Question

Some compounds have interatomic bonds that are partially ionic and partially covalent. Calculate the percentage ionic character of the following compounds: HF, HCl and HBr. Use this information to explain why HF has a higher boiling temperature than HCl (19.4 vs. -85^oC), even though HF has a lower molecular weight.

Answer 1

The below image provides the solution.

Answer 2

Step 1: Calculate % Ionic Character

Use Pauling's formula:

$$\mathrm{\%}\text{Ionic}=(1-e^{-0.25(\mathrm{\Delta }EN{)}^2})\times 100$$

Electronegativity (EN): H (2.20), F (3.98), Cl (3.16), Br (2.96).

• HF:

  $$\mathrm{\Delta }EN=3.98-2.20=1.78$$$$\mathrm{\%}\text{Ionic}=(1-e^{-0.25(1.78{)}^2})\times 100=43.2\mathrm{\%}$$

• HCl:

  $$\mathrm{\Delta }EN=3.16-2.20=0.96$$$$\mathrm{\%}\text{Ionic}=(1-e^{-0.25(0.96{)}^2})\times 100=19.7\mathrm{\%}$$

• HBr:

  $$\mathrm{\Delta }EN=2.96-2.20=0.76$$$$\mathrm{\%}\text{Ionic}=(1-e^{-0.25(0.76{)}^2})\times 100=12.4\mathrm{\%}$$

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Step 2: Why HF Boils Higher Than HCl

• HF has 43% ionic character → forms strong hydrogen bonds (H-F is very polar).

• HCl only has 20% ionic character → weaker dipole forces (no H-bonding).

• Even though HF is lighter, H-bonds require more energy to break, so it boils at 19.4°C vs. HCl at -85°C.

• 
  

Key Formula Used:

$$\mathrm{\%}\text{Ionic}=(1-e^{-0.25(\mathrm{\Delta }EN{)}^2})\times 100$$

Result: HF > HCl > HBr in ionic character and boiling point.