Question

Which of the following is the correct order of increasing number of lone pair of electrons on the ntral atom?

• A. ${IF_5}<{XeF_2}<{IF_7}<{ClF_3} $
• B. ${IF_7}<{ClF_3}<{XeF_2}<{IF_5}$
• C. ${IF_7}<{XeF_2}<{ClF_3}<{IF_5}$
• D. ${IF_7}<{IF_5}<{ClF_3}<{XeF_2}$

Hint:

Central Atom Lone Pair Calculation

• The number of lone pairs on the central atom is computed using:
  $\textLone Pairs = \frac12 \left( \textValence electrons of central atom - \textBonding pairs \right)$
• Group numbers help determine valence electrons:
  • Group 17: 7 valence electrons (e.g., Cl, I)
  • Group 18: 8 valence electrons (e.g., Xe)
• Fluorine is monovalent, so number of F atoms equals bonding pairs.
• Geometry and hybridization can further confirm these LP counts:
  • IF7: Pentagonal bipyramidal (0 LP)
  • IF5: Square pyramidal (1 LP)
  • ClF3: T-shaped (2 LP)
  • XeF2: Linear (3 LP)

Answer 1

The Correct Option is D

To solve this, we need to determine the number of lone pairs (LP) on the central atom in each molecule. The formula to calculate this is: \[ \text{Lone Pairs} = \frac{1}{2} (\text{Valence electrons of central atom} - \text{Number of bonded atoms}) \] Step-by-step analysis:

• {IF7 (Iodine Heptafluoride)}: 
  Iodine (Group 17) has 7 valence electrons. It forms 7 bonds with fluorine. 
  $\Rightarrow$ Lone pairs = $\frac{1}{2}(7 - 7) = 0$.
• {IF5 (Iodine Pentafluoride)}: 
  Iodine (Group 17) has 7 valence electrons and forms 5 bonds. 
  $\Rightarrow$ Lone pairs = $\frac{1}{2}(7 - 5) = 1$.
• {ClF3 (Chlorine Trifluoride)}: 
  Chlorine (Group 17) has 7 valence electrons and forms 3 bonds. 
  $\Rightarrow$ Lone pairs = $\frac{1}{2}(7 - 3) = 2$.
• {XeF2 (Xenon Difluoride)}: 
  Xenon (Group 18) has 8 valence electrons and forms 2 bonds. 
  $\Rightarrow$ Lone pairs = $\frac{1}{2}(8 - 2) = 3$.

Summary of Lone Pairs:

• ${IF_7}$ — 0 lone pairs
• ${IF_5}$ — 1 lone pair
• ${ClF_3}$ — 2 lone pairs
• ${XeF_2}$ — 3 lone pairs

Correct increasing order: ${IF_7}<{IF_5}<{ClF_3}<{XeF_2}$ This matches option (4).

Answer 2

Step 1: Identify the central atoms and their valence electrons
- IF₇: Central atom Iodine (I), group 17, has 7 valence electrons.
- IF₅: Central atom Iodine (I), group 17, has 7 valence electrons.
- ClF₃: Central atom Chlorine (Cl), group 17, has 7 valence electrons.
- XeF₂: Central atom Xenon (Xe), group 18, has 8 valence electrons.

Step 2: Determine the number of bonded atoms and electron pairs
- IF₇: Iodine bonded to 7 fluorines, uses 7 bonding pairs, no lone pairs left.
- IF₅: Iodine bonded to 5 fluorines, total valence electrons 7, bonding pairs 5, lone pairs = 7 - 5 = 2 pairs.
- ClF₃: Chlorine bonded to 3 fluorines, 7 valence electrons, bonding pairs 3, lone pairs = 7 - 3 = 4 electrons → 2 lone pairs (since 2 electrons = 1 pair), but careful: total electrons are shared differently, actual lone pairs on Cl is 2.
- XeF₂: Xenon bonded to 2 fluorines, 8 valence electrons, bonding pairs 2, lone pairs = 8 - 2 = 6 electrons → 3 lone pairs.

Step 3: Arrange in increasing order of lone pairs
- IF₇: 0 lone pairs
- IF₅: 1 lone pair (actually 1 lone pair, because 7 valence electrons minus 5 bonding pairs means 2 electrons = 1 lone pair)
- ClF₃: 2 lone pairs
- XeF₂: 3 lone pairs

Step 4: Conclusion
The correct order of increasing lone pairs on the central atom is:
IF₇ < IF₅ < ClF₃ < XeF₂.