Match List I with List II

List I (molecules/ions)	List II (No. of lone pairs of e$$^-$$ on central atom)
(A) IF$$_7$$	I. Three
(B) ICl$$_4^-$$	II. One
(C) XeF$$_6$$	III. Two
(D) XeF$$_2$$	IV. Zero

Choose the correct answer from the options given below:

We need to match each molecule/ion with the number of lone pairs on its central atom.

(A) IF$$_7$$:

Iodine has 7 valence electrons. It forms 7 bonds with 7 fluorine atoms, using all its valence electrons for bonding. The hybridization is $$sp^3d^3$$.

Lone pairs on I = $$\frac{7 - 7}{2} = 0$$ (Zero) --> matches IV

(B) ICl$$_4^-$$:

Iodine has 7 valence electrons, plus 1 from the negative charge = 8 electrons available. It forms 4 bonds with Cl atoms, leaving 4 non-bonding electrons.

Lone pairs on I = $$\frac{8 - 4}{2} = 2$$ (Two) --> matches III

The geometry is square planar ($$sp^3d^2$$, with 2 lone pairs in axial positions of the octahedron).

(C) XeF$$_6$$:

Xenon has 8 valence electrons. It forms 6 bonds with F atoms, leaving 2 non-bonding electrons.

Lone pairs on Xe = $$\frac{8 - 6}{2} = 1$$ (One) --> matches II

The geometry is distorted octahedral ($$sp^3d^3$$ hybridization with 1 lone pair).

(D) XeF$$_2$$:

Xenon has 8 valence electrons. It forms 2 bonds with F atoms, leaving 6 non-bonding electrons.

Lone pairs on Xe = $$\frac{8 - 2}{2} = 3$$ (Three) --> matches I

The geometry is linear ($$sp^3d$$ hybridization with 3 lone pairs in equatorial positions of a trigonal bipyramid).

The correct matching is: A-IV, B-III, C-II, D-I.

The correct answer is Option 2.