Question

Identify the more polar bond in each of the following pairs of bonds: (a) $\mathrm{H}\mathrm{F}\mathrm{o}\mathrm{r}\mathrm{H}\mathrm{C}\mathrm{l}$ (b) $\mathrm{N}\mathrm{O}\mathrm{o}\mathrm{r}\mathrm{C}\mathrm{O}$ (c) $\mathrm{S}\mathrm{H}\mathrm{o}\mathrm{r}\mathrm{O}\mathrm{H}$ (d) $\mathrm{P}\mathrm{C}\mathrm{l}\mathrm{o}\mathrm{r}\mathrm{H}\mathrm{C}\mathrm{l}$ (e) $\mathrm{C}\mathrm{H}\mathrm{o}\mathrm{r}\mathrm{N}\mathrm{H}$ (f) $\mathrm{S}\mathrm{O}\mathrm{o}\mathrm{r}\mathrm{P}\mathrm{O}$ (g) $\mathrm{C}\mathrm{N}\mathrm{o}\mathrm{r}\mathrm{N}\mathrm{N}$ .

Short answer

The more polar bond in each is:

1. $\mathrm{H}\mathrm{F}$ is more polar bond.
2. $\mathrm{C}\mathrm{O}$is more polar bond.
3. $\mathrm{O}\mathrm{H}$is more polar bond.
4. $\mathrm{P}\mathrm{C}\mathrm{l}$is more polar bond.
5. $\mathrm{N}\mathrm{H}$is more polar bond.
6. $\mathrm{P}\mathrm{O}$is more polar bond.
7. $\mathrm{C}\mathrm{N}$ is more polar bond.

Step-by-step solution

Define Chemical Bonding

A chemical bond is a long-term attraction between atoms, ions, or molecules that allows chemical compounds to form.

Explanation

The polarity of a bond is determined by the electronegativity difference between two atoms creating it.

We'll start by looking at the electronegativity values of each element in fig. $7.6$, and then compute the electronegativity difference.

Identifying the most polar bond in each pair

(a)

$\mathrm{H}\mathrm{F}$: Difference in electronegativity is: $4.0-2.1=1.9$.

$\mathrm{H}\mathrm{C}\mathrm{l}$: Difference in electronegativity is:$3.0-2.1=0.9$.

Therefore, $\mathrm{H}\mathrm{F}$ is more polar than $\mathrm{H}\mathrm{C}\mathrm{l}$.

Identifying the most polar bond in each pair

(b)

$\mathrm{N}\mathrm{O}$: The difference in electronegativity is:$3.5-3.0=0.5$.

$\mathrm{C}\mathrm{O}$: The difference in electronegativity is:$3.5-2.5=1.0$.

Therefore, $\mathrm{C}\mathrm{O}$ has a higher polarity than $\mathrm{N}\mathrm{O}$.

Identifying the most polar bond in each pair

(c)

$\mathrm{S}\mathrm{H}$: Difference in electronegativity is:$2.5-2.1=0.4$.

$\mathrm{O}\mathrm{H}$: Difference in electronegativity is:$3.5-2.1=1.4$.

Therefore, $\mathrm{O}\mathrm{H}$ is more polar than $\mathrm{S}\mathrm{H}$.

Identifying the most polar bond in each pair

(d)

$\mathrm{P}\mathrm{C}\mathrm{l}$: Difference in electronegativity is: $3.0-2.1=0.9$.

$\mathrm{S}\mathrm{C}\mathrm{l}$: Difference in electronegativity is:$3.0-2.5=0.5$.

Therefore, $\mathrm{P}\mathrm{C}\mathrm{l}$ is more polar than $\mathrm{S}\mathrm{C}\mathrm{l}$.

Identifying the most polar bond in each pair

(e)

$\mathrm{C}\mathrm{H}$: Difference in electronegativity is: $2.5-2.1=0.4$.

$\mathrm{N}\mathrm{H}$: Difference in electronegativity is:$3.0-2.1=0.9$.

Therefore, $\mathrm{N}\mathrm{H}$ is more polar than $\mathrm{C}\mathrm{H}$.

Identifying the most polar bond in each pair

(f)

$\mathrm{S}\mathrm{O}$: Difference in electronegativity is: $3.5-2.5=1.0$.

$\mathrm{P}\mathrm{O}$: Difference in electronegativity is:$3.5-2.1=1.4$.

Therefore, $\mathrm{P}\mathrm{O}$ is more polar than $\mathrm{S}\mathrm{O}$.

Identifying the most polar bond in each pair

(g)

$\mathrm{C}\mathrm{N}$: Difference in electronegativity is: $3.0-2.5=0.5$.

$\mathrm{N}\mathrm{N}$: As, both atoms have the same electronegativity, the difference in electronegativity is zero.

Therefore, $\mathrm{C}\mathrm{N}$ is more polar than $\mathrm{N}\mathrm{N}$.