Question

Arrange the bonds in order of increasing ionic character in the molecules. LiF, K\(_2\)O, N\(_2\), SO\(_2\), and ClF\(_3\).

• A. ClF\(_3\) \(<\) N\(_2\)\(<\) SO\(_2\) \(<\) K\(_2\)O \(<\) LiF
• B. LiF \(<\) K\(_2\)O \(<\) ClF\(_3\) \(<\) SO\(_2\) \(<\) N\(_2\)
• C. N\(_2\)$<$ClF\(_3\)$<$SO\(_2\)$<$K\(_2\)O$<$LiF
• D. N\(_2\)$<$SO\(_2\)$<$ClF\(_3\)$<$K\(_2\)O$<$LiF

Answer 1

The Correct Option is C

The problem asks to arrange the chemical bonds in the given molecules—LiF, K₂O, N₂, SO₂, and ClF₃—in order of increasing ionic character.

Concept Used:

The ionic character of a chemical bond is a measure of its polarity. It depends on the difference in electronegativity (\( \Delta \text{EN} \)) between the two atoms forming the bond. According to the Pauling scale, a larger difference in electronegativity corresponds to a greater charge separation and thus a higher degree of ionic character.

The relationship can be summarized as:

\[ \text{Ionic Character} \propto \Delta \text{EN} = | \text{EN}_{\text{atom 1}} - \text{EN}_{\text{atom 2}} | \]

A bond between identical atoms (\( \Delta \text{EN} = 0 \)) is purely covalent, while a bond with a large \( \Delta \text{EN} \) (typically > 1.7) is considered predominantly ionic.

Step-by-Step Solution:

Step 1: Identify the specific bonds within each molecule that need to be compared.

• In LiF, the bond is Li–F.
• In K₂O, the bond is K–O.
• In N₂, the bond is N≡N.
• In SO₂, the bond is S=O.
• In ClF₃, the bond is Cl–F.

Step 2: List the Pauling electronegativity (EN) values for each of the atoms involved.

• Nitrogen (N): 3.04
• Sulfur (S): 2.58
• Oxygen (O): 3.44
• Chlorine (Cl): 3.16
• Fluorine (F): 3.98
• Lithium (Li): 0.98
• Potassium (K): 0.82

Step 3: Calculate the electronegativity difference (\( \Delta \text{EN} \)) for each bond.

For N₂ (N≡N bond):

\[ \Delta \text{EN} = |3.04 - 3.04| = 0 \]

This is a purely covalent bond.

For ClF₃ (Cl–F bond):

\[ \Delta \text{EN} = |3.98 - 3.16| = 0.82 \]

This is a polar covalent bond.

For SO₂ (S=O bond):

\[ \Delta \text{EN} = |3.44 - 2.58| = 0.86 \]

This is also a polar covalent bond, slightly more polar than the Cl–F bond.

For K₂O (K–O bond):

\[ \Delta \text{EN} = |3.44 - 0.82| = 2.62 \]

This is a predominantly ionic bond.

For LiF (Li–F bond):

\[ \Delta \text{EN} = |3.98 - 0.98| = 3.00 \]

This is a highly ionic bond.

Final Computation & Result:

Step 4: Arrange the bonds in order of increasing \( \Delta \text{EN} \), which corresponds to increasing ionic character.

The calculated \( \Delta \text{EN} \) values in increasing order are:

\[ 0 \ (\text{for N}_2) < 0.82 \ (\text{for ClF}_3) < 0.86 \ (\text{for SO}_2) < 2.62 \ (\text{for K}_2\text{O}) < 3.00 \ (\text{for LiF}) \]

Therefore, the order of increasing ionic character for the bonds in the given molecules is:

N₂ < ClF₃ < SO₂ < K₂O < LiF