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How many nonbonding electron pairs are there in each of the following molecules: \((\mathrm{a})\left(\mathrm{CH}_{3}\right)_{2} \mathrm{S},(\mathbf{b}) \mathrm{HCN},(\mathbf{c}) \mathrm{C}_{2} \mathrm{H}_{2}\) \((\mathbf{d}) \mathrm{CH}_{3} \mathrm{F} ?\)

Short Answer

Expert verified
In summary, the number of nonbonding electron pairs in each molecule are as follows: (a) (CH3)2S: 3 nonbonding electron pairs (1 on each C atom and 1 on the S atom), (b) HCN: 1 nonbonding electron pair (on the N atom), (c) C2H2: 0 nonbonding electron pairs, and (d) CH3F: 3 nonbonding electron pairs (on the F atom).

Step by step solution

01

(a) (CH3)2S

The molecule (CH3)2S consists of two CH3 groups bonded to a sulfur (S) atom. Each carbon (C) atom has 4 valence electrons, each hydrogen (H) atom has 1 valence electron, and the sulfur (S) atom has 6 valence electrons. In each CH3 group, the carbon atom makes 3 single bonds with 3 hydrogen atoms (total of 6 bonding electrons) leaving 2 electrons as a nonbonding pair on carbon. The sulfur atom in (CH3)2S forms bonds with two carbon atoms from each CH3 group, and shares 2 electrons per bond (total of 4 bonding electrons). The sulfur atom has 2 remaining electrons, which form a nonbonding electron pair. In total, there are 3 non-bonding electron pairs in the molecule: 1 for each carbon atom and 1 for the sulfur atom.
02

(b) HCN

The molecule HCN consists of a hydrogen (H) atom, a carbon (C) atom, and a nitrogen (N) atom. Hydrogen has 1 valence electron, carbon has 4 valence electrons, and nitrogen has 5 valence electrons. The hydrogen atom forms a single bond with the carbon atom, with a total of 2 bonding electrons. The carbon atom forms a triple bond with the nitrogen atom, with a total of 6 bonding electrons. There are no nonbonding electron pairs on the hydrogen or carbon atoms, as all their valence electrons are involved in bonding. Nitrogen has one lone pair remaining after the triple bond with carbon, so there is 1 nonbonding electron pair in HCN.
03

(c) C2H2

The molecule C2H2 consists of two carbon (C) atoms and two hydrogen (H) atoms. Carbon has 4 valence electrons and hydrogen has 1 valence electron. Each hydrogen atom forms a single bond with a carbon atom, with a total of 2 bonding electrons for the hydrogen atoms. The two carbon atoms are joined by a triple bond, with a total of 6 bonding electrons. The two carbon atoms and the two hydrogen atoms have no remaining valence electrons, so there are no nonbonding electron pairs in the C2H2 molecule.
04

(d) CH3F

The molecule CH3F consists of a carbon (C) atom, three hydrogen (H) atoms, and a fluorine (F) atom. Carbon has 4 valence electrons, hydrogen has 1 valence electron, and fluorine has 7 valence electrons. The carbon atom forms 3 single bonds with hydrogen atoms, with a total of 6 bonding electrons. The carbon atom also forms a single bond with the fluorine atom, sharing 2 bonding electrons. There are no nonbonding electron pairs on the carbon or hydrogen atoms, as all their valence electrons are involved in bonding. Fluorine has 3 lone pairs remaining after the single bond with carbon, so there are 3 nonbonding electron pairs in the CH3F molecule.

