Question:medium

The Sum of number of lone pairs in central atom in IF5 and IF7 is:

Updated On: Mar 28, 2026
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Solution and Explanation

Step 1: Lone Pairs in IF5

Valence electrons:  

  • Iodine (I): 7 valence electrons
  • Fluorine (F): 7 valence electrons × 5 fluorine atoms = 35
  • Total valence electrons: \( 7 + 35 = 42 \)

Bonding electrons:

  • 5 I-F bonds = \( 5 × 2 = 10 \) bonding electrons

Remaining electrons:

  • \( 42 - 10 = 32 \) electrons used to complete the octet of fluorine atoms
  • \( 2 \) electrons left on iodine, forming 1 lone pair

Step 2: Lone Pairs in IF7

Valence electrons:

  • Iodine (I): 7 valence electrons
  • Fluorine (F): 7 valence electrons × 7 fluorine atoms = 49
  • Total valence electrons: \( 7 + 49 = 56 \)

Bonding electrons:

  • 7 I-F bonds = \( 7 × 2 = 14 \) bonding electrons

Remaining electrons:

  • \( 56 - 14 = 42 \) electrons used to complete the octet of fluorine atoms
  • No electrons left on iodine, so there are 0 lone pairs on iodine

Step 3: Total Lone Pairs on the Central Atom

The sum of lone pairs on iodine in IF5 and IF7 is:

\( 1 + 0 = 1 \)

Final Answer:

The total number of lone pairs on the central atoms in IF5 and IF7 is: 1.

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