Question

Bond pairs and lone pairs in \(IF_5\) are respectively:

• A. 4,2
• B. 5,1
• C. 6,0
• D. 4,1

Hint:

For \(IF_5\), six electron pairs surround iodine: five bond pairs and one lone pair.

Answer 1

The Correct Option is B

Concept: Bond pairs and lone pairs are determined from the number of valence electrons present on the central atom.

Step 1: Determine valence electrons of iodine. Iodine belongs to Group 17. Therefore, valence electrons in iodine: \[ =7 \]

Step 2: Formation of bonds with fluorine atoms. In \(IF_5\), iodine forms five covalent bonds with five fluorine atoms. Thus, five electrons are used for bond formation. Hence: \[ \text{Bond pairs}=5 \]

Step 3: Calculate remaining electrons. Remaining electrons on iodine: \[ 7-5=2 \] Two electrons constitute one lone pair. Therefore: \[ \text{Lone pairs}=1 \]

Step 4: Geometry of the molecule. Total electron pairs around iodine: \[ 5+1=6 \] Hybridization: \[ sp^3d^2 \] Electronic geometry is octahedral while molecular geometry becomes square pyramidal because of one lone pair. Hence, option (2) is correct.