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Chapter 20: Problem 52

Draw the \(d\) -orbital splitting diagrams for the octahedral complex ions of each of the following. a. \(\mathrm{Zn}^{2+}\) b. \(\mathrm{Co}^{2+}\) (high and low spin) c. \(\mathrm{Ti}^{3+}\)

Short Answer

Expert verified
a. Zn^2+: t2g: 鈫戔啌鈫戔啌鈫戔啌 eg: 鈫戔啌鈫戔啌 b. Co^2+ low spin: t2g: 鈫戔啌鈫戔啌鈫� eg: empty c. Co^2+ high spin: t2g: 鈫戔啈鈫戔啈鈫� eg: empty d. Ti^3+: t2g: 鈫� empty empty eg: empty empty

Step by step solution

01

(Step 1: Understand Octahedral d-Orbital Splitting)

In an octahedral complex, the d orbitals split into two groups with different energy levels when surrounded by ligands. The energy difference between these groups is called the crystal field splitting energy (鈭�). The three lower-energy orbitals are referred to as t2g orbitals (dxy, dyz, and dxz), while the two higher-energy orbitals are eg orbitals (dx^2-y^2 and dz^2).
02

(Step 2: Determine the d-electron configuration for each ion)

To draw the d-orbital splitting diagrams, we need to know the number of d-electrons for each ion: a. Zn^2+ (Zn: atomic number 30) - [Ar] 4s^2 3d^10 -> Zn^2+: [Ar] 3d^10 - 10 d electrons b. Co^2+ (Co: atomic number 27) - [Ar] 4s^2 3d^7 -> Co^2+: [Ar] 3d^5 - 5 d electrons c. Ti^3+ (Ti: atomic number 22) - [Ar] 4s^2 3d^2 -> Ti^3+: [Ar] 3d^1 - 1 d electron
03

(Step 3: Draw the d-orbital splitting diagram for Zn^2+)

Since the Zn^2+ ion has 10 d electrons, all of the d orbitals are fully occupied. The diagram will show all the t2g and eg orbitals filled with two electrons each: t2g: 鈫戔啌鈫戔啌鈫戔啌 eg: 鈫戔啌鈫戔啌
04

(Step 4: Draw the d-orbital splitting diagram for Co^2+ high spin and low spin)

Since the Co^2+ ion has 5 d electrons, the electrons can either occupy the t2g orbitals (low spin) or distribute themselves among both t2g and eg orbitals (high spin), depending on the strength of the crystal field. For Co^2+ low spin: t2g: 鈫� 鈫� 鈫� 鈫� 鈫� eg: empty For Co^2+ high spin: t2g: 鈫� 鈫� 鈫� 鈫� 鈫� eg: empty
05

(Step 5: Draw the d-orbital splitting diagram for Ti^3+)

Since the Ti^3+ ion has only 1 d electron, it will occupy the lowest energy t2g orbital. The diagram will show the following electron distribution: t2g: 鈫� empty empty eg: empty empty

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Most popular questions from this chapter

\(\mathrm{CoCl}_{4}^{2-}\) forms a tetrahedral complex ion and \(\mathrm{Co}(\mathrm{CN})_{6}^{3-}\) forms an octahedral complex ion. What is wrong about the following statements concerning each complex ion and the \(d\) orbital splitting diagrams? a. \(\mathrm{CoCl}_{4}^{2-}\) is an example of a strong-field case having two unpaired electrons. b. Because \(\mathrm{CN}^{-}\) is a weak-field ligand, \(\mathrm{Co}(\mathrm{CN})_{6}^{3-}\) will be a low-spin case having four unpaired electrons.

The following statements discuss some coordination compounds. For each coordination compound, give the complex ion and the counterions, the electron configuration of the transition metal, and the geometry of the complex ion. a. \(\mathrm{CoCl}_{2} \cdot 6 \mathrm{H}_{2} \mathrm{O}\) is a compound used in novelty devices that predict rain. b. During the developing process of black-and-white film, silver bromide is removed from photographic film by the fixer. The major component of the fixer is sodium thiosulfate. The equation for the reaction is: \(\begin{aligned} \operatorname{AgBr}(s)+2 \mathrm{Na}_{2} \mathrm{S}_{2} \mathrm{O}_{3}(a q) & \longrightarrow \\\ \mathrm{Na}_{3}\left[\mathrm{Ag}\left(\mathrm{S}_{2} \mathrm{O}_{3}\right)_{2}\right](a q)+& \mathrm{NaBr}(a q) \end{aligned}\) c. In the production of printed circuit boards for the electronics industry, a thin layer of copper is laminated onto an insulating plastic board. Next, a circuit pattern made of a chemically resistant polymer is printed on the board. The unwanted copper is removed by chemical etching, and the protective polymer is finally removed by solvents. One etching reaction is: \(\mathrm{Cu}\left(\mathrm{NH}_{3}\right)_{4} \mathrm{Cl}_{2}(a q)+4 \mathrm{NH}_{3}(a q)+\mathrm{Cu}(s) \longrightarrow\) \(2 \mathrm{Cu}\left(\mathrm{NH}_{3}\right)_{4} \mathrm{Cl}(a q)\)

The complex ion \(\mathrm{Cu}\left(\mathrm{H}_{2} \mathrm{O}\right)_{6}^{2+}\) has an absorption maximum at around \(800 \mathrm{nm} .\) When four ammonias replace water, \(\mathrm{Cu}\left(\mathrm{NH}_{3}\right)_{4}\left(\mathrm{H}_{2} \mathrm{O}\right)_{2}^{2+},\) the absorption maximum shifts to around \(600 \mathrm{nm} .\) What do these results signify in terms of the relative field splittings of \(\mathrm{NH}_{3}\) and \(\mathrm{H}_{2} \mathrm{O}\) ? Explain.

A coordination compound of cobalt(III) contains four ammonia molecules, one sulfate ion, and one chloride ion. Addition of aqueous \(\mathrm{BaCl}_{2}\) solution to an aqueous solution of the compound gives no precipitate. Addition of aqueous \(\mathrm{AgNO}_{3}\) to an aqueous solution of the compound produces a white precipitate. Propose a structure for this coordination compound.

How many unpaired electrons are present in the tetrahedral ion \(\mathrm{FeCl}_{4}^{-} ?\)
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