Predict the number of unpaired electrons in the following complex ions: (a) \(\left[\mathrm{Cr}(\mathrm{CN})_{6}\right]^{4-},(\mathrm{b})\left[\mathrm{Cr}\left(\mathrm{H}_{2} \mathrm{O}\right)_{6}\right]^{2+}\)

Short Answer

Expert verified
There are 4 unpaired electrons in both \(\left[\mathrm{Cr}(\mathrm{CN})_{6}\right]^{4-}\) and \(\left[\mathrm{Cr}\left(\mathrm{H}_{2}\mathrm{O}\right)_{6}\right]^{2+}\).

Step by step solution

01

Determine the Oxidation State of Chromium

Chromium is a transition metal and its oxidation state can be determined by charging the ligands. For \(\left[\mathrm{Cr}(\mathrm{CN})_{6}\right]^{4-}\), the oxidation state is calculated as follows: since CN-1 is -1 and there are 6 of these, and the overall charge of the ion is -4, the oxidation state of Cr is +2.
02

Electron Configuration of Cr in its Oxidation State

The ground state electron configuration of Cr is [Ar] 3d5 4s1. The oxidation state of +2 means 2 electrons are lost. The 4s orbital is emptied before 3d because it has higher energy. Therefore, the electron configuration of Cr+2 is [Ar] 3d4.
03

Determine Number of Unpaired Electrons

There are 4 in the 3d orbital. In accordance with Hund's rule, each electron occupies its own d-orbital before pairing, hence there are 4 unpaired electrons.
04

Repeat Steps 1-3 for Other Complex Ion

Repeat step 1-3 to find the number of unpaired electrons for \(\left[\mathrm{Cr}\left(\mathrm{H}_{2}\mathrm{O}\right)_{6}\right]^{2+}\). The oxidation state of Cr is +2, and the electron configuration is [Ar] 3d4. Therefore, there are 4 unpaired electrons.

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