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Chapter 24: Problem 57

Provide a valence bond description of the bonding in the \(\mathrm{Cr}\left(\mathrm{NH}_{3}\right)_{6}^{3+}\) ion. According to the valence bond description, how many unpaired electrons are there in the \(\mathrm{Cr}\left(\mathrm{NH}_{3}\right)_{6}^{3+}\) complex? How does this prediction compare with that of crystal field theory?

Short Answer

Expert verified
In Cr(NH3)6^3+ ion, as per valence bond theory, there will be no unpaired electrons. The prediction from valence bond is different from that of crystal field theory, which predicts that there are no unpaired electrons.

Step by step solution

01

Determine Chromium Electron Configuration

The atomic number of Chromium (Cr) is 24. So, its electron configuration is \([Ar] 3d^5 4s^1\). In the case of the \(\mathrm{Cr}\left(\mathrm{NH}_{3}\right)_{6}^{3+}\) ion, Chromium loses 3 electrons. This leads to an electron configuration of \([Ar] 3d^3\), which means it has 3 electrons in the 3d orbital.
02

Apply Valence Bond Theory

According to the valence bond theory, the presence of six NH3 ligands results in the hybridization of Cr. The hybridization is \(d^2sp^3\) since we have 6 ligands forming bonds with the Cr^3+ center. Three electrons in the d orbital get distributed among the hybrid orbitals, so one unpaired electron will be paired up with one ligand electron each, in three hybrid orbitals.
03

Determine the Number of Unpaired Electrons

As per the valence bond theory, unpaired electrons are the result of empty orbitals. Since three ligands each bonded with an electron in the hybrid orbitals, this leaves three hybrid orbitals with no unpaired electron.
04

Compare with Crystal Field Theory

If we apply crystal field theory instead, we would conclude there are no unpaired electrons. This theory suggests that strong field ligands like NH3 result in the pairing up of all the 3d electrons in lower t2g levels, leaving no unpaired electron.

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Most popular questions from this chapter

In both \(\left[\mathrm{Fe}\left(\mathrm{H}_{2} \mathrm{O}\right)_{6}\right]^{2+}\) and \(\left[\mathrm{Fe}(\mathrm{CN})_{6}\right]^{4-}\) ions, the iron is present as \(\mathrm{Fe}(\mathrm{II}) ;\) however, \(\left[\mathrm{Fe}\left(\mathrm{H}_{2} \mathrm{O}\right)_{6}\right]^{2+}\) is paramagnetic, whereas \(\left[\mathrm{Fe}(\mathrm{CN})_{6}\right]^{4-}\) is diamagnetic. Explain this difference.

Which of these general structures for a complex ion would you expect to exhibit cis and trans isomerism? Explain. (a) tetrahedral (b) square-planar (c) linear

Cyano complexes of transition metal ions (such as \(\mathrm{Fe}^{2+}\) and \(\mathrm{Cu}^{2+}\) ) are often yellow, whereas aqua complexes are often green or blue. Explain the basis for this difference in color.

Write simple chemical equations to show how the complex ion \(\left[\mathrm{CrOH}\left(\mathrm{H}_{2} \mathrm{O}\right)_{5}\right]^{2+}\) acts as \((\mathrm{a})\) an acid; (b) a base.

Estimate the total \(\left[\mathrm{Cl}^{-}\right]\) required in a solution that is initially \(0.10 \mathrm{M} \mathrm{CuSO}_{4}\) to produce a visible yellow color. \(\left[\mathrm{Cu}\left(\mathrm{H}_{2} \mathrm{O}\right)_{4}\right]^{2+}+4 \mathrm{Cl}^{-} \rightleftharpoons\left[\mathrm{CuCl}_{4}\right]^{2-}+4 \mathrm{H}_{2} \mathrm{O}\) \(K_{f}=4.2 \times 10^{5}\) Assume that \(99 \%\) conversion of \(\left[\mathrm{Cu}\left(\mathrm{H}_{2} \mathrm{O}\right)_{4}\right]^{2+}\) to \(\left[\mathrm{CuCl}_{4}\right]^{2-}\) is sufficient for this to happen, and ignore the presence of any mixed aqua- chloro complex ions.
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