, the sum of oxidation states will be x - 6 = -4. Solving for x, we get the oxidation state of Cr = +2. For the second complex, \(\left[\mathrm{Cr}\left(\mathrm{H}_{2} \mathrm{O}\right)_{6}{ }^{2+}\right]\): Let the oxidation state of Cr be x. Since water (H2O) is a neutral ligand with a charge of 0, the sum of oxidation states will be x + 0 = +2. Solving for x, we get the oxidation state of Cr = +2.

Write the electronic configuration for Chromium in both complexes

Cr has an atomic number of 24, so its ground-state electronic configuration is [Ar] 4s² 3d⁴. When it loses two electrons (forming Cr²⁺), its configuration becomes [Ar] 3d⁴. Given that the oxidation state of Chromium is the same in both complexes (+2), the electronic configuration for Cr in both complexes is [Ar] 3d⁴.

Determine the number of unpaired electrons in each complex

In both complexes, Chromium has the same electronic configuration, which is [Ar] 3d⁴. This configuration has four unpaired electrons. Therefore, both \(\left[\mathrm{Cr}(\mathrm{CN})_{6}{ }^{4-}\right]\) and \(\left[\mathrm{Cr}\left(\mathrm{H}_{2}\mathrm{O}\right)_{6}{ }^{2+}\right]\) complexes have the same number of unpaired electrons (4), and hence, they have the same paramagnetic properties. The given statement, "\(\left[\mathrm{Cr}(\mathrm{CN})_{6}{ }^{4-}\right]\) is less paramagnetic than \(\left[\mathrm{Cr}\left(\mathrm{H}_{2} \mathrm{O}\right)_{6}{ }^{2+}\right]\)" is incorrect, as we have just established that both complexes have the same paramagnetic properties.