Identify the incorrect statement from the following:

The statement “Nitrogen can form dπ− pπ bond with oxygen.” is incorrect because nitrogen, being a second‑period element, does not possess occupied d orbitals that can participate in π bonding.

**Why nitrogen lacks dπ orbitals**

• The valence shell of nitrogen is the 2nd shell (n = 2). The available orbitals are 2s and 2p. The 3d set belongs to the next principal quantum number (n��= 3) and is empty in the ground‑state atom.

• A dπ−pπ bond requires a filled or partially filled d orbital on one atom to overlap with a pπ orbital on the other atom. Since nitrogen has no d orbitals in its valence shell, it cannot donate a dπ electron pair.

**Contrast with the other statements**

1. $P(C$₂H₅)₃ and As(C₂H₅)₃ form dπ−dπ bonds with transition metals.
• Phosphorus (3rd period) and arsenic (4th period) have accessible 3d and 4d orbitals, respectively, allowing dπ‑dπ interactions with the d orbitals of transition metals.

2. $Nitrogen\; can\; form\; p\pi -p\pi \; multiple\; bonds\; with\; itself.$
• The classic example is the nitrogen‑nitrogen triple bond in N≡N (dinitrogen). Here each nitrogen contributes a pπ orbital to form two π bonds in addition to a σ bond.

3. $Phosphorus,\; arsenic\; and\; antimony\; show\; catenation\; property.$
• These elements form stable element‑element single bonds (P–P, As–As, Sb–Sb) because their larger atomic size and lower bond dissociation energies favor chain formation.

Since nitrogen cannot provide a dπ orbital, the claim that it can form a dπ−pπ bond with oxygen is false, making it the incorrect statement among the options.