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Key Concepts

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

Valence Electrons
Valence electrons are the outermost electrons of an atom. They play a crucial role in chemical bonding and reactions. Understanding them is essential for determining how atoms connect to form molecules. These electrons are found in the highest energy level of an atom and are the most loosely held. This makes them available for forming chemical bonds with other atoms.
  • Each element has a different number of valence electrons, which can usually be determined based on its position in the periodic table.
  • For example, carbon has four valence electrons, and hydrogen has one.
  • Sulfur, found in group 16, has six valence electrons, while fluorine has seven.
Knowing the number of valence electrons helps predict the types of bonds an atom can form. For instance, in CH3F, carbon uses its four valence electrons to create one bond with fluorine and three bonds with hydrogen atoms. The remaining electrons on fluorine form nonbonding electron pairs (or lone pairs). Understanding valence electrons is key in inferring stability and reactivity of molecules.
Molecular Structure
The molecular structure reveals how atoms are organized in a molecule. Each molecule has a distinct shape that directly influences its physical and chemical properties. Molecular structures are often represented in two dimensions via Lewis structures, which help visualize bonding patterns and nonbonding electrons.
  • For example, in the molecule \((\text{CH}_3)_2\text{S}\), the structure consists of a central sulfur atom bonded to two CH3 groups. The sulfur atom retains a nonbonding pair of electrons.
  • In contrast, the molecule \(\text{HCN}\) has a linear structure, where carbon is in the middle, bonded to hydrogen at one end and triple bonded to nitrogen at the other end. This linear alignment is due to the triple bond stabilizing the molecule.
  • The molecule \(\text{C}_2\text{H}_2\) consists of a triple bond between two carbon atoms, with single bonds to hydrogen atoms, forming a symmetrical linear shape.
  • Meanwhile, \(\text{CH}_3\text{F}\) has a tetrahedral structure due to the single bonds between each hydrogen and carbon, and between fluorine and carbon.
Understanding the molecular structure is crucial to infer how a molecule will interact with others, predicting behavior and the role of nonbonding electron pairs.
Electron Configuration
Electron configuration describes the arrangement of electrons in an atom, typically expressed in a sequence of numbers and letters. This configuration provides insights into the chemical behavior of an atom, dictating how it bonds and interacts with other atoms.
  • This configuration is pivotal in comprehending the distribution of valence electrons.
  • A sulfur atom’s electron configuration is \([Ne] 3s^2 3p^4\), indicating six electrons in the 3s and 3p orbitals, which classify as its valence electrons in molecules like \((\text{CH}_3)_2\text{S}\).
  • Fluorine’s electron configuration is \([He] 2s^2 2p^5\), showing that after forming a single bond, it maintains three nonbonding pairs of electrons in \(\text{CH}_3\text{F}\).
  • Nitrogen, in \(\text{HCN}\), uses its electron configuration \([He] 2s^2 2p^3\) to form a triple bond and still has one lone pair of electrons.
These configurations highlight the tendency of atoms to complete orbitals for stability, often following the Octet Rule, which states that atoms tend to have eight electrons in their valence shell. This principle influences the arrangement and number of nonbonding electron pairs in a molecule.

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Most popular questions from this chapter

Give the electron-domain and molecular geometries of a molecule that has the following electron domains on its central atom: (a) four bonding domains and no nonbonding domains, (b) three bonding domains and two nonbonding domains, (c) five bonding domains and one nonbonding domain, (d) four bonding domains and two nonbonding domains.

The lactic acid molecule, \(\mathrm{CH}_{3} \mathrm{CH}(\mathrm{OH}) \mathrm{COH},\) gives sour milk its unpleasant, sour taste. (a) Draw the Lewis structure for the molecule, assuming that carbon always forms four bonds in its stable compounds. (b) How many \(\pi\) and how many \(\sigma\) bonds are in the molecule? (c) Which CO bond is shortest in the molecule? (d) What is the hybridization of atomic orbitals around the carbon atom associated with that short bond? (e) What are the approximate bond angles around each carbon atom in the molecule?

The Lewis structure for allene is Make a sketch of the structure of this molecule that is analogous to Figure \(9.25 .\) In addition, answer the following three questions: (a) Is the molecule planar? (b) Does it have a nonzero dipole moment? (c) Would the bonding in allene be described as delocalized? Explain.

The structure of borazine, \(\mathrm{B}_{3} \mathrm{N}_{3} \mathrm{H}_{6},\) is a six-membered ring of alternating \(\mathrm{B}\) and \(\mathrm{N}\) atoms. There is one \(\mathrm{H}\) atom bonded to each \(\mathrm{B}\) and to each \(\mathrm{N}\) atom. The molecule is planar. (a) Write a Lewis structure for borazine in which the formal charge on every atom is zero. (b) Write a Lewis structure for borazine in which the octet rule is satisfied for every atom. (c) What are the formal charges on the atoms in the Lewis structure from part (b)? Given the electronegativities of \(\mathrm{B}\) and \(\mathrm{N},\) do the formal charges seem favorable or unfavorable? (d)Do either of the Lewis structures in parts (a) and (b) have multiple resonance structures? (e) What are the hybridizations at the B and N atoms in the Lewis structures from parts (a) and (b)? Would you expect the molecule to be planar for both Lewis structures? (f) The six \(B-N\) bonds in the borazine molecule are all identical in length at 1.44 A. Typical values for the bond lengths of \(\mathrm{B}-\mathrm{N}\) single and double bonds are 1.51 \(\mathrm{A}\) and \(1.31 \mathrm{A},\) respectively. Does the value of the \(\mathrm{B}-\mathrm{N}\) bond length seem to favor one Lewis structure over the other? (g) How many electrons are in the \(\pi\) system of borazine?

(a) The nitric oxide molecule, NO, readily loses one electron to form the \(\mathrm{NO}^{+}\) ion. Which of the following is the best explanation of why this happens: (i) Oxygen is more electronegative than nitrogen, (ii) The highest energy electron in NO lies in a \(\pi_{2 p}^{*}\) molecular orbital, or (iii) The \(\pi_{2 p}^{*}\) MO in NO is completely filled. (b) Predict the order of the \(\mathrm{N}-\mathrm{O}\) bond strengths in \(\mathrm{NO}, \mathrm{NO}^{+},\) and \(\mathrm{NO}^{-},\) and describe the magnetic properties of each.(c) With what neutral homonuclear diatomic molecules are the \(\mathrm{NO}^{+}\) and \(\mathrm{NO}^{-}\) ions isoelectronic (same number of electrons)?

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