P Block Elements — Concepts, Reactions & Solved Examples

What Are P Block Elements?

The p block is where chemistry gets genuinely interesting. These are the elements where the last electron enters a p orbital — covering Groups 13 to 18 of the periodic table. That’s 36 elements total, ranging from the very common (oxygen, carbon, nitrogen) to the exotic (xenon compounds, which shouldn’t even form bonds by old textbook logic).

The general electronic configuration is $ns^2np^{1-6}$ , where $n$ is the period number. This shared feature of valence electrons in p orbitals creates patterns — trends in oxidation states, melting points, reactivity — that the exam loves to test.

For your syllabus, p block is split across two years. Class 11 covers Groups 13 and 14 (Boron and Carbon families). Class 12 covers Groups 15, 16, 17, and 18 (Nitrogen, Oxygen, Halogen, and Noble Gas families). JEE Main asks 2-3 questions per paper; NEET asks 3-4. This is a high-weightage chapter you cannot afford to treat as rote memorisation.

Key Terms & Definitions

Inert pair effect — The reluctance of the 6s² electron pair (in heavier p block elements) to participate in bonding. This is why Tl shows +1, Pb shows +2, and Bi shows +3 as their more stable oxidation states, even though the group would predict +3, +4, +5 respectively. The 6s electrons are pulled closer to the nucleus by poor shielding from d and f electrons.

Anomalous behaviour of first element — The first member of each p block group behaves differently from the rest. Reasons: small size, high electronegativity, no d orbitals available for bonding, ability to form $p\pi$ - $p\pi$ multiple bonds.

Allotropy — The existence of an element in two or more physical forms in the same state. Carbon (diamond, graphite, fullerene), sulphur (rhombic, monoclinic), phosphorus (white, red, black) — all tested in board exams.

Disproportionation — A reaction where the same element simultaneously gets oxidised and reduced. Phosphorous acid $H_3PO_3$ disproportionates on heating:

4H_3PO_3 \rightarrow 3H_3PO_4 + PH_3\uparrow

Hybridisation — p block species show sp, sp², sp³, sp³d, sp³d² hybridisation. Knowing hybridisation directly gives you shape and bond angles — a question source that appears almost every year.

Oxoacids — Acids containing oxygen, a central atom, and OH groups. The p block forms a rich variety: $H_2SO_4$ , $HNO_3$ , $H_3PO_4$ , $HClO_4$ — their structures and acidic strength are heavily tested.

Group-by-Group Concepts

Group 13 — The Boron Family

Elements: B, Al, Ga, In, Tl

The dominant oxidation state is +3. But as we go down, the +1 state stabilises (inert pair effect). Tl⁺ is more stable than Tl³⁺.

Boron stands apart from the rest. It’s a metalloid, forms covalent compounds, and has no d orbitals — so its maximum covalency is 4 (not higher). Boron forms an electron-deficient compound $BF_3$ (sp² hybridised, trigonal planar), which acts as a Lewis acid because it readily accepts an electron pair.

BF_3 + :NH_3 \rightarrow F_3B \leftarrow NH_3

The nitrogen lone pair fills the empty p orbital of boron.

Aluminium is the star of Group 13 for reactions. Key ones to know:

Reacts with NaOH (aluminium is amphoteric): $2Al + 2NaOH + 2H_2O \rightarrow 2NaAlO_2 + 3H_2\uparrow$
Thermite reaction: $Fe_2O_3 + 2Al \rightarrow Al_2O_3 + 2Fe + \text{heat}$
Alums: $KAl(SO_4)_2 \cdot 12H_2O$ — double salt, used in water purification

Group 14 — The Carbon Family

Elements: C, Si, Ge, Sn, Pb

Carbon is unique: it forms $p\pi$ - $p\pi$ bonds with itself (chain formation, graphite’s delocalised system) and with N, O. No other Group 14 element can do this.

Allotropes of carbon are a guaranteed 1-mark question in CBSE:

Diamond: sp³, tetrahedral network, hardest natural substance, non-conductor
Graphite: sp², hexagonal layers, conductor (delocalised π electrons), used in electrodes
Fullerene (C₆₀): sp² like graphite, soccer-ball structure, hollow cage

Silicon forms $SiO_2$ (silica), which unlike $CO_2$ is a giant covalent solid (Si uses d orbitals to form extended Si–O–Si chains). This is why $CO_2$ is a gas at room temperature but $SiO_2$ is a hard solid.

