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Chapter 9: Problem 83

Write Lewis structures for \(\mathrm{BrF}_{3}, \mathrm{ClF}_{5},\) and \(\mathrm{IF}_{7} .\) Identify those in which the octet rule is not obeyed.

Short Answer

Expert verified
All the given compounds, \(\mathrm{BrF}_{3}, \(\mathrm{ClF}_{5}\) and \(\mathrm{IF}_{7}\ neither obey the octet rule because their central atoms Bromine, Chlorine, and Iodine respectively possess more than eight valence electrons in their Lewis structures.

Step by step solution

01

Draw the Lewis Structure for \(\mathrm{BrF}_{3}\)

Start by determining the valence electrons for each atom. Bromine (Br) has 7 and each Fluorine (F) atom has 7, so in total there will be 28 valence electrons. First, place the least electronegative atom in the center, which here is Bromine (Br), then put three Fluorine (F) atoms around it. Connect each Fluorine to Bromine with a single bond, accounting for 6 valence electrons. Distribute the remaining 22 electrons to the Fluorine atoms, adding 6 to each Fluorine and placing the remaining 4 electrons on the Bromine atom. This structure shows Bromine having more than eight valence electrons and therefore not obeying the octet rule.
02

Draw the Lewis Structure for \(\mathrm{ClF}_{5}\)

Starting with Chlorine (Cl) with 7 valence electrons and each Fluorine (F) atom with 7 valence electrons, the total becomes 42 valence electrons. Place Chlorine in the center and surround it with 5 Fluorine atoms. Form single bonds between Chlorine and each Fluorine, accounting for ten valence electrons. Distribute the remaining 32 electrons to the Fluorine atoms, adding 6 to each, and place the remaining 2 electrons on the Chlorine atom. The structure shows Chlorine having more than 8 valence electrons which means it's not obeying the octet rule.
03

Draw the Lewis Structure for \(\mathrm{IF}_{7}\)

Start with Iodine (I) which has 7 valence electrons and each Fluorine (F) atom with 7 valence electrons. In total there will be 56 valence electrons. Put Iodine in the center and surround it with 7 Fluorine atoms. Form single bonds between Iodine and each Fluorine, accounting for 14 valence electrons. Distribute the remaining 42 electrons to the Fluorine atoms, adding 6 to each. This structure doesn't require any electrons to be placed on the Iodine atom. The structure shows Iodine having more than 8 valence electrons, meaning Iodine is not obeying the octet rule.

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Most popular questions from this chapter

Write Lewis structures for the following molecules: (a) \(\mathrm{BrF}_{3},\) (b) \(\mathrm{H}_{2} \mathrm{Te}\) (c) \(\mathrm{NH}_{2} \mathrm{OH}\) (d) \(\mathrm{POCl}_{3}\) (P is bonded to \(\mathrm{O}\) and \(\mathrm{Cl}\) atoms \(),\) (e) \(\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{~F}\), (f) \(\mathrm{NF}_{3}\), (g) \(\mathrm{CH}_{3} \mathrm{NH}_{2}\).

Hydrazine borane, \(\mathrm{NH}_{2} \mathrm{NH}_{2} \mathrm{BH}_{3}\), has been proposed as a hydrogen storage material. When reacted with lithium hydride (LiH), hydrogen gas is released: $$ \mathrm{NH}_{2} \mathrm{NH}_{2} \mathrm{BH}_{3}+\mathrm{LiH} \longrightarrow \mathrm{LiNH}_{2} \mathrm{NHBH}_{3}+\mathrm{H}_{2} $$ Write Lewis structures for \(\mathrm{NH}_{2} \mathrm{NH}_{2} \mathrm{BH}_{3}\) and \(\mathrm{NH}_{2} \mathrm{NHBH}_{3}^{-}\) and assign all formal charges.

Write Lewis structures for \(\mathrm{SeF}_{4}\) and \(\mathrm{SeF}_{6}\). Is the octet rule satisfied for Se?

In the gas phase, aluminum chloride exists as a dimer (a unit of two) with the formula \(\mathrm{Al}_{2} \mathrm{Cl}_{6}\). Its skeletal structure is given by Complete the Lewis structure and indicate the coordinate covalent bonds in the molecule.

Explain why the bond enthalpy of a molecule is usually defined in terms of a gas-phase reaction. Why are bond-breaking processes always endothermic and bond-forming processes always exothermic?
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