Subtopics - s and p Block Elements (NEET)
Seven topic blocks: s-block alkali and alkaline earth metals, hydrogen and its compounds, Group 13 boron family, Group 14 carbon family, Group 15 nitrogen family, Group 16 oxygen family, and Groups 17-18 halogens and noble gases.
1) s-Block Elements: Alkali and Alkaline Earth Metals
Group 1 (ns1) alkali metals and Group 2 (ns2) alkaline earth metals. Properties: low IE, high electropositive character, soft metals with low density. Flame colours: Li crimson, Na golden yellow, K pale violet, Ca brick red, Sr blood red, Ba apple green. Lithium shows diagonal relationship with magnesium. Oxides: Li forms Li2O, Na forms Na2O2 (peroxide), K forms KO2 (superoxide). Hydration energy decreases down group; Li+ most hydrated.
2) Hydrogen: Isotopes, Hydrides, and Hydrogen Peroxide
Hydrogen has three isotopes: protium (1H, 99.985%), deuterium (2H), tritium (3H, radioactive). Three types of hydrides: ionic (NaH, CaH2), covalent (NH3, H2O), metallic/interstitial (PdH). Hydrogen peroxide (H2O2): open-book structure, O-O bond 1.48 A, bond angle 101.5 degrees. Acts as both oxidising and reducing agent. Decomposes to H2O + O2.
3) Group 13: Boron Family and Group 14: Carbon Family
Group 13 (ns2np1): B is non-metal, Al-Tl are metals. Inert pair effect: stability of +1 oxidation state increases from Ga to Tl. Boron compounds: borax (Na2B4O7.10H2O) and diborane (B2H6, 3-centre 2-electron banana bonds). Group 14 (ns2np2): C is non-metal, Si/Ge metalloids, Sn/Pb metals. Carbon allotropes (diamond sp3, graphite sp2, fullerene). Silicon compounds: silicones, silicates, SiO2.
4) Group 15: Nitrogen Family
N, P are non-metals; As, Sb metalloids; Bi metal. Nitrogen exists as N2 (triple bond). Phosphorus as P4 (tetrahedral). Oxidation states: +5, +3, -3. N cannot show +5 (no d-orbitals). Ammonia (sp3, tetrahedral with lone pair, 107.8 degrees), nitric acid (Ostwald process), and phosphorus oxyacids (H3PO4 tribasic, H3PO3 dibasic, H3PO2 monobasic).
5) Group 16: Oxygen Family and Sulphur Compounds
O is gas, S/Se/Te solids, Po radioactive. Oxygen is paramagnetic (unpaired electrons in pi-antibonding MO). Ozone (O3): bent, 116.8 degrees, powerful oxidiser. Sulphur allotropes: rhombic S8 (crown), monoclinic. H2SO4 (contact process), Na2S2O3 (hypo, photography). Sulphur has high catenation tendency.
6) Group 17: Halogens and Interhalogen Compounds
F, Cl, Br, I have ns2np5 configuration. F is the most electronegative element and shows -1 oxidation state only. Other halogens show variable oxidation states (+1, +3, +5, +7). Reactivity: F2 > Cl2 > Br2 > I2. Bond energy: F-F unusually low (159 kJ/mol) due to lone pair repulsion. Oxyacids of chlorine: HClO4 > HClO3 > HClO2 > HOCl (acid strength). Interhalogen compounds: AB, AB3, AB5, AB7 types.
7) Group 18: Noble Gases and Xenon Compounds
Noble gases have stable ns2np6 configuration (He: 1s2). Very high IE, zero EA. Weak van der Waals forces give low MP/BP. Xenon forms compounds with F and O because Xe has low IE and F is highly electronegative. XeF2 (linear, sp3d), XeF4 (square planar, sp3d2), XeF6 (distorted octahedral, sp3d3).
s and p Block Elements Download Notes & Weightage Plan
For each topic in the s and p Block Elements chapter below, you get (2) the exact resources to download and how to use them, and (3) a simple importance & time plan so NEET students know what to do first and what to revise last.
s-Block Elements: Alkali and Alkaline Earth Metals
Group 1 and 2 properties, trends, anomalies, diagonal relationship, oxide types.
1) Download Packs For This Topic (And How To Use Them)
Don't download everything and forget it. Use these like a small "attack kit": read → highlight → test → revise the same sheet again.
2) Importance, Weightage & Time Allocation (Practical)
Use this to avoid over-studying. This topic is usually low effort, quick return if your recall is clean.