Lead shows the inert pair effect clearly: $PbCl_4$ is a weaker oxidiser than $SnCl_4$ , and $PbO$ is more stable than $PbO_2$ .

Group 15 — The Nitrogen Family

Elements: N, P, As, Sb, Bi

Electronic configuration: $ns^2np^3$ — half-filled p subshell, which gives extra stability. This is why nitrogen is relatively inert and $N_2$ has a very high bond dissociation energy (941 kJ/mol, the highest for any diatomic molecule).

Nitrogen can show oxidation states from −3 to +5. The +5 state in nitrogen compounds is only possible without d orbitals involvement — it achieves +5 through ionic bonds, not expanded octet. This is a common exam trap.

JEE frequently asks: why does $NH_3$ have higher boiling point than $PH_3$ ? Answer: $NH_3$ forms stronger hydrogen bonds (N is more electronegative and smaller). $PH_3$ shows only weak van der Waals forces.

Phosphorus allotropes:

White phosphorus (P₄, tetrahedral, sp³, very reactive, stored under water)
Red phosphorus (polymeric, less reactive, used in matchboxes)
Black phosphorus (most stable, layered structure like graphite)

Oxoacids of phosphorus — tested very specifically:

Acid	Formula	Oxidation state of P	Basicity
Phosphoric acid	$H_3PO_4$	+5	3
Phosphorous acid	$H_3PO_3$	+3	2
Hypophosphorous acid	$H_3PO_2$	+1	1

The basicity (number of ionisable OH groups) decreases as P–H bonds replace P–OH bonds. $H_3PO_3$ and $H_3PO_2$ are reducing agents because they have P–H bonds.

Group 16 — The Oxygen Family (Chalcogens)

Elements: O, S, Se, Te, Po

Oxygen forms only two bonds (no d orbitals), while sulphur can expand its octet. This explains why $SF_6$ exists but $OF_6$ does not.

Sulphur allotropes tested in boards:

Rhombic sulphur ( $S_8$ , stable below 96°C, yellow crystals)
Monoclinic sulphur ( $S_8$ , stable 96–119°C, needle-shaped)

Oxoacids of sulphur — a favourite in JEE:

$H_2SO_3$ (sulphurous acid, +4) — reducing agent
$H_2SO_4$ (sulphuric acid, +6) — dehydrating agent, oxidising agent
$H_2S_2O_7$ (oleum/pyrosulphuric acid) — formed when $SO_3$ dissolves in $H_2SO_4$
$H_2S_2O_8$ (peroxodisulphuric acid / Marshall’s acid, +7 technically, has –O–O– linkage)

Group 17 — The Halogens

Elements: F, Cl, Br, I, At

All have the configuration $ns^2np^5$ — one electron short of a noble gas. They are the most reactive non-metals.

Fluorine is unique: highest electronegativity (3.98), smallest size, no d orbitals, so it shows only −1 and 0 oxidation states. It cannot form interhalogen compounds like $ClF_3$ where the central atom needs to expand its octet.

Interhalogen compounds: $XX'$ , $XX'_3$ , $XX'_5$ , $XX'_7$ — the heavier halogen is always the central atom. $IF_7$ has pentagonal bipyramidal geometry (sp³d³).

Oxyacids of chlorine — increasing oxidation state, increasing strength:

HOCl < HClO_2 < HClO_3 < HClO_4

Higher oxidation state = more electronegative Cl = pulls electron density from O–H = stronger acid.

Group 18 — Noble Gases

Elements: He, Ne, Ar, Kr, Xe, Rn

Xenon compounds are the only truly stable noble gas compounds. They form because xenon is large enough and polarisable enough to use d orbitals.

Key xenon fluorides:

$XeF_2$ : linear, sp³d (3 lone pairs), 2 bonds
$XeF_4$ : square planar, sp³d² (2 lone pairs), 4 bonds
$XeF_6$ : distorted octahedral, sp³d³ (1 lone pair), 6 bonds

$XeO_3$ has a pyramidal structure (sp³, 1 lone pair) — similar to $NH_3$ structurally but very different in chemistry. $XeO_3$ is a powerful oxidising agent.

