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Chapter 10: Problem 6

Which of the following have Lewis structures that \(d o\) not obey the octet rule: \(\mathrm{NF}_{3}, \mathrm{AlCl}_{3}, \mathrm{SiF}_{6}^{2-}, \mathrm{SO}_{3}, \mathrm{PH}_{4}^{+}\) \(\mathrm{PO}_{4}^{3-}, \mathrm{ClO}_{2} ?\)

Short Answer

Expert verified
The molecules that do not obey the octet rule are AlCl3, SiF6^2-, and PH4^+.

Step by step solution

01

Draw Lewis Structures for Each Molecule

For each molecule, determine the total number of valence electrons by summing up the group numbers of each constituent atom. Draw a hypothetical structure by first sketching a skeletal structure, then filling up the valence shells of outer atoms, and finally allocating any remaining electrons to the central atoms. Remember to add or subtract electrons for ionic compounds to account for their charges.
02

Check for Octet Rule

For each molecule, check if the central atom obeys the octet rule by ensuring it is surrounded by eight electrons (including shared ones). For atoms in the second period of the periodic table (e.g., Nitrogen in NF3), there cannot be more than eight electrons because the s and p subshells can hold a maximum of eight electrons only. For atoms in the third period and beyond (e.g., Aluminum in AlCl3, Silicon in SiF6^2-, Sulfur in SO3, Phosphorus in PH4^+, PO4^3-), there can be more than eight electrons because they can use their vacant d-orbitals for expanded octets.
03

Identify Exceptions

For NF3, SO3 and PO4^3-, each central atom (Nitrogen in NF3, Sulfur in SO3, Phosphorus in PO4^3-) follows the octet rule. For AlCl3, the central atom Aluminum is surrounded by only 6 electrons. For SiF6^2-, the central atom Silicon is surrounded by 12 electrons. For PH4^+, the central atom Phosphorus is surrounded by 8 electrons, but as a ion it is surrounded by 10 electrons. For ClO2, the central atom Chlorine can be surrounded by more than 8 electrons as it forms an expanded octet. Therefore, the molecules that do not obey the octet rule are AlCl3, SiF6^2-, and PH4^+.

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Most popular questions from this chapter

Use the \(\mathrm{NH}_{3}\) molecule as an example to explain the difference between molecular geometry and electron-group geometry.

Acetone \(\left(\mathrm{CH}_{3}\right) \mathrm{C}=\mathrm{O},\) a ketone, will react with a strong base \((\mathrm{A})\) to produce the enolate anion, \(\mathrm{CH}_{3}(\mathrm{C}=\mathrm{O}) \mathrm{CH}_{2}^{-} .\) Draw the Lewis structure of the enolate anion, and describe the relative contributions of any resonance structures.

Use the VSEPR theory to predict the shape of (a) the molecule OSF \(_{2} ;\) (b) the molecule \(\mathrm{O}_{2} \mathrm{SF}_{2} ;\) (c) the ion \(\mathrm{SF}_{5}^{-} ;\) (d) the ion \(\mathrm{ClO}_{4}^{-} ;\) (e) the ion \(\mathrm{ClO}_{3}^{-}\).

Hydrogen azide, \(\mathrm{HN}_{3}\), is a liquid that explodes violently when subjected to physical shock. In the \(\mathrm{HN}_{3}\) molecule, one nitrogen- to-nitrogen bond length is \(113 \mathrm{pm},\) and the other is \(124 \mathrm{pm} .\) The \(\mathrm{H}-\mathrm{N}-\mathrm{N}\) bond angle is \(112^{\circ} .\) Draw Lewis structures and a sketch of the molecule consistent with these facts.

Use VSEPR theory to predict the geometric shapes of the following molecules and ions: (a) \(\mathrm{PCl}_{3} ;\) (b) \(\mathrm{SO}_{4}^{2-}\); (c) \(\mathrm{SOCl}_{2} ;\) (d) \(\mathrm{SO}_{3} ;\) (e) \(\mathrm{BrF}_{4}^{+}\).
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