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Chapter 19: Problem 60

Describe the bonding in \(\mathrm{SO}_{2}\) and \(\mathrm{SO}_{3}\) using the localized electron model (hybrid orbital theory). How would the molecular orbital model describe the \(\pi\) bonding in these two compounds?

Short Answer

Expert verified
In both SO鈧� and SO鈧�, the sulfur atom is sp虏 hybridized, and 蟺 bonding occurs through the axial overlap of p orbitals. In SO鈧�, the bond order between sulfur and each oxygen is 1.5, while in SO鈧�, the bond order is 4/3. Both the localized electron model (hybrid orbital theory) and the molecular orbital model describe the 蟺 bonding consistently in these compounds.

Step by step solution

01

Analyzing SO鈧� using hybrid orbital theory

To describe the bonding in SO鈧� using hybrid orbital theory, we first need to determine the hybridization of the central sulfur atom. Sulfur has 6 valence electrons, and will form 2 single bonds with 2 oxygen atoms. The lone pair of electrons on sulfur participates in resonance, and thus, the hybridization of sulfur will be sp虏.
02

Describing 蟺 bonding in SO鈧� using hybrid orbital theory

In SO鈧�, the 蟺 bonding between sulfur (sp虏 hybridized) and the oxygen atom (also sp虏 hybridized) takes place by the axial overlap between the two p orbitals that are perpendicular to the plane formed by the three atoms (sulfur and two oxygens). The 蟺 electron cloud is distributed above and below this plane, forming a 蟺 bond. Due to resonance taking place, the bond order between sulfur and each oxygen atom is 1.5.
03

Analyzing SO鈧� using hybrid orbital theory

To describe the bonding in SO鈧� using hybrid orbital theory, we first need to determine the hybridization of the central sulfur atom. Sulfur has 6 valence electrons and will form 3 single bonds with 3 oxygen atoms. The hybridization of sulfur will be sp虏.
04

Describing 蟺 bonding in SO鈧� using hybrid orbital theory

In SO鈧�, the 蟺 bonding between the central sulfur atom (sp虏 hybridized) and each of the oxygen atoms (also sp虏 hybridized) occurs by the axial overlap between their respective p orbitals, which are perpendicular to the plane formed by the four atoms (sulfur and three oxygens). The 蟺 electron cloud is distributed above and below this plane, forming 蟺 bonds. Due to resonance taking place among the sulfur-oxygen bonds, each sulfur-oxygen bond has an equal bond order of 4/3.
05

Describing 蟺 bonding in SO鈧� using molecular orbital theory

According to the molecular orbital model, the 蟺 bonding in SO鈧� results from the combination of three 2p orbitals鈥攐ne from the central sulfur atom and two from the oxygen atoms. This combination creates three molecular orbitals: one bonding 蟺 orbital and two antibonding 蟺* orbitals. Because there are two of sulfur's electrons participating in the 蟺 bonding, the bonding order for each sulfur-oxygen bond in SO鈧� is 1.5.
06

Describing 蟺 bonding in SO鈧� using molecular orbital theory

In molecular orbital theory, the 蟺 bonding in SO鈧� results from the combination of four 2p orbitals鈥攐ne from the central sulfur atom and three from the oxygen atoms. This combination creates four molecular orbitals: two bonding 蟺 orbitals and two antibonding 蟺* orbitals. With four electrons taking part in the 蟺 bonding, the bond order for each sulfur-oxygen bond in SO鈧� is 4/3. In conclusion, both the localized electron model (hybrid orbital theory) and the molecular orbital model describe the 蟺 bonding in SO鈧� and SO鈧� in a consistent way, with a bond order of 1.5 in SO鈧� and 4/3 in SO鈧�.

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Most popular questions from this chapter

The Group \(5 \mathrm{A}\) (15) elements can form molecules or ions that involve three, five, or six covalent bonds; \(\mathrm{NH}_{3}, \mathrm{AsCl}_{5},\) and \(\mathrm{PF}_{6}^{-}\) are examples. Draw the Lewis structure for each of these substances, and predict the molecular structure and hybridization for each. Why doesn't \(\mathrm{NF}_{5}\) or \(\mathrm{NCl}_{6}^{-}\) form?

Give the Lewis structure, molecular structure, and hybridization of the oxygen atom for OF \(_{2}\). Would you expect \(\mathrm{OF}_{2}\) to be a strong oxidizing agent like \(\mathrm{O}_{2} \mathrm{F}_{2}\) discussed in Exercise \(61 ?\)

Write balanced equations describing the reaction of lithium metal with each of the following: \(\mathrm{O}_{2}, \mathrm{S}, \mathrm{Cl}_{2}, \mathrm{P}_{4}, \mathrm{H}_{2}, \mathrm{H}_{2} \mathrm{O},\) and HCl.

While selenic acid has the formula \(\mathrm{H}_{2} \mathrm{SeO}_{4}\) and thus is directly related to sulfuric acid, telluric acid is best visualized as \(\mathrm{H}_{6} \mathrm{TeO}_{6}\) or \(\mathrm{Te}(\mathrm{OH})_{6}\) a. What is the oxidation state of tellurium in \(\operatorname{Te}(\mathrm{OH})_{6} ?\) b. Despite its structural differences with sulfuric and selenic acid, telluric acid is a diprotic acid with \(\mathrm{p} K_{\mathrm{a}_{1}}=7.68\) and \(\mathrm{p} K_{\mathrm{a}_{2}}=11.29 .\) Telluric acid can be prepared by hydrolysis of tellurium hexafluoride according to the equation $$\operatorname{TeF}_{6}(g)+6 \mathrm{H}_{2} \mathrm{O}(l) \longrightarrow\operatorname{Te}(\mathrm{OH})_{6}(a q)+6 \mathrm{HF}(a q)$$ Tellurium hexafluoride can be prepared by the reaction of elemental tellurium with fluorine gas:$$\operatorname{Te}(s)+3 \mathrm{F}_{2}(g) \longrightarrow \operatorname{Te} \mathrm{F}_{6}(g)$$.If a cubic block of tellurium (density \(=6.240 \mathrm{g} / \mathrm{cm}^{3}\) ) measuring \(0.545 \mathrm{cm}\) on edge is allowed to react with 2.34 L fluorine gas at 1.06 atm and \(25^{\circ} \mathrm{C}\), what is the \(\mathrm{pH}\) of a solution of \(\mathrm{Te}(\mathrm{OH})_{6}\) formed by dissolving the isolated \(\operatorname{Te} \mathrm{F}_{6}(g)\) in \(115 \mathrm{mL}\) solution? Assume \(100 \%\) yield in all reactions.

The oxyanion of nitrogen in which it has the highest oxidation state is the nitrate ion \(\left(\mathrm{NO}_{3}^{-}\right) .\) The corresponding oxyanion of phosphorus is \(\mathrm{PO}_{4}^{3-} .\) The \(\mathrm{NO}_{4}^{3-}\) ion is known but not very stable. The \(\mathrm{PO}_{3}^{-}\) ion is not known. Account for these differences in terms of the bonding in the four anions.
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