Q21 E - Chemistry (OpenStax)

Question

Identify the more polar bond in each of the following pairs of bonds: (a) \({\rm{HF or HCl}}\) (b) \({\rm{NO or CO}}\) (c) \({\rm{SH or OH}}\) (d) \({\rm{PCl or HCl}}\) (e) \({\rm{CH or NH}}\) (f) \({\rm{SO or PO}}\) (g) \({\rm{CN or NN}}\) .

Step-by-Step Solution

Answer

The more polar bond in each is:

1. \({\rm{HF}}\) is more polar bond.
2. \({\rm{CO}}\) is more polar bond.
3. \({\rm{OH}}\) is more polar bond.
4. \({\rm{PCl}}\) is more polar bond.
5. \({\rm{NH}}\) is more polar bond.
6. \({\rm{PO}}\) is more polar bond.
7. \({\rm{CN}}\) is more polar bond.

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. \({\rm{7}}{\rm{.6}}\), and then compute the electronegativity difference.

Identifying the most polar bond in each pair

(a)

\({\rm{HF}}\): Difference in electronegativity is: \({\rm{4}}{\rm{.0  -   2}}{\rm{.1  =  1}}{\rm{.9}}\).

\({\rm{HCl}}\): Difference in electronegativity is: \({\rm{3}}{\rm{.0  -   2}}{\rm{.1  =  0}}{\rm{.9}}\).

Therefore, \({\rm{HF}}\) is more polar than \({\rm{HCl}}\).

Identifying the most polar bond in each pair

(b)

\({\rm{NO}}\): The difference in electronegativity is: \({\rm{3}}{\rm{.5  -   3}}{\rm{.0  =  0}}{\rm{.5}}\). 

\({\rm{CO}}\): The difference in electronegativity is: \({\rm{3}}{\rm{.5  -   2}}{\rm{.5  =  1}}{\rm{.0}}\). 

Therefore, \({\rm{CO}}\) has a higher polarity than \({\rm{NO}}\). 

Identifying the most polar bond in each pair

(c)

\({\rm{SH}}\): Difference in electronegativity is: \({\rm{2}}{\rm{.5  -   2}}{\rm{.1  =  0}}{\rm{.4}}\).

\({\rm{OH}}\): Difference in electronegativity is: \({\rm{3}}{\rm{.5  -   2}}{\rm{.1  =  1}}{\rm{.4}}\).

Therefore, \({\rm{OH}}\) is more polar than \({\rm{SH}}\).

Identifying the most polar bond in each pair

(d)

\({\rm{PCl}}\): Difference in electronegativity is: \({\rm{3}}{\rm{.0  -   2}}{\rm{.1  =  0}}{\rm{.9}}\).

\({\rm{SCl}}\): Difference in electronegativity is: \({\rm{3}}{\rm{.0  -   2}}{\rm{.5  =  0}}{\rm{.5}}\).

Therefore, \({\rm{PCl}}\) is more polar than \({\rm{SCl}}\).

Identifying the most polar bond in each pair

(e)

\({\rm{CH}}\): Difference in electronegativity is: \({\rm{2}}{\rm{.5  -   2}}{\rm{.1  =  0}}{\rm{.4}}\).

\({\rm{NH}}\): Difference in electronegativity is: \({\rm{3}}{\rm{.0  -   2}}{\rm{.1  =  0}}{\rm{.9}}\).

Therefore, \({\rm{NH}}\) is more polar than \({\rm{CH}}\).

Identifying the most polar bond in each pair

(f)

\({\rm{SO}}\): Difference in electronegativity is: \({\rm{3}}{\rm{.5  -   2}}{\rm{.5  =  1}}{\rm{.0}}\).

\({\rm{PO}}\): Difference in electronegativity is: \({\rm{3}}{\rm{.5  -   2}}{\rm{.1  =  1}}{\rm{.4}}\).

Therefore, \({\rm{PO}}\) is more polar than \({\rm{SO}}\).

Identifying the most polar bond in each pair

(g)

\({\rm{CN}}\): Difference in electronegativity is: \({\rm{3}}{\rm{.0  -   2}}{\rm{.5  =  0}}{\rm{.5}}\).

\({\rm{NN}}\): As, both atoms have the same electronegativity, the difference in electronegativity is zero.

Therefore, \({\rm{CN}}\) is more polar than \({\rm{NN}}\).