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Chapter 22: Problem 125

All of the following have a tetrahedral shape except (a) \(\mathrm{SO}_{4}^{2-} ;\) (b) \(\mathrm{XeF}_{4} ;\) (c) \(\mathrm{CCl}_{4} ;\) (d) \(\mathrm{XeO}_{4} ;\) (e) \(\mathrm{NH}_{4}^{+}\)..

Short Answer

Expert verified
All the given options except (b) \( \mathrm{XeF}_{4} \) have a tetrahedral shape.

Step by step solution

01

Analysis of SO4-2

The sulfur atom in \( \mathrm{SO}_{4}^{2-} \) has six valence electrons. Each of the four oxygen atoms contributes two more for bonding, making a total of 14 valence electrons. Two electrons will go to fill the S-O bonding orbitals and the remaining ten electrons fill the non-bonding orbitals. Thus, there are four S-O bonding pairs and no lone pairs on the center sulfur atom, which is indicative of a tetrahedral shape.
02

Analysis of XeF4

Xenon (Xe) has eight valence electrons and each of the four fluorine (F) atoms contributes one for bonding. This makes a total of 12 electrons. Two electrons will go to fill the Xe-F bonding orbitals and the remaining electrons fill the lone pairs. This results in four XeF bonding pairs and 2 lone pairs on the central atom, falling in the octahedral category, but due to the existence of lone pairs, it forms a square planar structure, not a tetrahedral.
03

Analysis of CCl4

Carbon (C) in \( \mathrm{CCl}_{4} \) has four valence electrons, and each of the four chlorine (Cl) atoms contributes one more for bonding, making a total of eight electrons. These electrons will fill the C-Cl bonding orbitals. Therefore, there are four C-Cl bonding pairs and no lone pairs on carbon, which matches the criteria for a tetrahedral shape.
04

Analysis of XeO4

Xenon (Xe) also has eight valence electrons, and each of the four oxygen (O) atoms contributes two more for bonding. This forms a total of 16 valence electrons. These electrons will fill the Xe-O bonding orbitals. Therefore, there are four Xe-O bonding pairs and no lone pairs on Xenon, fitting the criteria for a tetrahedral shape.
05

Analysis of NH4+

Nitrogen (N) in \( \mathrm{NH}_{4}^{+} \) has five valence electrons, and each of the four hydrogen (H) atoms contributes one more for bonding, but one electron is lost due to the +1 charge, making a total of eight electrons. These electrons will fill the N-H bonding orbitals. Hence, there are four N-H bonding pairs and no lone pair on Nitrogen, which also matches the criteria for a tetrahedral shape.

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Most popular questions from this chapter

The amide anion \(\mathrm{NH}_{2}^{-}\) is a very strong base. On the basis of molecular orbital theory, would you expect \(\mathrm{NH}_{2}^{-}\) to be linear or bent?

Show how you would use elemental sulfur, chlorine gas, metallic sodium, water, and air to produce aqueous solutions containing (a) \(\mathrm{Na}_{2} \mathrm{SO}_{3} ;\) (b) \(\mathrm{Na}_{2} \mathrm{SO}_{4}\) (c) \(\mathrm{Na}_{2} \mathrm{S}_{2} \mathrm{O}_{3} .\) [Hint: You will have to use information from other chapters as well as this one.].

Predict the geometric structures of (a) \(\mathrm{BrF}_{3} ;\) (b) IF \(_{5}\); (c) \(\mathrm{Cl}_{3} \mathrm{IF}^{-}\). (Central atom underlined.).

One reaction for the production of adipic acid, \(\mathrm{HOOC}\left(\mathrm{CH}_{2}\right)_{4} \mathrm{COOH},\) used in the manufacture of nylon, involves the oxidation of cyclohexanone, \(\mathrm{C}_{6} \mathrm{H}_{10} \mathrm{O},\) in a nitric acid solution. Assume that dinitrogen monoxide is also formed, and write a balanced equation for this reaction.

Write chemical equations for the following reactions: (a) the displacement of \(\mathrm{H}_{2}(\mathrm{g})\) from \(\mathrm{HCl}(\mathrm{aq})\) by \(\mathrm{Al}(\mathrm{s})\) (b) the re-forming of propane gas \(\left(\mathrm{C}_{3} \mathrm{H}_{8}\right)\) with steam (c) the reduction of \(\mathrm{MnO}_{2}(\mathrm{s})\) to \(\mathrm{Mn}(\mathrm{s})\) with \(\mathrm{H}_{2}(\mathrm{g})\)
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