1. We need to find the wrong statement among A, B, C, and D.

2. Option A: Nitrogen cannot form $d\pi - p\pi$ bond.

• Nitrogen belongs to the 2nd period.
• It has no vacant $d$-orbitals in its valence shell.
• Therefore, nitrogen cannot form $d\pi - p\pi$ bonding.

So, A is correct.

3. Option B: Single N–N bond is weaker than the single P–P bond.

• The $N$ atom is very small.
• In an $N-N$ single bond, lone pair-lone pair repulsions are significant because the atoms are close.
• Hence the $N-N$ single bond is unusually weak.
• $P-P$ bond is comparatively stronger than $N-N$ single bond.

So, B is correct.

4. Option C: $N_2O_4$ has two resonance structures.

Structure of $N_2O_4$ can be represented with resonance involving the $N-O$ bonds in each $NO_2$ unit. A simplified textbook statement is that it shows two equivalent major resonance forms.

So, C is correct.

5. Option D: The stability of hydrides increases from $NH_3$ to $BiH_3$ in group 15 of the periodic table.

For group 15 hydrides: $NH_3,\ PH_3,\ AsH_3,\ SbH_3,\ BiH_3$

Down the group:

• atomic size increases,
• $E-H$ bond strength decreases,
• therefore thermal stability decreases.

Hence the order of stability is: $NH_3 > PH_3 > AsH_3 > SbH_3 > BiH_3$

So the statement "stability increases from $NH_3$ to $BiH_3$" is wrong.

Therefore, D is the wrong statement.

6. Final Answer

The wrong statement is: $\boxed{D}$

This matches the stored correct answer.