1. We need to count the species in which the central atom has exactly one lone pair.

The given compounds are: $\mathrm{O}_3,\ \mathrm{H_2O},\ \mathrm{SF_4},\ \mathrm{ClF_3},\ \mathrm{NH_3},\ \mathrm{BrF_5},\ \mathrm{XeF_4}$

2. Check each compound one by one.

(i) $\mathrm{O_3}$

Structure of ozone is bent with central O atom.

• Central O forms bonds with two O atoms.
• Central O has one lone pair.

So, $\mathrm{O_3}$ has one lone pair on central atom.

(ii) $\mathrm{H_2O}$

• Central atom = O
• Oxygen has 6 valence electrons.
• It forms 2 bonds with H, leaving 2 lone pairs.

So, $\mathrm{H_2O}$ does not have one lone pair.

(iii) $\mathrm{SF_4}$

• Central atom = S
• Sulfur has 6 valence electrons.
• It forms 4 bonds with F.
• One lone pair remains.

So, $\mathrm{SF_4}$ has one lone pair.

(iv) $\mathrm{ClF_3}$

• Central atom = Cl
• Chlorine has 7 valence electrons.
• It forms 3 bonds with F.
• Remaining electrons give 2 lone pairs.

So, $\mathrm{ClF_3}$ does not have one lone pair.

(v) $\mathrm{NH_3}$

• Central atom = N
• Nitrogen has 5 valence electrons.
• It forms 3 bonds with H.
• One lone pair remains.

So, $\mathrm{NH_3}$ has one lone pair.

(vi) $\mathrm{BrF_5}$

• Central atom = Br
• Bromine has 7 valence electrons.
• It forms 5 bonds with F.
• One lone pair remains.

So, $\mathrm{BrF_5}$ has one lone pair.

(vii) $\mathrm{XeF_4}$

• Central atom = Xe
• Xenon has 8 valence electrons.
• It forms 4 bonds with F.
• Remaining electrons correspond to 2 lone pairs.

So, $\mathrm{XeF_4}$ does not have one lone pair.

3. Compounds with exactly one lone pair on central atom are: $\mathrm{O_3},\ \mathrm{SF_4},\ \mathrm{NH_3},\ \mathrm{BrF_5}$

Total number: $4$

4. Comparison with stored answer:

• Derived answer = $4$
• Stored correct answer = $4$

Hence, the answer agrees with the stored correct answer.