Solved Examples

Example 1 — Easy (CBSE Level)

Q: Arrange the following in increasing order of bond angle: $NH_3$ , $NF_3$ , $NCl_3$ .

Solution:

The central atom N is sp³ in all three. The bond angle depends on the electronegativity of substituents — more electronegative atoms pull bonding pairs away from nitrogen, reducing repulsion between bond pairs, which decreases the bond angle.

Electronegativity order: F > Cl > H

So F pulls bond pairs farthest → greatest compression of bond angle in $NF_3$ .

N–H bonds stay close to N → less compression → largest bond angle in $NH_3$ .

Answer: $NF_3 < NCl_3 < NH_3$

Bond angles approximately: 102.5° < 107° < 107.3°

Example 2 — Medium (JEE Main Level)

Q: $H_3PO_3$ is a diprotic acid, not triprotic. Explain and write the equilibria. (JEE Main 2023 pattern)

Solution:

First, draw the structure. $H_3PO_3$ has phosphorus in +3 state. The structure shows:

Two –OH groups (ionisable)
One P–H bond (non-ionisable — H directly on P cannot leave as H⁺)

H_3PO_3 \rightleftharpoons H^+ + H_2PO_3^-

H_2PO_3^- \rightleftharpoons H^+ + HPO_3^{2-}

The third H cannot ionise because it’s bonded directly to phosphorus, not to oxygen. Only O–H bonds are ionisable in oxoacids.

Since it has P–H bond, $H_3PO_3$ is also a reducing agent — this is how NEET and JEE distinguish it from $H_3PO_4$ .

Example 3 — Hard (JEE Advanced Level)

Q: In the reaction of $XeF_6$ with water, one product is an acid and another is a noble gas compound. Write the balanced equation and identify the hybridisation of both xenon-containing species.

Solution:

$XeF_6$ undergoes hydrolysis:

XeF_6 + 3H_2O \rightarrow XeO_3 + 6HF

$XeF_6$ : 6 bonding pairs + 1 lone pair = 7 electron pairs → sp³d³ hybridisation, distorted octahedral shape (the lone pair makes it non-regular).

$XeO_3$ : 3 bonding pairs + 1 lone pair = 4 electron pairs → sp³ hybridisation, pyramidal shape (bond angle ~103°).

The HF produced makes the solution acidic. $XeO_3$ itself is a powerful oxidising agent and explodes on contact with organic material — this is tested as a one-line fact in NEET.

Exam-Specific Tips

CBSE Class 12 Marking Scheme: The p block II chapter (Groups 15-18) carries ~5 marks in theory. Expect one 3-mark question on structure/properties of $SF_4$ , $ClF_3$ , or $XeF_4$ , plus one 2-mark on oxyacid comparison. Always draw the structure — CBSE gives 1 mark just for the correct Lewis structure.

JEE Main Weightage: In the last 3 years, p block has appeared in 2-3 questions per paper consistently. High-frequency subtopics: oxyacids of S, N, P; interhalogen compounds; xenon fluorides; anomalous behaviour of first elements. NEET asks 3-4 questions; Group 17 (halogens) and Group 16 are most favoured.

SAT Chemistry: The SAT Subject Test focuses on trends — electronegativity, ionisation energy, atomic radius across the p block. Know which property increases left-to-right vs. top-to-bottom. The periodic trends questions here are conceptual, not calculation-heavy.

For NEET: Memorise the structures of $SF_4$ (see-saw), $ClF_3$ (T-shaped), $I_3^-$ (linear), $XeF_4$ (square planar) using VSEPR. These are 1-mark direct questions in NEET.

For JEE Advanced: Expect multi-step reasoning. “Which of the following does NOT show disproportionation?” requires knowing oxidation states AND whether intermediate states exist for that element.

Common Mistakes to Avoid

Mistake 1: Thinking $H_3PO_3$ is triprotic because it has three H atoms. Always count only the H atoms bonded to oxygen (P–H bonds don’t ionise). Draw the structure first, always.

Mistake 2: Writing $N_2O_5$ as the anhydride of $HNO_2$ . The anhydride of $HNO_2$ is $N_2O_3$ , and the anhydride of $HNO_3$ is $N_2O_5$ . Memorise: higher acid → higher anhydride oxide.

