Related Formula
Bond Length ∝ Atomic Size$$\text{Bond Length} \propto \text{Atomic Size}$$
Bond Strength ∝ 1Lone Pair-Lone Pair Repulsion (for small atoms)$$\text{Bond Strength} \propto \frac{1}{\text{Lone Pair-Lone Pair Repulsion (for small atoms)}}$$
Core Logic
Statement I: The N-N$\mathrm{N-N}$ single bond is weaker than the P-P$\mathrm{P-P}$ single bond due to strong non-bonding lone pair-lone pair repulsions arising from the small size of nitrogen. However, because nitrogen is smaller in atomic radius than phosphorus, the bond length of N-N$\mathrm{N-N}$ is shorter than P-P$\mathrm{P-P}$ (dN-N < dP-P$d_{\mathrm{N-N}} < d_{\mathrm{P-P}}$). Hence, Statement I is false.
Statement II: In group 15, only N$\mathrm{N}$ and P$\mathrm{P}$ in +3$+3$ oxidation state readily undergo disproportionation. Heavier elements (As, Sb, Bi$\mathrm{As}, \mathrm{Sb}, \mathrm{Bi}$) in +3$+3$ oxidation state are increasingly stable due to the inert pair effect and do not readily disproportionate. Hence, Statement II is false.
Step 1: Final Conclusion
Both Statement I and Statement II are false.
Pattern Recognition
N-N single bond: Weaker due to lp-lp repulsion, but SHORTER due to small atomic radius.
Group 15 +3 state: Disproportionation is prominent for N and P, not for heavier elements.
Chapter Mix
Class 12 Chemistry: p-Block Elements