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From their Lewis structures, determine the number of \(\sigma\) and \(\pi\) bonds in each of the following molecules or ions: (a) hydrazine, \(\mathrm{N}_{2} \mathrm{H}_{4}\) (b) hydrogen cyanide, HCN; (c) sulphur trioxide, \(\mathrm{SO}_{3} ;\) (d) ozone, \(\mathrm{O}_{3}\).

Short Answer

Expert verified
In summary: (a) Hydrazine (N2H4) has 5 \(\sigma\) bonds and no \(\pi\) bonds. (b) Hydrogen cyanide (HCN) has 2 \(\sigma\) bonds and 2 \(\pi\) bonds. (c) Sulphur trioxide (SO3) has 3 \(\sigma\) bonds and 3 \(\pi\) bonds. (d) Ozone (O3) has 2 \(\sigma\) bonds and 1 \(\pi\) bond.

Step by step solution

01

Drawing Lewis structures for each molecule or ion

Draw the Lewis structures for hydrazine (N2H4), hydrogen cyanide (HCN), sulfur trioxide (SO3), and ozone (O3) For N2H4: H H \ / N - N / \ H H For HCN: H - C ≡ N For SO3: O || O - S - O For O3: O - O - O
02

Identifying the type of bonds in each structure

Determine whether the bonds within each structure are single, double, or triple For N2H4: N-H bonds: single N-N bond: single For HCN: H-C bond: single C-N bond: triple For SO3: S-O bonds: double For O3: O-O bonds: single (between left and central O atoms) O-O bond: double (between central and right O atoms)
03

Count the number of \(\sigma\) and \(\pi\) bonds in each structure

Count the number of \(\sigma\) and \(\pi\) bonds according to the following rules: - For single bonds, there is 1 \(\sigma\) bond - For double bonds, there is 1 \(\sigma\) bond and 1 \(\pi\) bond - For triple bonds, there is 1 \(\sigma\) bond and 2 \(\pi\) bonds For N2H4: - 4 N-H single bonds: 4 \(\sigma\) bonds - 1 N-N single bond: 1 \(\sigma\) bond Total: 5 \(\sigma\) bonds For HCN: - 1 H-C single bond: 1 \(\sigma\) bond - 1 C-N triple bond: 1 \(\sigma\) bond and 2 \(\pi\) bonds Total: 2 \(\sigma\) bonds and 2 \(\pi\) bonds For SO3: - 3 S-O double bonds: 3 \(\sigma\) bonds and 3 \(\pi\) bonds Total: 3 \(\sigma\) bonds and 3 \(\pi\) bonds For O3: - 1 O-O single bond: 1 \(\sigma\) bond - 1 O-O double bond: 1 \(\sigma\) bond and 1 \(\pi\) bond Total: 2 \(\sigma\) bonds and 1 \(\pi\) bond In conclusion: (a) Hydrazine (N2H4) has 5 \(\sigma\) bonds and no \(\pi\) bonds. (b) Hydrogen cyanide (HCN) has 2 \(\sigma\) bonds and 2 \(\pi\) bonds. (c) Sulphur trioxide (SO3) has 3 \(\sigma\) bonds and 3 \(\pi\) bonds. (d) Ozone (O3) has 2 \(\sigma\) bonds and 1 \(\pi\) bond.

