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Chapter 23: Problem 63

Draw the crystal-field energy-level diagrams and show the placement of \(d\) electrons for each of the following: (a) \(\left[\mathrm{Cr}\left(\mathrm{H}_{2} \mathrm{O}\right)_{6}\right]^{2+}\) (four unpaired electrons), \((\mathbf{b})\left[\operatorname{Mn}\left(\mathrm{H}_{2} \mathrm{O}\right)_{6}\right]^{2+}\) (a high-spin complex), (c) \(\left[\mathrm{Ru}\left(\mathrm{NH}_{3}\right)_{5}\left(\mathrm{H}_{2} \mathrm{O}\right)\right]^{2+}\) (a low-spin complex) \((\mathbf{d})\left[\operatorname{Ir} \mathrm{Cl}_{6}\right]^{2-}\) (a low-spin complex) \((\mathbf{e})\left[\mathrm{Cr}(\mathrm{en})_{3}\right]^{3+}\) \((\mathbf{f})\left[\mathrm{NiF}_{6}\right]^{4-}.\)

Short Answer

Expert verified
(a) For [Cr(H2O)6]2+ (high-spin): t2g: _↑_ _↑_ _↑_ eg: _↑_ _↓_ Four unpaired electrons. (b) For [Mn(H2O)6]2+ (high-spin): t2g: _↑_ _↑_ _↑_ eg: _↑_ _↑_ Five unpaired electrons. (c) For [Ru(NH3)5(H2O)]2+ (low-spin): t2g: _↑↓_ _↑↓_ _↑↓_ eg: _↓_ _↓_ No unpaired electrons. (d) For [IrCl6]2- (low-spin): t2g: _↑↓_ _↑↓_ _↑↓_ eg: _↑_ _↑↓_ One unpaired electron. (e) For [Cr(en)3]3+ (low-spin): t2g: _↑↓_ _↑↓_ _↑_ eg: _↓_ _↓_ One unpaired electron. (f) For [NiF6]4- (high-spin): t2g: _↑_ _↑_ _↑_ eg: _↑_ _↑_ Four unpaired electrons.

Step by step solution

01

(a) Cr(H2O)6 2+

: For the complex [Cr(H2O)6]2+, the central metal ion is Cr2+. The electronic configuration of Cr2+ is [Ar] 3d4. As the ligand is H2O, a weak field ligand, the complex will be high-spin, hence the filling of d electrons will be according to the increasing energy levels. The crystal field energy level diagram would look like this: t2g: _↑_ _↑_ _↑_ eg: _↑_ _↓_ Thus, we can see there are four unpaired electrons.
02

(b) Mn(H2O)6 2+

: Similarly, for the complex [Mn(H2O)6]2+, the central metal ion is Mn2+. The electronic configuration of Mn2+ is [Ar] 3d5. As H2O is a weak field ligand, the complex is high-spin. The crystal field energy level diagram would look like this: t2g: _↑_ _↑_ _↑_ eg: _↑_ _↑_ There are five unpaired electrons.
03

(c) Ru(NH3)5(H2O) 2+

: For the complex [Ru(NH3)5(H2O)]2+, the central metal ion is Ru2+. The electronic configuration of Ru2+ is [Kr] 4d6. As NH3 is a strong field ligand, the complex is low-spin. The crystal field energy level diagram would look like this: t2g: _↑↓_ _↑↓_ _↑↓_ eg: _↓_ _↓_ There are no unpaired electrons.
04

(d) IrCl6 2-

: For the complex [IrCl6]2-, the central metal ion is Ir2-. The electronic configuration of Ir2- is [Xe] 5d9. Since Cl- is a weak field ligand, the complex is low-spin. The crystal field energy level diagram would look like this: t2g: _↑↓_ _↑↓_ _↑↓_ eg: _↑_ _↑↓_ There is just one unpaired electron.
05

(e) Cr(en)3 3+

: For the complex [Cr(en)3]3+, the central metal ion is Cr3+. The electronic configuration of Cr3+ is [Ar] 3d3. Since en is a strong field ligand, the complex is low-spin. The crystal field energy level diagram would look like this: t2g: _↑↓_ _↑↓_ _↑_ eg: _↓_ _↓_ There is only one unpaired electron.
06

(f) NiF6 4-

: For the complex [NiF6]4-, the central metal ion is Ni4-. The electronic configuration of Ni4- is [Ar] 3d6. Since F- is a weak field ligand, the complex is high-spin. The crystal field energy level diagram would look like this: t2g: _↑_ _↑_ _↑_ eg: _↑_ _↑_ There are four unpaired electrons.

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Most popular questions from this chapter

For each of the following pairs, identify the molecule or ion that is more likely to act as a ligand in a metal complex: (a) acetonitrile \(\left(\mathrm{CH}_{3} \mathrm{CN}\right)\) or ammonium \(\left(\mathrm{NH}_{4}^{+}\right)\) (b) hydride \(\left(\mathrm{H}^{-}\right)\) or hydronium \(\left(\mathrm{H}_{3} \mathrm{O}^{+}\right),(\mathbf{c})\) carbon monoxide \((\mathrm{CO})\) or methane \(\left(\mathrm{CH}_{4}\right)\) .

Solutions of \(\left[\mathrm{Co}\left(\mathrm{NH}_{3}\right)_{6}\right]^{2+},\left[\mathrm{Co}\left(\mathrm{H}_{2} \mathrm{O}\right)_{6}\right]^{2+}(\) both octahedral \()\) and \(\left[\mathrm{CoCl}_{4}\right]^{2-}\) (tetrahedral) are colored. One is pink, one is blue, and one is yellow. Based on the spectrochemical series and remembering that the energy splitting in tetrahedral complexes is normally much less that that in octahedral ones, assign a color to each complex.

Determine if each of the following complexes exhibits geometric isomerism. If geometric isomers exist, determine how many there are. (a) tetrahedral \(\left[\operatorname{Cd}\left(\mathrm{H}_{2} \mathrm{O}\right)_{2} \mathrm{Cl}_{2}\right],(\mathbf{b})\) square-planar \(\left[\operatorname{IrCl}_{2}\left(\mathrm{PH}_{3}\right)_{2}\right]^{-},(\mathbf{c})\) octahedral \(\left[\mathrm{Fe}(o-\mathrm{phen})_{2} \mathrm{Cl}_{2}\right]^{+}.\)

For a given metal ion and set of ligands, is the crystal-field splitting energy larger for a tetrahedral or an octahedral geometry?

(a) A complex absorbs photons with an energy of 4.51 \(\times 10^{-19} \mathrm{J}\) . What is the wavelength of these photons? (b) If this is the only place in the visible spectrum where the complex absorbs light, what color would you expect the complex to be?
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