Disproportionation and Bonding Properties of Nitrogen Group Elements

To determine the correct statements among the given options, let us analyze each statement individually based on standard inorganic chemistry principles (NCERT Group 15 elements):

1. Statement A: Nitrogen in oxidation states from $+1$ to $+4$ disproportionates in acid medium.

   • Analysis: Nitrogen exhibits oxidation states ranging from $-3$ to $+5$. In acidic solution, all nitrogen species with oxidation states from $+1$ to $+4$ tend to disproportionate into species with higher and lower oxidation states. For example, nitrous acid ($\text{HNO}_2$, where $\text{N}$ is in $+3$ state) disproportionates as: $3\text{HNO}_2 \rightarrow \text{HNO}_3 + \text{H}_2\text{O} + 2\text{NO}$
   • Status: Correct

2. Statement B: Nitrogen has the ability to form $d\pi - p\pi$ multiple bonds with itself and other elements with small size and high electronegativity.

   • Analysis: Nitrogen belongs to the second period and lacks vacant $d$-orbitals in its valence shell ($n=2$). Thus, nitrogen forms $p\pi - p\pi$ multiple bonds with itself ($\text{N}\equiv\text{N}$) and with other small, highly electronegative elements like carbon ($\text{C}\equiv\text{N}$) and oxygen ($\text{N}=\text{O}$), not $d\pi - p\pi$ bonds.
   • Status: Incorrect

3. Statement C: N-N single bond is stronger than P-P single bond.

   • Analysis: The single $\text{N-N}$ bond is weaker than the single $\text{P-P}$ bond because the short $\text{N-N}$ bond length leads to high interelectronic repulsion between the non-bonding lone pair electrons on the small nitrogen atoms.
   • Status: Incorrect

4. Statement D: Nitrogen has highest density in its group due to small size.

   • Analysis: Down Group 15 ($\text{N, P, As, Sb, Bi}$), density increases due to a significant increase in atomic mass relative to atomic volume. Nitrogen, being a gas at room temperature, has the lowest density in its group.
   • Status: Incorrect

5. Statement E: The maximum covalency of nitrogen is four since it has only four valence orbitals for bonding.

   • Analysis: Nitrogen has a valence electron configuration of $2s^2 2p^3$. The four available valence orbitals (one $2s$ and three $2p$) limit its maximum covalency to $4$ (as seen in $\text{NH}_4^+$), because it cannot expand its octet due to the absence of $d$-orbitals.
   • Status: Correct

Therefore, the correct statements are A and E Only.

Correct Option: D