1. Identify what is being asked

We need the pair in which:

• one molecule is an odd electron molecule, and
• the other is an expanded octet molecule.

2. Recall definitions

(i) Odd electron molecule

A molecule having an odd total number of valence electrons is called an odd electron molecule. Example: $\mathrm{NO}$

(ii) Expanded octet molecule

A molecule in which the central atom has more than 8 electrons in its valence shell is called an expanded octet molecule. Examples: $\mathrm{SF_6}$, $\mathrm{H_2SO_4}$ (sulfur can have expanded valency in the usual Lewis description used in JEE level chemistry).

3. Check each relevant molecule

$\mathrm{BCl_3}$

Total valence electrons:

Boron forms 3 bonds and has only 6 electrons around it. So, $\mathrm{BCl_3}$ is electron-deficient, not odd-electron, and not expanded octet.

$\mathrm{NO}$

Total valence electrons:

This is odd, so $\mathrm{NO}$ is an odd electron molecule.

$\mathrm{SF_6}$

Sulfur forms 6 bonds with fluorine, so sulfur has 12 electrons around it. Hence, $\mathrm{SF_6}$ is an expanded octet molecule.

$\mathrm{H_2SO_4}$

In the standard Lewis structure used at this level, sulfur is bonded such that it accommodates more than 8 electrons. Hence, $\mathrm{H_2SO_4}$ is also treated as an expanded octet molecule.

4. Evaluate options

Option A: $\mathrm{BCl_3}$ and $\mathrm{SF_6}$

• $\mathrm{BCl_3}$: not odd-electron
• $\mathrm{SF_6}$: expanded octet

So this does not satisfy the pair.

Option B: $\mathrm{NO}$ and $\mathrm{H_2SO_4}$

• $\mathrm{NO}$: odd-electron molecule
• $\mathrm{H_2SO_4}$: expanded octet molecule

This satisfies the requirement.

Option C: $\mathrm{SF_6}$ and $\mathrm{H_2SO_4}$

• both are expanded octet molecules
• no odd-electron molecule

So this is incorrect.

Option D: $\mathrm{BCl_3}$ and $\mathrm{NO}$

• $\mathrm{BCl_3}$: electron-deficient
• $\mathrm{NO}$: odd-electron

No expanded octet molecule here, so incorrect.

5. Final answer

The correct pair is:

So, Option B is correct.