Which of the following orders is not correct?

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(NONE) $1$. In $H_2O_2$ $(HO-OH)$, the $O-O$ bond length is $148 \ pm$, while in $F_2O_2$ $(FO-OF)$, the $O-O$ bond length is $122 \ pm$ due to strong

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$N-N < P-P$ (Single bond strength)

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(NONE) $1$. In $H_2O_2$ $(HO-OH)$, the $O-O$ bond length is $148 \ pm$, while in $F_2O_2$ $(FO-OF)$, the $O-O$ bond length is $122 \ pm$ due to strong repulsion between lone pairs on $F$ atoms. Thus, $H_2O_2 > F_2O_2$ is correct.$2$. In $PO_4^{3-}$, $HPO_3^{2-}$, and $H_2PO_2^-$, the number of $P-O$ bonds decreases, and the resonance stabilization changes. The bond order follows the sequence $PO_4^{3-} (1.25) $3$. The $N-N$ single bond is weaker than the $P-P$ single bond because of the high inter-electronic repulsion between the lone pairs of the small $N$ atoms. Thus, $N-N $4$. The extent of axial overlapping follows the order $2p-2p > 2s-2p > 2s-2s$ based on the directional nature and energy levels of orbitals. This is correct.Wait, re-evaluating: All options provided are actually correct statements. However, if forced to choose the most common 'incorrect' textbook statement, option $C$ is often debated due to bond energy values, but $N-N$ is indeed weaker. Given the standard chemistry context, all statements are factually correct. If this is a multiple-choice question where one must be wrong, there may be a typo in the source. Based on standard data, all are correct.

List-$I$	List-$II$
$(A)$ $C_{2}H_{4}$	$(I)$ $3\sigma$ bonds,$2\pi$ bonds
$(B)$ $C_{2}H_{2}$	$(II)$ $3\sigma$ bonds,$1$ lone pair
$(C)$ $CH_{4}$	$(III)$ $4\sigma$ bonds
$(D)$ $NH_{3}$	$(IV)$ $5\sigma$ bonds,$1\pi$ bond

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