1. We need to count how many of the given species have exactly two lone pairs of electrons on the central atom.

The species are:

• $SF_4$
• $XeF_4$
• $CF_4$
• $H_2O$

2. Analyze each species one by one.

(i) $SF_4$

• Central atom: $S$
• Sulfur has 6 valence electrons.
• It forms 4 bonds with 4 fluorine atoms.
• So, after making 4 bond pairs, sulfur has 1 lone pair left.

Thus, $SF_4$ has one lone pair, not two.

(ii) $XeF_4$

• Central atom: $Xe$
• Xenon has 8 valence electrons.
• It forms 4 bonds with 4 fluorine atoms.
• After forming 4 bond pairs, xenon has 2 lone pairs left.

Thus, $XeF_4$ has two lone pairs.

(iii) $CF_4$

• Central atom: $C$
• Carbon has 4 valence electrons.
• It forms 4 bonds with 4 fluorine atoms.
• No electrons remain as lone pairs on carbon.

Thus, $CF_4$ has zero lone pairs.

(iv) $H_2O$

• Central atom: $O$
• Oxygen has 6 valence electrons.
• It forms 2 bonds with 2 hydrogen atoms.
• So, oxygen has 2 lone pairs left.

Thus, $H_2O$ has two lone pairs.

3. Species with exactly two lone pairs are:

• $XeF_4$
• $H_2O$

Hence, the number of such species is $2$

4. Comparison with stored answer:

• Derived answer: $2$
• Stored correct answer: $1$

These do not match. The stored answer appears to be incorrect, because both $XeF_4$ and $H_2O$ have two lone pairs on the central atom.