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Draw the Lewis structures for each of the following molecules or ions. Which do not obey the octet rule? (a) \(\mathrm{NO},\) (b) \(\mathrm{BF}_{3}\), (c) \(\mathrm{ICl}_{2}^{-}\) (d) \(\mathrm{OPBr}_{3}\) (the \(\mathrm{P}\) is the central atom), (e) XeF 4 .

Short Answer

Expert verified
The Lewis structures for the given molecules or ions are as follows: (a) NO: ``` O //\\ N - O ``` (b) BF3: ``` F | F-B-F ``` (c) ICl2-: ``` Cl-I-Cl // \\ ``` (d) OPBr3: ``` O \ Br-P-Br | Br ``` (e) XeF4: ``` F F | | Xe | | F F ``` NO and BF3 do not obey the octet rule, while ICl2-, OPBr3, and XeF4 do.

Step by step solution

01

Calculating total valence electrons

For each molecule or ion, we calculate the total number of valence electrons by adding up the valence electrons for each atom. (a) NO: N (5) + O (6) = 11 valence electrons (b) BF3: B (3) + 3F (7) = 3 + 21 = 24 valence electrons (c) ICl2-: I (7) + 2Cl (7) + 1 extra electron (for negative charge) = 7 + 14 + 1 = 22 valence electrons (d) OPBr3: O (6) + P (5) + 3Br (7) = 6 + 5 + 21 = 32 valence electrons (e) XeF4: Xe (8) + 4F (7) = 8 + 28 = 36 valence electrons
02

Drawing Lewis structures

For each molecule or ion, we arrange the atoms and electrons so that each atom satisfies the octet rule, with a few exceptions. (a) NO: ``` O //\\ N - O ``` (b) BF3: ``` F | F-B-F ``` (c) ICl2-: ``` Cl-I-Cl // \\ ``` (d) OPBr3: ``` O \ Br-P-Br | Br ``` (e) XeF4: ``` F F | | Xe | | F F ```
03

Check for octet rule compliance

Now, we will determine if the molecules or ions obey the octet rule. (a) NO: Nitrogen has only 7 electrons, so it does not obey the octet rule. (b) BF3: Boron has only 6 electrons, so it does not obey the octet rule. (c) ICl2-: This ion obeys the octet rule, as all atoms have eight valence electrons. (d) OPBr3: This molecule obeys the octet rule, as all atoms have eight valence electrons. (e) XeF4: This molecule obeys the octet rule, as all atoms have eight valence electrons. In conclusion, NO and BF3 do not obey the octet rule, while ICl2-, OPBr3, and XeF4 do.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Octet Rule
Understanding the octet rule is essential for drawing Lewis structures. The octet rule is based on the observation that atoms of main-group elements tend to bond in such a way that each atom has eight electrons in its valence shell, giving it the same electron configuration as a noble gas. This rule explains why elements form the number of bonds that they do. For example, in the exercise, \textbf{NO} and \textbf{BF\(_{3}\)} do not obey the octet rule since nitrogen in NO ends up with seven electrons, one shy of an octet, and boron in BF\(_{3}\) has only six. However, it is noteworthy that there are exceptions to the octet rule, which include molecules with an odd number of electrons, molecules in which one or more atoms possess more or fewer than eight electrons, and molecules with atoms that have expanded valence shells. Boron trifluoride is such an exception, where boron is stable with six valence electrons.
Valence Electrons
Valence electrons play the central role in chemical bonding and molecule formation. They are the electrons in the outermost shell of an atom and are involved in forming bonds to adjacent atoms. In the provided exercise, counting valence electrons is the first step and key to correctly drawing the Lewis structures. For instance, nitrogen has five valence electrons while oxygen has six, making a total of 11 for \textbf{NO}. Similarly, boron has three and fluorine has seven, totaling 24 for \textbf{BF\(_{3}\)}. The number of valence electrons determines how atoms bond and what molecular structure they form. In ions, it's important to account for gained or lost electrons as well, such as with the \textbf{ICl\(_{2}^{-}\)} ion, which includes an extra electron due to its negative charge.
Molecular Geometry
Molecular geometry refers to the three-dimensional arrangement of atoms within a molecule. Understanding the geometry is crucial for predicting both the physical and chemical properties of the molecule. Lewis structures can be used to predict molecular geometry by applying VSEPR (Valence Shell Electron Pair Repulsion) theory, which says that electron pairs around a central atom will assume a geometry that minimizes the repulsion between them. In terms of the exercise, \textbf{BF\(_{3}\)} is a trigonal planar molecule since it has three bonding pairs and no lone pairs, whereas \textbf{XeF\(_{4}\)} forms a square planar structure due to the four bonding pairs and two lone pairs on xenon. When predicting molecular geometries, it's essential to consider both the number of bonding pairs and lone pairs on the central atom.

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Most popular questions from this chapter

(a) Use the concept of resonance to explain why all six \(\mathrm{C}-\mathrm{C}\) bonds in benzene are equal in length. (b) Are the \(\mathrm{C}-\mathrm{C}\) bond lengths in benzene shorter than \(\mathrm{C}-\mathrm{C}\) single bonds? Are they shorter than \(\mathrm{C}=\mathrm{C}\) double bonds?

Which of the following bonds are polar: (a) \(\mathrm{B}-\mathrm{F},\) (b) \(\mathrm{Cl}-\mathrm{Cl}\), $$ \text { (c) } \mathrm{Se}-\mathrm{O} $$ (d) \(\mathrm{H}-\mathrm{I} ?\) Which is the more electronegative atom in each polar bond?

The \(\mathrm{Ti}^{2+}\) ion is isoelectronic with the Ca atom. (a) Are there any differences in the electron configurations of \(\mathrm{Ti}^{2+}\) and Ca? (b) With reference to Figure 6.24 , comment on the changes in the ordering of the \(4 s\) and \(3 d\) subshells in Ca and \(\mathrm{Ti}^{2+}\), (c) Will Ca and \(\mathrm{Ti}^{2+}\) have the same number of unpaired electrons? Explain.

(a) Write the electron configuration for the element titanium, Ti. How many valence electrons does this atom possess? (b) Hafnium, Hf, is also found in group \(4 \mathrm{~B}\). Write the electron configuration for Hf. (c) Ti and Hf behave as though they possess the same number of valence electrons. Which of the subshells in the electron configuration of Hf behave as valence orbitals? Which behave as core orbitals?

Some chemists believe that satisfaction of the octet rule should be the top criterion for choosing the dominant Lewis structure of a molecule or ion. Other chemists believe that achieving the best formal charges should be the top criterion. Consider the dihydrogen phosphate ion, \(\mathrm{H}_{2} \mathrm{PO}_{4}^{-},\) in which the \(\mathrm{H}\) atoms are bonded to \(\mathrm{O}\) atoms. (a) What would be the predicted dominant Lewis structure if satisfying the octet rule is the top criterion? (b) What would it be if achieving the best formal charges is the top criterion? (c) Is there another Lewis structure you can draw that satisfies neither of these criteria?

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