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

In order to calculate the number of unpaired electrons, we need to determine the oxidation state of the Chromium ion in each complex. For \(\left[\mathrm{Cr}(\mathrm{CN})_{6}\right]^{4-}\), the overall charge of the complex is -4 and the Cyanide ion (CN) has a charge of -1. Thus, the charge on Cr in this complex is: $$6\times (-1)+x= -4$$ $$x=+2$$ For \(\left[\mathrm{Cr}\left(\mathrm{H}_{2} \mathrm{O}\right)_{6}\right]^{2+}\), the overall charge of the complex is +2, and the water molecule (H₂O) is neutral with a charge of 0. Thus, the charge on Cr in this complex is: $$6\times (0)+x= 2$$ $$x=+2$$ In both complexes, the oxidation state of Chromium is +2.

Chromium has the atomic number 24, and its ground state electron configuration is \([\mathrm{Ar}]\, 3d^5 4s^1\). When Chromium gains an oxidation state of +2, it loses 2 electrons from the \(4s\) and \(3d\) orbitals. This results in the electron configuration of \(\mathrm{Cr}^{2+}\) as \([\mathrm{Ar}]\, 3d^4\).

Crystal field theory predicts that the electronic configuration of a metal ion in a complex can change due to the electrostatic interactions with the ligands. In both complexes, since the Chromium ion has the same oxidation state, the primary difference in their paramagnetism will come from the crystal field splitting caused by the different ligands. Cyanide (CN) is a strong field ligand, while water (H₂O) is considered a weak field ligand. In the presence of a strong field ligand like Cyanide, the crystal field splitting is large enough to cause the \(3d\) orbitals to split into two sets: \(t_{2g}\) (lower energy) and \(e_g\) (higher energy). This results in a low-spin complex with paired electrons, as the electrons will occupy the lower energy orbitals first. The electron configuration of \(\left[\mathrm{Cr}(\mathrm{CN})_{6}\right]^{4-}\) will be \(t_{2g}^4 \, e_g^0\) with no unpaired electrons. In contrast, the crystal field splitting caused by water, a weak field ligand, will be smaller. This results in a high-spin complex with unpaired electrons. The electron configuration of \(\left[\mathrm{Cr}\left(\mathrm{H}_{2} \mathrm{O}\right)_{6}\right]^{2+}\) will be \(t_{2g}^3 \, e_g^1\) with one unpaired electron.

The complex \(\left[\mathrm{Cr}(\mathrm{CN})_{6}\right]^{4-}\) is a low-spin complex with no unpaired electrons. In contrast, the complex \(\left[\mathrm{Cr}\left(\mathrm{H}_{2} \mathrm{O}\right)_{6}\right]^{2+}\) is a high-spin complex with one unpaired electron. Paramagnetism is directly related to the number of unpaired electrons, with more unpaired electrons resulting in greater paramagnetism. Therefore, the complex \(\left[\mathrm{Cr}(\mathrm{CN})_{6}\right]^{4-}\) is less paramagnetic than \(\left[\mathrm{Cr}\left(\mathrm{H}_{2} \mathrm{ O}\right)_{6}\right]^{2+}\), as it has fewer unpaired electrons.