Mistake 3: Saying fluorine shows +1 oxidation state in $OF_2$ . Fluorine is always −1 (or 0 in $F_2$ ) — it’s the most electronegative element and NEVER positive. In $OF_2$ , oxygen is +2 and fluorine is −1.

Mistake 4: Confusing bond angle in $H_2O$ (104.5°) and $H_2S$ (92°). Students often say “H₂S has higher bond angle because S is bigger.” Wrong — S is bigger, so S–H bonds are longer, electron pairs are farther from nucleus, repulsion is LESS, so bond angle is smaller, not larger.

Mistake 5: Writing the structure of $HNO_3$ without the lone pair on N, then getting the oxidation state of N wrong. Draw $HNO_3$ properly: one N=O, one N–OH, one N→O (coordinate bond). Oxidation state of N = +5. If you draw two N=O bonds, you’ll get confused.

Practice Questions

Q1. Why is the bond angle in $H_2O$ (104.5°) greater than in $H_2S$ (92°)?

Oxygen is smaller than sulphur. The O–H bonds are short and the bonding electron pairs in $H_2O$ are close to the central atom, creating strong bond-pair–bond-pair repulsions. This pushes the bond angle above the tetrahedral 109.5° baseline (it’s reduced from 109.5° by lone pairs, but stays at 104.5°). In $H_2S$ , the S–H bonds are longer, bonding pairs are farther from S, repulsion between bond pairs is weaker, so the angle compresses to 92°.

Q2. Which of $H_3PO_4$ , $H_3PO_3$ , $H_3PO_2$ is the strongest reducing agent and why?

$H_3PO_2$ (hypophosphorous acid) is the strongest reducing agent. It has two P–H bonds (and only one P–OH group, so it’s monobasic). More P–H bonds = more easily oxidised = stronger reducing agent. The order is: $H_3PO_2 > H_3PO_3 > H_3PO_4$ . $H_3PO_4$ has no P–H bonds and is not a reducing agent at all.

Q3. Arrange $HF$ , $HCl$ , $HBr$ , $HI$ in increasing order of acidic strength in aqueous solution.

HF < HCl < HBr < HI

Even though $HF$ has the most polar bond, it’s the weakest acid in water because the H–F bond is so strong (bond dissociation enthalpy = 569 kJ/mol) that it doesn’t dissociate easily. As we go down the group, bond strength decreases more than the decrease in polarity, so acid strength increases. $HI$ is the strongest hydrohalic acid.

Q4. $SF_4$ has a see-saw shape. Explain using VSEPR theory.

S in $SF_4$ : sulphur has 6 valence electrons. It forms 4 bonds with F using 4 electrons. Remaining 2 electrons form 1 lone pair. Total electron pairs = 5 → sp³d hybridisation → trigonal bipyramidal arrangement of electron pairs. The lone pair occupies an equatorial position (where repulsion is minimised — equatorial lone pair has only 2 × 90° interactions vs. 3 × 90° for axial). Removing the lone pair from the shape picture, we get see-saw (or sawhorse) geometry with bond angles ~173° (axial F–S–F) and ~102° (equatorial F–S–F).

Q5. Explain why noble gases have very high ionisation energies compared to halogens.

Noble gases have completely filled valence shells ( $ns^2np^6$ ), which is an extremely stable configuration. The electrons are held very tightly by the nucleus with no half-filled or partially-filled orbitals to provide any “tendency” to lose an electron. Halogens, on the other hand, have one electron short of a noble gas configuration — they actually have a strong tendency to GAIN electrons. Their ionisation energies are high too, but noble gas ionisation energies are higher because removal of an electron from a fully-filled, spherically symmetric electron cloud requires breaking into a very stable arrangement.

Q6. $PCl_5$ exists but $NCl_5$ does not. Why?

To form 5 bonds, the central atom needs 5 unpaired electrons in its valence shell. Nitrogen’s valence shell is $n=2$ , which has only 2s and 2p orbitals — no d orbitals available. Nitrogen cannot expand its octet beyond 4 bonds. Phosphorus is in period 3 ( $n=3$ ), where 3d orbitals are available and energetically accessible for bonding. P can promote electrons to 3d and achieve 5 unpaired electrons, forming 5 bonds in $PCl_5$ .

