Chemistry Liquids MDCAT 2012
PMDC Verified Question 69 of 73
In 'H-F' bond, electronegativity difference is 2.0. What is the type of this bond?
A
Polar covalent bond
B
pi (\( \pi \)) bond
C
Non-polar covalent bond
D
Co-ordinate covalent bond
Tap any option to test your recall and reveal the step-by-step Propolis autopsy.

Propolis Cognitive Error Autopsy

Official Correct Choice:
Option A: Polar covalent bond
Concept:

Bond polarity is determined by the electronegativity difference (\( \Delta \text{E.N.} \)) between the two bonded atoms. However, bonding in Hydrogen Halides acts slightly unusually due to their molecular nature.

Formula:

$$ \Delta \text{E.N.} = 4.0 (\text{F}) - 2.1 (\text{H}) = 1.9 \approx 2.0 $$

Solution:

  • Generally, an electronegativity difference > 1.7 suggests an ionic bond.


  • However, HF is a distinct exception. It exists as discrete molecules and is a gas at room temperature, which are characteristic properties of covalent compounds.


  • Therefore, despite the large electronegativity difference, HF strictly forms a highly polar covalent bond (with significant ionic character, but primarily covalent).


Why other options are incorrect:

It is not non-polar (as electrons are unequal). It is a sigma bond, not a pi bond. It forms via mutual sharing, not through the dative donation of a lone pair.

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