- Scoring Focus: Li-Mg diagonal relationship: both form nitrides directly, both carbonates decompose on heating (unlike Na2CO3 which does not decompose). Flame colour of Na is golden yellow (most tested).
- High-risk Area: Mixing up which alkali metal forms which oxide. Li = monoxide only, Na = peroxide (Na2O2), K = superoxide (KO2). Students reverse Na and K.
- Best Practice Style: Small cation (Li) stabilises small anion (O2-). Large cation (K, Rb, Cs) stabilises large anion (O2-).
Hydrogen: Isotopes, Hydrides, and Hydrogen Peroxide
Three isotopes, three hydride types, H2O2 structure and dual nature.
1) Download Packs For This Topic (And How To Use Them)
Don't download everything and forget it. Use these like a small "attack kit": read → highlight → test → revise the same sheet again.
2) Importance, Weightage & Time Allocation (Practical)
Use this to avoid over-studying. This topic is usually low effort, quick return if your recall is clean.
- Scoring Focus: H2O2 is both an oxidising and reducing agent depending on the reagent. With KMnO4 it acts as reducing agent (decolourises). With KI it acts as oxidising agent.
- High-risk Area: Forgetting that H2O2 has an open-book (non-planar) structure. Students often draw it as planar. The two O-H bonds are in different planes with a dihedral angle of about 111.5 degrees in the gas phase.
- Best Practice Style: H2O2 = non-planar, open book. The O-O bond is a single bond (1.48 A, longer than O=O at 1.21 A).
Group 13: Boron Family and Group 14: Carbon Family
Boron compounds (borax, diborane), inert pair effect, carbon allotropes, silicones.
1) Download Packs For This Topic (And How To Use Them)
Don't download everything and forget it. Use these like a small "attack kit": read → highlight → test → revise the same sheet again.
2) Importance, Weightage & Time Allocation (Practical)
Use this to avoid over-studying. This topic is usually low effort, quick return if your recall is clean.
- Scoring Focus: Diborane (B2H6) has 12 valence electrons but needs 14 for 7 conventional bonds. The deficit is resolved by 2 three-centre two-electron banana bonds. Each bridge bond uses one H and two B atoms sharing 2 electrons across three nuclei.
- High-risk Area: Thinking boron has d-orbitals for expansion. B has no d-orbitals and maximum covalency is 4 (not 5). Electron deficiency in B compounds is resolved by multicentre bonding, not octet expansion.
- Best Practice Style: B: electron deficient, forms 3c-2e bonds. C: forms 4 bonds, maximum catenation. Si: 3d available, can expand octet (SiF6 2-).
Nitrogen compounds (NH3, HNO3), phosphorus allotropes and oxyacids.
1) Download Packs For This Topic (And How To Use Them)
Don't download everything and forget it. Use these like a small "attack kit": read → highlight → test → revise the same sheet again.
2) Importance, Weightage & Time Allocation (Practical)
Use this to avoid over-studying. This topic is usually low effort, quick return if your recall is clean.
- Scoring Focus: Basicity of phosphorus oxyacids is determined by the number of P-OH bonds, not total H atoms. H3PO3 has 3 H atoms but only 2 are ionisable (two P-OH groups). The third H is directly bonded to P (P-H bond, non-ionisable). So H3PO3 is dibasic.
- High-risk Area: Assuming all H atoms in phosphorus oxyacids are ionisable. Only the H atoms in P-OH bonds contribute to basicity. H atoms in P-H bonds do not ionise. This is the single most tested point from Group 15.
- Best Practice Style: Draw the structure. Count P-OH bonds. That number = basicity.
Group 16: Oxygen Family and Sulphur Compounds
O2 paramagnetism, ozone, H2SO4 contact process, Na2S2O3.
1) Download Packs For This Topic (And How To Use Them)
Don't download everything and forget it. Use these like a small "attack kit": read → highlight → test → revise the same sheet again.
2) Importance, Weightage & Time Allocation (Practical)
Use this to avoid over-studying. This topic is usually low effort, quick return if your recall is clean.
- Scoring Focus: Contact process catalyst is V2O5 (vanadium pentoxide) at 450 C. SO3 is NOT dissolved directly in water (forms dense mist). Instead SO3 is absorbed in H2SO4 to form oleum (H2S2O7), which is then diluted.
- High-risk Area: Thinking SO3 is absorbed directly in water to make H2SO4. In practice, SO3 + H2O creates a mist that is very difficult to condense. Instead, SO3 + H2SO4 -> H2S2O7 (oleum), then H2S2O7 + H2O -> 2H2SO4.