Q7. Write the disproportionation reaction of chlorine in hot concentrated NaOH.

In hot concentrated NaOH, chlorine forms chlorate (not hypochlorite):

3Cl_2 + 6NaOH \xrightarrow{\Delta} 5NaCl + NaClO_3 + 3H_2O

Here Cl₂ (0 state) disproportionates: 5 atoms go to −1 (in NaCl) and 1 atom goes to +5 (in NaClO₃). In cold dilute NaOH, the product is NaOCl (hypochlorite, +1).

Q8. Why does $BiF_3$ exist but $BiF_5$ is less stable compared to $AsF_5$ ?

This is the inert pair effect. Bismuth is in Period 6. Its 6s² electron pair is very difficult to involve in bonding because d and f electrons below it provide poor shielding, making the 6s electrons more tightly bound. So the +3 state (using only p electrons) is more stable for Bi than the +5 state (using both s and p electrons). Arsenic is in Period 4 — no such strong inert pair effect — so $AsF_5$ (+5 state) is quite stable. $BiF_5$ does exist but is a strong fluorinating agent because the +5 state is unstable for Bi and it readily reverts to +3.

Frequently Asked Questions

Q: What is the difference between a Lewis acid and a Brønsted acid? How does boron relate?

A Lewis acid accepts an electron pair; a Brønsted acid donates a proton. $BF_3$ is a Lewis acid but NOT a Brønsted acid — it has no H to donate. This distinction matters because JEE asks “which of the following is a Lewis acid but not a Brønsted acid?” and $BF_3$ , $AlCl_3$ are the classic answers.

Q: Why is fluorine always −1 in its compounds?

Fluorine is the most electronegative element (Pauling scale: 3.98) and has no d orbitals to expand its octet. It cannot be the central atom in interhalogen compounds and cannot show positive oxidation states. Every compound fluorine forms, it pulls electron density towards itself to achieve −1. The only exception is $F_2$ itself where it’s 0.

Q: How do I remember the shape of all xenon compounds?

Use electron pairs: $XeF_2$ = 5 pairs (3 LP, 2 BP) = linear. $XeF_4$ = 6 pairs (2 LP, 4 BP) = square planar. $XeF_6$ = 7 pairs (1 LP, 6 BP) = distorted octahedral. $XeO_3$ = 4 pairs (1 LP, 3 BP) = pyramidal. Pattern: more lone pairs → more distortion.

Q: What is the difference between $H_2SO_3$ and sulphite ion?

$H_2SO_3$ (sulphurous acid) has S in +4 state. It’s a weak acid and also a reducing agent (gets oxidised to +6 in sulphate). The sulphite ion $SO_3^{2-}$ is its conjugate base. In reactions, $SO_3^{2-}$ acts as a reducing agent in acidic solution. Don’t confuse $H_2SO_3$ (sulphurous) with $H_2SO_4$ (sulphuric) in oxidation state questions.

Q: Is $H_3PO_3$ a diprotic acid or triprotic acid? This always confuses me.

Diprotic. The trick: only O–H bonds ionise in oxoacids. Draw the structure of $H_3PO_3$ — it has two O–H groups and one P–H bond. The P–H bond does not ionise. So only 2 protons are released → diprotic. Basicity = number of ionisable OH groups = 2.

Q: Why does $O_2$ exist as a diatomic molecule but sulphur exists as $S_8$ ?

Oxygen forms a stable $O=O$ double bond ( $p\pi$ - $p\pi$ ) because oxygen’s small size allows effective overlap of p orbitals. Sulphur is too large for effective $p\pi$ - $p\pi$ overlap, so forming a double bond is energetically unfavourable. Instead, sulphur satisfies its valency by forming single S–S bonds in a ring structure ( $S_8$ ), which is more stable than a double-bonded $S_2$ .

Q: In board exams, how many marks does p block typically carry?

In CBSE Class 12 boards, p block II (Groups 15-18) is part of Unit 3, which carries 9 marks. Expect 2-3 short answer questions (2-3 marks each). Common question types: structure of interhalogen compound, explain anomalous behaviour, write oxyacid structures. Class 11 p block I (Groups 13-14) appears in Unit 3 of Class 11 boards.