How many lone pair (s) of electrons is present in HF?

Analyze the question: We need the number of lone pairs (non-bonding electron pairs) in a hydrogen fluoride (HF) molecule.

Concept: A lone pair is a pair of valence electrons not involved in bonding. To find them:

1. Count the total valence electrons.
2. Subtract the electrons used in bonds.
3. Divide what remains by 2.

Solution:

• Hydrogen (Group 1) has 1 valence electron.
• Fluorine (Group 17) has 7 valence electrons.
• Total = 1 + 7 = 8 valence electrons (4 pairs).

Hydrogen only needs 2 electrons (duplet), and fluorine needs 8 (octet).

• H and F share one pair of electrons, giving one single covalent bond (H-F). This uses 2 electrons.
• Remaining electrons = 8 - 2 = 6 electrons = 3 pairs.
• All of these go on fluorine to complete its octet.

Structure:

    ..
H - F:
    ..

Fluorine has 3 lone pairs plus 1 bonding pair, which gives 8 electrons around it. Hydrogen has no lone pairs.

Answer: 3 lone pairs, which is option D.

Tip: For a simple molecule with a halogen bonded to one atom, the halogen always has 3 lone pairs and 1 bonding pair (7 valence electrons - 1 used in bonding = 6 electrons = 3 pairs). The same applies to HCl, HBr, and HI.