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Group 15 Oxidation States & Reactivity | chemca

Group 15 Oxidation States & Reactivity | chemca
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Inorganic Chemistry • Group 15

Oxidation States & Reactivity Trends

The Inert Pair Effect, N-N single bonds, and restricted covalency.

By chemca Team • Updated Sep 2026

Group 15 elements have five electrons in their valence shell ($ns^2 np^3$). They can either gain 3 electrons to achieve an octet or share/lose up to 5 electrons. This flexibility results in a wide variety of oxidation states, but their stability varies drastically as you move down the group due to atomic size and orbital shielding.

1. Common Oxidation States (-3, +3, +5)

A. The Negative State: -3

To show a -3 oxidation state, the atom must be small and highly electronegative to attract 3 electrons.

  • Trend: The tendency to exhibit a -3 oxidation state decreases rapidly down the group due to increasing atomic size and increasing metallic character.
  • Nitrogen: Readily shows -3 in compounds like Ammonia ($NH_3$) and nitrides ($Mg_3N_2$).
  • Bismuth: Hardly forms any compound in the -3 oxidation state.

B. The Positive States (+3 and +5) & The Inert Pair Effect

Group 15 elements can utilize their $p$ electrons (+3) or both their $s$ and $p$ electrons (+5) in bonding. Here lies one of the most frequently tested trends in the p-block.

The Trend down the group:
  • The stability of the +5 oxidation state decreases.
  • The stability of the +3 oxidation state increases.

Why? The Inert Pair Effect

As you move down the group (especially from As to Bi), the intervening $d$ and $f$ orbitals are filled. These orbitals provide very poor shielding of the nuclear charge. The nucleus pulls strongly on the $ns^2$ electrons, holding them tightly and making them reluctant to participate in bond formation. Thus, the $ns^2$ pair becomes "inert," and only the three $np^3$ electrons are easily available, making the +3 state highly stable for Bismuth.

Because Bi(V) is unstable, the only well-characterized Bi(V) compound is $BiF_5$. Bi(V) acts as a very strong oxidizing agent!
Stability Trends: The Inert Pair Effect N P As Sb Bi Stability of +5 State Increases UP Stability of +3 State Increases DOWN For Bi, the ns² electrons are pulled strongly by the poorly-shielded nucleus. They become inert, leaving only the np³ electrons available for bonding.

Figure 1: Opposing stability trends of the +3 and +5 oxidation states down Group 15.

2. Anomalous Properties of Nitrogen

Like the first member of every p-block group, Nitrogen differs drastically from the rest of its family. This is due to its small size, high electronegativity, high ionization enthalpy, and non-availability of d-orbitals.

A. $p\pi - p\pi$ Multiple Bonding

Nitrogen has the unique ability to form $p\pi - p\pi$ multiple bonds with itself and with other elements having small size and high electronegativity (like C, O). Thus, Nitrogen exists as a diatomic gas ($N \equiv N$).

Heavier elements of this group (like P, As) cannot form effective $p\pi - p\pi$ bonds because their atomic orbitals are too large and diffuse to have effective overlapping. Phosphorus exists as a solid $P_4$ molecule with single bonds.

B. The Weak N-N Single Bond & Catenation

While the $N \equiv N$ triple bond is exceptionally strong, the $N-N$ single bond is weaker than the $P-P$ single bond.

Why? Because the nitrogen atom is extremely small, the non-bonding electron pairs (lone pairs) on the two bonded nitrogen atoms are forced very close to each other. The resulting high interelectronic repulsion weakens the bond. Consequently, Nitrogen shows very little catenation (chain-forming) ability compared to Phosphorus.

C. Restricted Covalency (Why $NCl_5$ doesn't exist)

Nitrogen belongs to the second period, so its valence shell ($n=2$) contains only $2s$ and $2p$ orbitals. It completely lacks vacant d-orbitals.

  • The maximum covalency of Nitrogen is restricted to 4 (one s and three p orbitals), for example in the Ammonium ion ($NH_4^+$).
  • It cannot expand its octet. Therefore, while Phosphorus can form $PCl_5$ (by utilizing its 3d orbitals), Nitrogen cannot form $NCl_5$.

D. Disproportionation Reactions

Nitrogen exhibits all oxidation states from -3 to +5. In acid solution, almost all oxidation states from +1 to +4 tend to disproportionate (simultaneously oxidize and reduce).

$3HNO_2 \rightarrow HNO_3 + H_2O + 2NO$
($N$ goes from +3 in $HNO_2$ to +5 in $HNO_3$ and +2 in $NO$)

3. Reactivity Trends: Hydrides

All Group 15 elements form hydrides of the type $EH_3$ (like $NH_3, PH_3, AsH_3, SbH_3, BiH_3$). Down the group:

  • Bond dissociation enthalpy decreases: Size of E increases, so the E-H bond becomes longer and weaker.
  • Thermal Stability decreases: Because the bond is weaker, $BiH_3$ is the least stable.
  • Reducing Character increases: The E-H bond breaks easier, releasing Hydrogen, making $BiH_3$ the strongest reducing agent.
  • Basic Character decreases: $NH_3$ is the most basic. As you go down, the electron density of the lone pair is dispersed over a much larger volume, making it less available for donation.

Mastery Check: Oxidation States & Reactivity

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