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Key Concepts

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

Sigma Bonds
In the world of molecular chemistry, sigma (\( \sigma \)) bonds are fundamental and represent the strongest type of covalent bond. They occur when two atomic orbitals overlap directly between the nuclei of two atoms. This head-on overlap maximizes the bonding, allowing electron density to be concentrated along the axis connecting the two nuclei.
Sigma bonds are often present in all types of bonds, whether they're single, double, or triple. In fact, the first bond that is formed between any two atoms is always a sigma bond.
  • Single bond: 1 sigma bond (e.g., the N-H bond in hydrazine \( \mathrm{N}_{2} \mathrm{H}_{4} \))
  • Double bond: 1 sigma bond and 1 pi bond (e.g., the S-O bond in sulfur trioxide \( \mathrm{SO}_{3} \)
  • Triple bond: 1 sigma bond and 2 pi bonds (e.g., the C-N bond in hydrogen cyanide HCN)
Understanding sigma bonds is crucial as they not only provide structural integrity but also impact the rotational capability around the bond. Sigma bonds allow for free rotation, unlike their pi bond counterparts.
Pi Bonds
Pi (\( \pi \)) bonds add an additional layer of connection between atoms in molecules, often coexisting with sigma bonds. These bonds form due to the sideways overlap of p-orbitals. Unlike sigma bonds, pi bonds concentrate electron density above and below the bond axis.
Pi bonds are crucial in contributing to the double or triple bond nature of a molecular structure. An essential property of pi bonds is that they are weaker than sigma bonds. Thus, they allow less flexibility and do not permit the rotation around the bond axis as sigma bonds do.
  • In double bonds: They feature 1 pi bond in addition to the sigma bond (e.g., \( \mathrm{SO}_{3} \)
  • In triple bonds: They consist of 2 pi bonds on top of the sigma bond (e.g., HCN)
Pi bonds influence the chemical reactivity and physical properties of molecules. For example, the pi bonds in the ozone (\( \mathrm{O}_{3} \)) molecule are responsible for its resonance stability.
Bond Types
The nature of chemical bonds significantly defines a molecule's properties. Broadly, bonds are categorized into three types: single, double, and triple bonds. Each of these bonds has distinct characteristics due to the combination of sigma and pi bonds.
Single bonds are made entirely of one sigma bond, which allows for free rotation around the bond axis. This is seen in molecules such as hydrazine (\( \mathrm{N}_{2} \mathrm{H}_{4} \)).
Double bonds, consisting of one sigma and one pi bond, restrict rotational movement due to the pi bond. This bond type is seen in sulfur trioxide (\( \mathrm{SO}_{3} \)), which has three double bonds contributing to its planar structure.
Triple bonds, which are the strongest being made of one sigma and two pi bonds, are seen in hydrogen cyanide (HCN). These bonds are shorter than single and double bonds, reflecting their increased bond strength and reduced bond length.
Each bond type has unique chemical and physical properties, which impact molecular behavior and reactivity.
Molecular Structure
The molecular structure is a three-dimensional arrangement of atoms within a molecule. Understanding molecular structure through Lewis structures helps to predict molecular behavior and properties. The arrangement is dictated by the types and numbers of bonds (sigma and pi bonds), as well as the spatial arrangement due to atom size and geometry.
For example, the shape of sulfur trioxide (\( \mathrm{SO}_{3} \)) is trigonal planar, influenced by its three double bonds. Each bond generally causes the molecule to have a flat, 120-degree separation between the oxygen atoms.
Ozone (\( \mathrm{O}_{3} \)) showcases a bent shape rather than a linear form due to the presence of one double and one single bond influencing bond angles and electron arrangements.
Molecular structures are vital for predicting the interactions of molecules with others, their physical state at room temperature, and even their color and solubility. By studying molecular structures, chemists can deduce a great deal about a substance’s chemical nature and potential reactions.

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Most popular questions from this chapter

How would we describe a substance that contains only paired electrons and is weakly repelled by a magnetic field? Which of the following ions would you expect to possess similar characteristics: \(\mathrm{H}_{2}^{-}, \mathrm{Ne}_{2}^{+}, \mathrm{F}_{2}, \mathrm{O}_{2}^{2+} ?\)

Explain the following: (a) The peroxide ion, \(\mathrm{O}_{2}^{2-}\), has a longer bond length than the superoxide ion, \(\mathrm{O}_{2}^{-}\). (b) The magnetic properties of \(\mathrm{B}_{2}\) are consistent with the \(\pi_{2 \mathrm{p}}\) MOs being lower in energy than the \(\sigma_{2 p}\) MO. (c) The \(\mathrm{O}_{2}^{2+}\) ion has a stronger O- \(O\) bond than \(\mathrm{O}_{2}\) itself.

Butadiene, \(\mathrm{C}_{4} \mathrm{H}_{6},\) is a planar molecule that has the following carbon-carbon bond lengths: $$ \mathrm{H}_{2} \mathrm{C}=\mathrm{CH}_{134 \mathrm{pm}} \mathrm{CH}=\mathrm{CH}_{2} $$ (a) Predict the bond angles around each of the carbon atoms and sketch the molecule. (b) From left to right, what is the hybridization of each carbon atom in butadiene? (c) The middle \(\mathrm{C}-\mathrm{C}\) bond length in butadiene \((148 \mathrm{pm})\) is a little shorter than the average \(\mathrm{C}-\mathrm{C}\) single bond length (154 pm). Does this imply that the middle \(\mathrm{C}-\mathrm{C}\) bond in butadiene is weaker or stronger than the average \(\mathrm{C}-\mathrm{C}\) single bond? (d) Based on your answer for part (c), discuss what additional aspects of bonding in butadiene might support the shorter middle \(\mathrm{C}-\mathrm{C}\) bond.

(a) Draw a picture showing how two \(p\) orbitals on two different atoms can be combined to make a \(\sigma\) bond. (b) Sketch a \(\pi\) bond that is constructed from \(p\) orbitals. (c) Which is generally stronger, a \(\sigma\) bond or a \(\pi\) bond? Explain. (d) Can two s orbitals combine to form a \(\pi\) bond? Explain.

(a) Predict the electron-domain geometry around the central \(\mathrm{S}\) atom in \(\mathrm{SF}_{2}, \mathrm{SF}_{4}\), and \(\mathrm{SF}_{6}\). ( \(\mathbf{b}\) ) The anion \(\mathrm{IO}_{4}^{-}\) has a tetrahedral structure: three oxygen atoms form double bonds with the central iodine atom and one oxygen atom which carries a negative charge forms a single bond. Predict the molecular geometry of \(\mathrm{IO}_{6}{ }^{5-}\).

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