Analyze Statement (II) first

We know the directing effects in electrophilic aromatic substitution:

• $-\mathrm{CH_3}$ is an electron-donating group by hyperconjugation and inductive effect, so it is ortho/para-directing.
• $-\mathrm{NO_2}$ is a strong electron-withdrawing group, so it is meta-directing.

Therefore, Statement (II) is true.

Now analyze Statement (I)

The substrate is m-xylene, i.e. $1,3$-dimethylbenzene.

Its structure has methyl groups at positions 1 and 3:

$\mathrm{C_6H_4(CH_3)_2}$

We now nitrate it using $\mathrm{HNO_3/H_2SO_4}$.

Directing effects during nitration

Each $-\mathrm{CH_3}$ group is ortho/para-directing.

For m-xylene, possible nitration positions are 2, 4, 5, and 6.

• Position 2: ortho to both methyl groups → sterically crowded
• Position 4: para to one methyl and ortho to the other
• Position 6: similar to position 2 by symmetry considerations
• Position 5: meta to one methyl and ortho/para relation less favored overall

The major nitration product is formed at position 4, giving 4-nitro-m-xylene (equivalently 2-nitro-1,3-dimethylbenzene depending on numbering convention).

Oxidation step

Strong oxidation converts each benzylic $-\mathrm{CH_3}$ group into $-\mathrm{COOH}$.

So:

$\text{4-nitro-1,3-dimethylbenzene} \xrightarrow[\text{oxidation}]{} \text{4-nitrobenzene-1,3-dicarboxylic acid}$

Hence, Statement (I) is also true.

Conclusion

• Statement (I): True
• Statement (II): True

Therefore, the correct option is:

$\boxed{\text{B}}$

Comparison with stored correct answer

Stored correct answer = B

Our derived answer = B

So, they agree.