JEE Mains Chemistry · The p-Block Elements

Group 15: Periodic Trends and Hydrides

Nitrogen, phosphorus, arsenic, antimony and bismuth (ns² np³) change from non-metal to metal down the group; nitrogen is the exception in almost every property, and the hydrides EH₃ lose stability and basicity while gaining reducing power.

All The p-Block Elements notes JEE Mains Chemistry strategy

Why this matters

Twenty-one PYQs, twenty of them multiple choice, and six from 2026, more than any other page of the chapter. Twelve test the group trends and nitrogen's anomalies: electronegativity, the N–N bond, maximum covalency, oxidation states, the halides and the oxides; nine compare the hydrides from NH₃ to BiH₃ by stability, basicity, reducing power, bond angle and boiling point.

Concept 1 of 2: Group 15 trends and the anomalous behaviour of nitrogen

Nitrogen is small, very electronegative and has no d orbitals. Each of these explains an exception. Being small, it forms strong pπ–pπ multiple bonds, so it exists as N≡N while phosphorus exists as P₄. Its lone pairs sit close together, so the N–N single bond is weaker than P–P even though it is shorter. Having only four valence orbitals (2s and three 2p), it can form at most four bonds, so NCl₅ cannot exist while PCl₅ does. Down the group the atoms grow, but from arsenic to bismuth the growth is small because filled d and f shells shield poorly.

Definition

Radius: a big jump from N to P, only a small rise from As to Bi.
Oxidation states: −3, +3, +5 are common. Down the group +5 becomes less stable and +3 more stable (inert pair). Bismuth is the only metal.
Nitrogen: maximum covalency 4; forms $p\pi$ – $p\pi$ multiple bonds with itself and with small electronegative atoms (C, O); no pentahalides; NF₃ is its only stable trihalide.
N–N single bond is weaker than P–P (lone-pair repulsion) but shorter.
Disproportionation: nitrogen in +1 to +4 disproportionates in acid. For As, Sb and Bi the +3 state is stable to disproportionation.
P and As (not N) form $d\pi$ – $d\pi$ bonds with transition metals, for example as $\mathrm{P(C_2H_5)_3}$ ligands.
Oxides: for one element $\mathrm{E_2O_5}$ is more acidic than $\mathrm{E_2O_3}$ . Down the group the acidity falls: oxides of N and P acidic, As and Sb amphoteric, Bi basic.
Halides: $\mathrm{SbCl_5}$ is more covalent than $\mathrm{SbCl_3}$ ; the higher oxidation state polarises chloride more.

Element	Covalent radius (pm)	First ionisation enthalpy (kJ/mol)	Electronegativity	Character
N	70	1402	3.0	Non-metal, diatomic gas $\mathrm{N_2}$ No d orbitals: maximum covalency 4.
P	110	1012	2.1	Non-metal, $\mathrm{P_4}$ molecules
As	121	947	2.0	Metalloid
Sb	141	834	1.9	Metalloid
Bi	148	703	1.9	Metal, the only one in the group

The biggest steps are between N and P in every column; below arsenic the changes are small.

Practice this conceptself-check · 4 quick reps

The same idea in a real exam question:

JEE Mains · 2026 · 5 Apr 2026 Shift 1 · Q34Moderate

Example 1 · The p-Block Elements · Group 15: Periodic Trends and Hydrides

Correct statements from the following are : (A) Nitrogen in oxidation states from +1 to +4 disproportionate in acid medium. (B) Nitrogen has the ability of form

d\pi - p\pi

multiple bonds with itself and other elements with small size and high electronegativity. (C) N-N single bond is stronger than P-P single bond. (D) Nitrogen has highest density in its group due to small size. (E) The maximum covalency of nitrogen is four since it has only four valence orbitals for bonding. Choose the correct answer from the options given below :

The N–N single bond is weaker AND shorter than P–P

Nitrogen's small size makes the N–N bond short, but the lone pairs on the two nitrogens are then close and repel, so the bond is weaker than P–P. A statement that it is 'weaker and longer' or 'stronger' is false.

The +5 state becomes LESS stable down group 15

Because of the inert pair effect, +3 grows more stable and +5 less stable from N to Bi. Bismuth(V) is a strong oxidising agent.

Nitrogen's multiple bonds are pπ–pπ

Nitrogen has no d orbitals, so it cannot form

d\pi

–

p\pi

bonds. Its maximum covalency is 4, never 5 or 6.

Drill 11 more on group 15 trends and the anomalous behaviour of nitrogen

Concept 2 of 2: Hydrides of group 15 from NH₃ to BiH₃

Down the group the central atom grows, so the E–H bond gets longer and weaker. A weaker bond means a less stable hydride that gives up hydrogen more easily, so the reducing power rises from NH₃ to BiH₃. The lone pair also spreads over a bigger atom, so the hydride holds a proton less well and basicity falls. The bond angle shrinks towards 90° because the heavier atoms use almost pure p orbitals. Boiling points follow size, except that ammonia is lifted by hydrogen bonding.