- Best Practice Style: Contact process: S -> SO2 -> SO3 (V2O5) -> oleum -> H2SO4.
Group 17: Halogens and Interhalogen Compounds
Halogen reactivity, F-F bond weakness, oxyacids of chlorine, interhalogen types.
1) Download Packs For This Topic (And How To Use Them)
Don't download everything and forget it. Use these like a small "attack kit": read → highlight → test → revise the same sheet again.
2) Importance, Weightage & Time Allocation (Practical)
Use this to avoid over-studying. This topic is usually low effort, quick return if your recall is clean.
- Scoring Focus: Oxyacids of chlorine: more O atoms = more electron withdrawal from O-H bond = weaker O-H = stronger acid. HClO4 (3 extra O on Cl) is the strongest. HOCl (no extra O) is the weakest. This pattern applies to oxyacids of all nonmetals.
- High-risk Area: Thinking HF is a strong acid because F is the most electronegative element. HF is actually a weak acid in dilute aqueous solution because the H-F bond (568 kJ/mol) is very strong and difficult to break. HI is the strongest HX acid because the H-I bond is weakest.
- Best Practice Style: Acid strength of HX: inversely proportional to bond strength. HI > HBr > HCl >> HF.
Group 18: Noble Gases and Xenon Compounds
Noble gas properties and xenon fluoride structures.
1) Download Packs For This Topic (And How To Use Them)
Don't download everything and forget it. Use these like a small "attack kit": read → highlight → test → revise the same sheet again.
2) Importance, Weightage & Time Allocation (Practical)
Use this to avoid over-studying. This topic is usually low effort, quick return if your recall is clean.
- Scoring Focus: XeF4 is square planar (not tetrahedral) because of 2 lone pairs on Xe in axial positions of an octahedral arrangement (sp3d2). XeF2 is linear (not bent) because 3 lone pairs occupy the equatorial positions of a TBP arrangement (sp3d).
- High-risk Area: Predicting XeF4 as tetrahedral. XeF4 has 6 electron pairs around Xe (4 bond pairs + 2 lone pairs). The 2 lone pairs go trans to each other (axial), leaving 4 F atoms in the equatorial plane. Result: square planar, NOT tetrahedral.
- Best Practice Style: Count total electron pairs on Xe. Subtract bond pairs to get lone pairs. Apply VSEPR: lone pairs go where they minimise repulsion.
s and p Block Elements Chapter NEET Traps & Common Mistakes (Topic-Wise)
Each subtopic below is of the s and p Block Elements chapter and shows what NEET students usually do wrong in NEET examination, a short example of the mistake, and how NEET frames the question to trick you with close options are given below.
Mistake Snapshot (What Students Do Wrong)
- Counting all H atoms as ionisable: H3PO3 has 3 hydrogen atoms but only 2 are ionisable because only 2 are in P-OH bonds. The third H is directly bonded to P (P-H bond) and does not ionise.
- Thinking H3PO2 is tribasic: H3PO2 (hypophosphorous acid) has 3 H atoms but only 1 P-OH bond. The other 2 H atoms are in P-H bonds. It is monobasic.
H3PO3 (phosphorous acid) has 2 P-OH bonds and 1 P-H bond. Only the 2 P-OH hydrogens are replaceable by a base. Therefore H3PO3 is dibasic, not tribasic. When neutralised with NaOH: H3PO3 + 2NaOH -> Na2HPO3 + 2H2O.
How NEET Frames The Trap
NEET asks the basicity of a phosphorus oxyacid.
Q. The basicity of H3PO3 (phosphorous acid) is
A. 2 B. 3 C. 1 D. 0
Trick: 2 (Option A): H3PO3 has 2 P-OH bonds and 1 P-H bond. Only the H atoms in P-OH groups are ionisable. Option B incorrectly counts all 3 H atoms as ionisable. The P-H bond does not dissociate to give H+.
Mistake Snapshot (What Students Do Wrong)
- Predicting XeF4 as tetrahedral: XeF4 has 4 bond pairs + 2 lone pairs = 6 electron pairs (octahedral arrangement). The 2 lone pairs go trans to minimise repulsion, making the shape square planar, not tetrahedral.
- Predicting XeF2 as bent: XeF2 has 2 bond pairs + 3 lone pairs = 5 electron pairs (trigonal bipyramidal). The 2 F atoms go in axial positions (lone pairs equatorial), giving a linear shape.
XeF4: Xe has 8 valence electrons. 4 form bonds with F (4 bond pairs). 4 remain as 2 lone pairs. Total 6 electron pairs = octahedral geometry. The 2 lone pairs occupy opposite axial positions to minimise repulsion. The 4 F atoms lie in the equatorial plane: square planar shape.
