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Chapter 8: Problem 64

Draw the Lewis structures for each of the following molecules or ions. Identify instances where the octet rule is not obeyed; state which atom in each compound does not follow the octet rule; and state how many electrons surround these atoms: (a) \(\mathrm{NO},(\mathbf{b}) \mathrm{BF}_{3},(\mathbf{c}) \mathrm{ICl}_{2}^{-},(\mathbf{d}) \mathrm{OPBr}_{3}(\) the \(\mathrm{P}\) is the central atom), (e) XeF.

Short Answer

Expert verified
In summary: - NO does not obey the octet rule as nitrogen is surrounded by six electrons. - BF3 does not obey the octet rule as boron is surrounded by six electrons. - ICl2- does not obey the octet rule as iodine is surrounded by ten electrons. - OPBr3 does not obey the octet rule as phosphorus is surrounded by ten electrons. - XeF does not obey the octet rule as xenon is surrounded by ten electrons.

Step by step solution

01

(a) Drawing Lewis Structure for NO

To draw the Lewis structure for the NO molecule, find the total number of valence electrons. Nitrogen has five valence electrons, and oxygen has six. The total count is 11 valence electrons. Next, place nitrogen and oxygen atoms next to each other since nitrogen has fewer lone pairs. Connect them with a single bond and distribute remaining electrons as lone pairs. NO Lewis Structure: O=N:\( \,\,\,\) The nitrogen atom is surrounded by six electrons, violating the octet rule. Therefore, NO does not obey the octet rule.
02

(b) Drawing Lewis Structure for BF3

Boron has three valence electrons, and each fluorine atom has seven valence electrons. Overall, there are 24 valence electrons. Place boron at the center and surround it with three fluorine atoms. Connect boron and each fluorine atom with single bonds and distribute the remaining electrons. BF3 Lewis Structure: F | B - F | F The boron atom is surrounded by six electrons, violating the octet rule. Therefore, BF3 does not obey the octet rule.
03

(c) Drawing Lewis Structure for ICl2-

Iodine has seven valence electrons, chlorine has seven valence electrons, and there is one extra electron from the negative charge, so the total count is 22 valence electrons. Place iodine at the center and surround it with two chlorine atoms. Connect iodine and each chlorine atom with single bonds and distribute the remaining electrons as lone pairs. ICl2- Lewis Structure: Cl | I - Cl ｜ - The iodine atom is surrounded by ten electrons, violating the octet rule. Therefore, ICl2- does not obey the octet rule.
04

(d) Drawing Lewis Structure for OPBr3

Oxygen has six valence electrons, phosphorus has five valence electrons, and each bromine atom has seven valence electrons. Overall, there are 32 valence electrons. Place phosphorus at the center, with oxygen above and three bromine atoms surrounding it. Connect each atom with single bonds and distribute the remaining electrons as lone pairs. OPBr3 Lewis Structure: O ｜ Br - P - Br ｜ Br The phosphorus atom is surrounded by ten electrons, violating the octet rule. Therefore, OPBr3 does not obey the octet rule.
05

(e) Drawing Lewis Structure for XeF

Xenon has eight valence electrons, and fluorine has seven valence electrons. The total count is 15 valence electrons. Place xenon at the center and connect it to the fluorine atom with single bonds and distribute the remaining electrons as lone pairs. XeF Lewis Structure: F ｜ Xe The xenon atom is surrounded by ten electrons, violating the octet rule. Therefore, XeF does not obey the octet rule.

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Most popular questions from this chapter

Using only the periodic table as your guide, select the most electronegative atom in each of the following sets: (a) Na, \(\mathrm{Mg}, \mathrm{K}, \mathrm{Ca} ;(\mathbf{b}) \mathrm{P}, \mathrm{S},\) As, \(\mathrm{Se} ;(\mathbf{c}) \mathrm{Be}, \mathrm{B}, \mathrm{C}, \mathrm{Si} ;(\mathbf{d}) \mathrm{Zn}, \mathrm{Ge}, \mathrm{Ga}, \mathrm{Gs}\)

The substance chlorine monoxide, ClO(g), is important in atmospheric processes that lead to depletion of the ozone layer. The ClO molecule has an experimental dipole moment of \(1.24 \mathrm{D},\) and the \(\mathrm{Cl}-\) O bond length is 1.60 \(\mathrm{A}\) . (a) Determine the magnitude of the charges on the Cl and O atoms in units of the electronic charge, \(e\) (b) Based on the electronegativities of the elements, which atom would you expect to have a partial negative charge in the Clo molecule? (c) Using formal charges as a guide, propose the dominant Lewis structure for the molecule. (d) The anion \(\mathrm{ClO}^{-}\) exists. What is the formal charge on the Cl for the best Lewis structure for \(\mathrm{ClO}^{-}\) ?

There are many Lewis structures you could draw for sulfuric acid, \(\mathrm{H}_{2} \mathrm{SO}_{4}\) (each \(\mathrm{H}\) is bonded to an O). (a) What Lewis structure(s) would you draw to satisfy the octet rule? (b) What Lewis structure(s) would you draw to minimize formal charge?

(a) Draw the Lewis structure for hydrogen peroxide, \(\mathrm{H}_{2} \mathrm{O}_{2}\) . (b) What is the weakest bond in hydrogen peroxide? (c) Hydrogen peroxide is sold commercially as an aqueous solution in brown bottles to protect it from light. Calculate the longest wavelength of light that has sufficient energy to break the weakest bond in hydrogen peroxide.

By referring only to the periodic table, select (a) the most electronegative element in group \(6 \mathrm{A} ;(\mathbf{b})\) the least electronegative element in the group Al, Si, P; (c) the most electronegative element in the group \(\mathrm{Ga}, \mathrm{P}, \mathrm{Cl}, \mathrm{Na} ;(\mathbf{d})\) the element in the group \(\mathrm{K}\) \(\mathrm{C}, \mathrm{Zn}, \mathrm{F}\) that is most likely to form an ionic compound with Ba.
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