Definition

Thermal stability: $\mathrm{NH_3 > PH_3 > AsH_3 > SbH_3 > BiH_3}$ .
Reducing power: $\mathrm{NH_3 < PH_3 < AsH_3 < SbH_3 < BiH_3}$ ; $\mathrm{BiH_3}$ is the strongest reducing agent.
Basicity: $\mathrm{NH_3 > PH_3 > AsH_3 > SbH_3 > BiH_3}$ .
Bond angle falls from 107.8° in $\mathrm{NH_3}$ to about 91° in $\mathrm{SbH_3}$ .
Boiling point: $\mathrm{PH_3 < AsH_3 < NH_3 < SbH_3}$ . Ammonia is hydrogen bonded in the liquid; phosphine is not.

Hydride	H–E–H angle (°)	Boiling point (K)	E–H bond enthalpy (kJ/mol)	Character
$\mathrm{NH_3}$	107.8	238.5	389	Most stable and most basic; weakest reducing agent; hydrogen bonded
$\mathrm{PH_3}$	93.6	185.5	322	Lowest boiling point in the group: no hydrogen bonding and a small molar mass The lowest boiling point is PH₃, not NH₃.
$\mathrm{AsH_3}$	91.8	210.6	297	Less basic and more reducing than $\mathrm{PH_3}$
$\mathrm{SbH_3}$	91.3	254.6	255	Highest boiling point of the four: the largest dispersion forces

BiH₃, not listed because it is too unstable to measure well, continues every trend: least stable, least basic, strongest reducing agent.

Practice this conceptself-check · 4 quick reps

The same idea in a real exam question:

JEE Mains · 2026 · 8 Apr 2026 Shift 2 · Q34Moderate

Example 2 · The p-Block Elements · Group 15: Periodic Trends and Hydrides

Find the correct statements related to group 15 hydrides. (A) Reducing nature increases from

{NH}_{3}

{BiH}_{3}

(B) Tendency to donate lone pair of electrons decreases from

{NH}_{3}

{BiH}_{3}

{NH}_{3}

{BiH}_{3}

(D) HEH bond angle decreases from

{NH}_{3}

{SbH}_{3}

(

E =

Elements of group 15) Choose the correct answer from the options given below :

Boiling point does not rise steadily down group 15

Phosphine boils lowest. Ammonia is raised by hydrogen bonding, so the order is

\mathrm{PH_3 < AsH_3 < NH_3 < SbH_3}

, not a steady rise from NH₃.

Ammonia is the weakest reducing agent, not the strongest

The N–H bond is the strongest E–H bond in the group, so ammonia gives up hydrogen least easily. Reducing power rises down the group to

\mathrm{BiH_3}

Basicity decreases down group 15

The lone pair on a large atom such as Sb or Bi is spread out and held loosely in a large orbital, so it binds a proton poorly. Basicity falls from

\mathrm{NH_3}

\mathrm{BiH_3}

Drill 8 more on hydrides of group 15 from nh₃ to bih₃

Summary — formulas & gotchas at a glance

A revision cheat-sheet for the formulas and gotchas above. Click any concept name to jump back to its full explanation.

Reference tables (2)

Group 15 trends and the anomalous behaviour of nitrogen5 rows

Element	Covalent radius (pm)	First ionisation enthalpy (kJ/mol)	Electronegativity	Character
N	70	1402	3.0	Non-metal, diatomic gas $\mathrm{N_2}$ No d orbitals: maximum covalency 4.
P	110	1012	2.1	Non-metal, $\mathrm{P_4}$ molecules
As	121	947	2.0	Metalloid
Sb	141	834	1.9	Metalloid
Bi	148	703	1.9	Metal, the only one in the group

The biggest steps are between N and P in every column; below arsenic the changes are small.

Hydrides of group 15 from NH₃ to BiH₃4 rows

Hydride	H–E–H angle (°)	Boiling point (K)	E–H bond enthalpy (kJ/mol)	Character
$\mathrm{NH_3}$	107.8	238.5	389	Most stable and most basic; weakest reducing agent; hydrogen bonded
$\mathrm{PH_3}$	93.6	185.5	322	Lowest boiling point in the group: no hydrogen bonding and a small molar mass The lowest boiling point is PH₃, not NH₃.
$\mathrm{AsH_3}$	91.8	210.6	297	Less basic and more reducing than $\mathrm{PH_3}$
$\mathrm{SbH_3}$	91.3	254.6	255	Highest boiling point of the four: the largest dispersion forces

BiH₃, not listed because it is too unstable to measure well, continues every trend: least stable, least basic, strongest reducing agent.

Watch out for (6)

The N–N single bond is weaker AND shorter than P–P
→ Group 15 trends and the anomalous behaviour of nitrogen
The +5 state becomes LESS stable down group 15
→ Group 15 trends and the anomalous behaviour of nitrogen
Nitrogen's multiple bonds are pπ–pπ
→ Group 15 trends and the anomalous behaviour of nitrogen
Boiling point does not rise steadily down group 15
→ Hydrides of group 15 from NH₃ to BiH₃
Ammonia is the weakest reducing agent, not the strongest
→ Hydrides of group 15 from NH₃ to BiH₃
Basicity decreases down group 15
→ Hydrides of group 15 from NH₃ to BiH₃

Test yourself on The p-Block Elements

20 past JEE Mains questions from this chapter, timed at 48 minutes and marked the way the exam marks it. You see your score and every answer the moment you finish. Free to start.

Take the The p-Block Elements chapter test Drill all 21 questions