How NEET Frames The Trap
NEET asks the geometry or hybridisation of XeF2, XeF4, or XeF6.
Q. The shape of XeF4 is
A. Square planar B. Tetrahedral C. See-saw D. Square pyramidal
Trick: Square planar (Option A): sp3d2 hybridisation, 4 bond pairs + 2 lone pairs in octahedral arrangement. Lone pairs trans to each other. Tetrahedral (Option B) would require no lone pairs. See-saw (Option C) is for 4 BP + 1 LP.
Mistake Snapshot (What Students Do Wrong)
- Thinking HOCl is the strongest because Cl is most electronegative: Acid strength depends on how easily the O-H bond breaks. More oxygen atoms on Cl pull electron density away from O-H, weakening it. HOCl has no extra O atoms, so the O-H bond is strongest (least acidic).
- Confusing oxidation state with acid strength: Higher oxidation state of Cl correlates with more O atoms and thus stronger acid, but the reason is electron withdrawal, not oxidation state directly.
HOCl has Cl bonded to one O (carrying H). HClO4 has Cl bonded to 4 O atoms (one carrying H). The 3 extra O atoms in HClO4 pull electron density from the O-H bond via inductive effect, making H+ easier to release. Therefore HClO4 is the strongest and HOCl is the weakest.
How NEET Frames The Trap
NEET asks to arrange oxyacids in order of acid strength.
Q. The correct order of acid strength of oxyacids of chlorine is
A. HClO4 > HClO3 > HClO2 > HOCl B. HOCl > HClO2 > HClO3 > HClO4 C. HClO3 > HClO4 > HClO2 > HOCl D. HClO2 > HOCl > HClO3 > HClO4
Trick: HClO4 > HClO3 > HClO2 > HOCl (Option A): acid strength increases with the number of oxygen atoms bonded to chlorine. Option B reverses the order. More oxygen = more electron withdrawal = weaker O-H bond = stronger acid.
Mistake Snapshot (What Students Do Wrong)
- Assuming HF is the strongest hydrohalic acid because F is most electronegative: Acid strength of binary acids in a group depends on bond strength, not electronegativity. The H-F bond (568 kJ/mol) is very strong and difficult to break. HI has the weakest H-X bond and is the strongest acid.
- Confusing electronegativity trend with acid strength trend: Electronegativity increases F > Cl > Br > I, but acid strength of HX goes in the opposite direction: HI > HBr > HCl > HF. Down the group, bond length increases and bond strength decreases, making the acid stronger.
HF has the strongest H-X bond (568 kJ/mol) because F is very small and holds the bonding electrons tightly. In water, HF does not fully dissociate because the H-F bond is too strong to break completely. HI with the weakest H-I bond (295 kJ/mol) dissociates most readily and is the strongest acid.
How NEET Frames The Trap
NEET asks which hydrohalic acid is the strongest or the weakest.
Q. Among the hydrohalic acids, the weakest acid in aqueous solution is
A. HF B. HCl C. HBr D. HI
Trick: HF (Option A) is the weakest because the H-F bond (568 kJ/mol) is the strongest among all H-X bonds. Acid strength in aqueous solution follows bond dissociation energy (lower BDE = stronger acid): HI > HBr > HCl >> HF.
Mistake Snapshot (What Students Do Wrong)
- Thinking all alkali metals form the same type of oxide: Li forms mainly Li2O (monoxide), Na forms Na2O2 (peroxide), K/Rb/Cs form KO2 (superoxide). The type of oxide depends on the size of the cation: larger cations stabilise larger anions.
- Saying Na forms a superoxide: Na forms a peroxide (Na2O2), not a superoxide. Only K, Rb, and Cs form superoxides as the predominant product when burned in excess oxygen.
When alkali metals burn in excess oxygen: Li gives Li2O, Na gives Na2O2, K gives KO2. Larger cations (K+, Rb+, Cs+) stabilise larger superoxide anion (O2-) through lattice energy. Small Li+ stabilises only the small oxide anion (O2-).
How NEET Frames The Trap
NEET asks which oxide is formed when a given alkali metal burns in air.
Q. When potassium burns in excess oxygen, the major product formed is
A. KO2 B. K2O C. K2O2 D. KOH
Trick: KO2 (potassium superoxide, Option A): K is large enough to stabilise the superoxide ion O2-. K2O (Option B) is the monoxide, formed only with limited oxygen. K2O2 (Option C) is the peroxide, predominantly formed by Na.