Problem 79 - Chemistry: The Central Science

Question

From their Lewis structures, determine the number of \(\sigma\) and \(\pi\) bonds in each of the following molecules or ions: (a) \(\mathrm{CO}_{2} ;\) (b) thiocyanate ion, \(\mathrm{NCS}^{-}\); (c) formaldehyde, \(\mathrm{H}_{2} \mathrm{CO} ;\) (d) formic acid, HCOOH, which has one \(\mathrm{H}\) and two \(\mathrm{O}\) atoms attached to \(\mathrm{C}\).

Step-by-Step Solution

Answer

The number of σ and π bonds for each molecule or ion are: (a) CO₂: 2 σ bonds and 2 π bonds (b) \(\mathrm{NCS}^{-}\): 2 σ bonds and 2 π bonds (c) H₂CO: 3 σ bonds and 1 π bond (d) HCOOH: 4 σ bonds and 1 π bond

Step 1: Draw Lewis structures

For each of the given molecules or ions, draw the Lewis structure. The Lewis structure represents all the valence electrons and bonding patterns in each molecule. (a) CO₂ (b) \(\mathrm{NCS}^{-}\) (c) H₂CO (d) HCOOH

Step 2: Determine σ and π bonds

Now, using the drawn Lewis structures, count the number of single bonds, double bonds, and triple bonds present in each molecule or ion. A single bond is a σ bond, a double bond is made up of one σ bond and one π bond, and a triple bond is made up of one σ bond and two π bonds. (a) CO₂: The CO₂ molecule has two double bonds between the carbon atom and the oxygen atoms. (b) \(\mathrm{NCS}^{-}\): The thiocyanate ion has a single bond between C and N atoms, and a triple bond between C and S atoms. (c) H₂CO: The H₂CO molecule has two single bonds between H and C atoms, and a double bond between C and O atoms. (d) HCOOH: The formic acid molecule has a single bond between H and C, a double bond between C and O, and two single bonds between C and the other O and between that O and H.

Step 3: Count the σ and π bonds

Now, we can count the σ and π bonds for each molecule or ion based on the number of single, double, and triple bonds. (a) CO₂: 2 double bonds = 2 σ bonds + 2 π bonds (b) \(\mathrm{NCS}^{-}\): 1 single bond + 1 triple bond = (1 σ bond) + (1 σ bond + 2 π bonds) = 2 σ bonds + 2 π bonds (c) H₂CO: 2 single bonds + 1 double bond = 2 σ bonds + (1 σ bond + 1 π bond) = 3 σ bonds + 1 π bond (d) HCOOH: 3 single bonds + 1 double bond = 3 σ bonds + (1 σ bond + 1 π bond) = 4 σ bonds + 1 π bond As a result, we find the following number of σ and π bonds for each molecule or ion: (a) CO₂: 2 σ bonds and 2 π bonds (b) \(\mathrm{NCS}^{-}\): 2 σ bonds and 2 π bonds (c) H₂CO: 3 σ bonds and 1 π bond (d) HCOOH: 4 σ bonds and 1 π bond

Key Concepts

Sigma Bond

A sigma (\(\sigma\)) bond is the strongest type of covalent chemical bond. It is formed by the direct overlapping of atomic orbitals. Imagine two atoms getting so close to each other that their electron clouds start to overlap. This overlap forms a strong bond along the axis that connects the two nuclei. This axis is where the sigma bond takes place.
- Most single bonds in molecules are sigma bonds.
- Sigma bonds allow for free rotation of the bonded atoms, which means that the atoms can spin around the bond axis without breaking the bond.
- Because they are end-to-end overlaps, sigma bonds are at the core of the molecule's structure, providing stability and a strong foundation.
In the molecules discussed in the exercise, each single bond between atoms is a sigma bond. Additionally, each double bond and triple bond also contains one sigma bond.

Pi Bond

Pi (\(\pi\)) bonds are formed when parallel orbitals overlap, allowing electron clouds from adjacent atoms to come into close contact. These bonds are typically found in conjunction with sigma bonds, especially in double and triple bonds.
- Double bonds consist of one sigma bond and one pi bond, while triple bonds consist of one sigma and two pi bonds.
- Pi bonds are generally weaker than sigma bonds because the side-to-side overlap is not as substantial as the head-on overlap of sigma bonds.
- Since pi bonds exist above and below the bond axis, they prevent the atoms involved from rotating freely around the bond. This restricts the geometry of the molecule.
Thus, in the context of the exercise, every double bond contains one pi bond and every triple bond contains two pi bonds. This is crucial when identifying and counting the different types of bonds in a molecule.

Bonding Patterns

Bonding patterns describe how atoms connect or link to each other in a molecule, forming different types of chemical bonds. Understanding these patterns helps in predicting the molecule's structure and properties.
- Lewis structures are tools chemists use to represent the bonding patterns in a molecule. They show how electrons are arranged among the atoms, including both bonding electrons (which form bonds) and lone pairs.
- The strategy is typically to pair atoms so that each reaches a stable electronic configuration, often involving both single, double, and sometimes triple bonds.
- A bonding pattern can already tell you a lot about the behavior and reactivity of a molecule.
In the exercise, each of the molecules and ions given has its unique bonding pattern—ranging from double bonds in carbon dioxide to a triple bond in the thiocyanate ion. Recognizing these leads directly to understanding how many sigma and pi bonds are present.

Valence Electrons

Valence electrons are the outermost electrons of an atom and play a critical role in chemical bonding. They're the electrons involved when atoms form bonds with each other.
- Typically, atoms are most stable when they have eight valence electrons, known as the "octet rule." Hydrogen is an exception, as it is stable with two.
- Counting valence electrons is the first step in drawing a Lewis structure, where these electrons are arranged to satisfy the bonding needs of each atom.
- The number of valence electrons informs us on how many bonds an atom can form. This is visible in its tendency to achieve a full outer shell through sharing, gaining, or losing electrons.
In the original exercise, each step to solve how many sigma and pi bonds are present begins by accounting for the number of valence electrons in each atom. This forms the basis of understanding what types of bonds (single, double, or triple) will form in the molecule.