Drawn structures of key inorganic compounds — JEE Advanced 2026

Key:
Single bond
Double bond (= O)
P
S
N
O
Cl/Xe
Oxyacids of Phosphorus
H₃PO₂
Hypophosphorous acid
P O H H OH = O
Tetrahedral P · sp³
Basicity = 1 Strong reducer
2 direct P–H bonds (non-acidic). Only 1 P–OH → monobasic. Reduces Cu²⁺, Ag⁺.
H₃PO₃
Phosphorous acid
P O OH OH H = O
Tetrahedral P · sp³
Basicity = 2 Reducer
1 direct P–H (non-acidic), 2 P–OH → dibasic. Reducing but weaker than H₃PO₂.
H₃PO₄
Orthophosphoric acid
P O OH OH OH = O
Tetrahedral P · sp³
Basicity = 3 Non-reducing
No P–H bonds → non-reducing. 3 P–OH → tribasic/triprotic.
H₄P₂O₇
Pyrophosphoric acid (formed by condensation of 2 H₃PO₄)
O P P O O OH OH OH OH = O = O bridge
2 tetrahedral P · P–O–P bridge · sp³
Basicity = 4 Non-reducing
Two H₃PO₄ joined by –O– bridge losing H₂O. No P–H → non-reducing. 4 P–OH groups → tetrabasic.
Oxyacids of Sulphur
H₂SO₃
Sulphurous acid
S O OH OH = O LP
Pyramidal S · sp³ · 1 lone pair
Basicity = 2 Reducer
S has 1 lone pair. Pyramidal shape. Reducing agent — S goes from +4 to +6.
H₂SO₄
Sulphuric acid
S O O OH OH = O = O
Tetrahedral S · sp³ · S(+6)
Basicity = 2 Oxidiser (conc.)
2 S=O (non-acidic) + 2 S–OH (acidic). Strong diprotic acid. Conc. H₂SO₄ is strong oxidiser and dehydrating agent.
H₂S₂O₇
Pyrosulphuric acid (Oleum) — H₂SO₄ + SO₃
O S S O O O O OH OH bridge
2 tetrahedral S · S–O–S bridge · both S(+6)
Basicity = 2 Fuming H₂SO₄
Formed when SO₃ dissolves in H₂SO₄. Each S has 2 S=O + 1 S–OH + 1 bridging S–O–S. Very strong oxidiser.
H₂S₂O₈
Peroxodisulphuric acid (Marshall's acid)
O O S S O O O O OH OH –O–O–
–O–O– peroxy bridge · S(+6) · both sp³
Basicity = 2 Contains O–O Strong oxidiser
Has O–O peroxide linkage (like H₂O₂). This is what makes it a strong oxidiser. S is +6 in both halves.
Oxyacids of Chlorine (HClO → HClO₄)
HClO
Hypochlorous acid
H O Cl : :
Cl(+1) · no Cl=O
0 Cl=O bonds Weakest acid
No Cl=O. Cl has 3 lone pairs. Weakest of the 4 chlorine oxyacids. Cl oxidation state: +1.
HClO₂
Chlorous acid
: : Cl O OH = O
Cl(+3) · 1 Cl=O · bent
1 Cl=O bond 2 lone pairs on Cl
Cl(+3). 1 Cl=O + 1 Cl–OH. Cl has 2 lone pairs. Bent geometry. JEE 2015 asked: Cl=O bonds in HClO₂ + HClO₃ = 2 ✓
HClO₃
Chloric acid
: Cl O O OH
Cl(+5) · 2 Cl=O · pyramidal
2 Cl=O bonds 1 lone pair on Cl
Cl(+5). 2 Cl=O + 1 Cl–OH. Pyramidal (1 lone pair). JEE 2015: lone pairs on HClO₂+HClO₃ = 2+1 = 3 ✓
HClO₄
Perchloric acid
Cl O O O OH = O
Cl(+7) · 3 Cl=O · tetrahedral · sp³
3 Cl=O bonds 0 lone pairs on Cl Strongest acid
Cl(+7). sp³ hybridised (JEE 2015 ✓). 3 Cl=O + 1 Cl–OH. Strongest oxyacid — no lone pairs on Cl, perfect tetrahedral.
Key Molecular Shapes (VSEPR) — Frequently Asked
XeF₄
Xenon tetrafluoride
Xe F F F F LP LP
Square planar · sp³d² · 2 axial LP
6 electron pairs Non-polar
sp³d² (6 pairs total = 4 bonds + 2 lone pairs). Square planar shape. Lone pairs occupy axial positions → non-polar molecule.
ClF₃
Chlorine trifluoride
Cl F F F LP LP
T-shaped · sp³d · 2 equatorial LP
5 electron pairs Polar
sp³d (trig bipyramidal electron geometry). 2 lone pairs occupy equatorial → T-shaped. Polar molecule.
H₂O₂
Hydrogen peroxide
O O H H 111.5° ∠HOO = 96.5° non-planar (open book)
Non-planar · dihedral 111.5° · sp³ O
O–O single bond Oxidiser & reducer O in –1 state
Open-book structure. NOT planar. Dihedral angle 111.5° (gas), ~94° (solid). O is sp³. Acts as both oxidiser and reducer.
N₂O₃
Dinitrogen trioxide
N N O O O : N–N bond + N=O
Contains N–N bond · N(+2) & N(+4)
Has N–N bond JEE 2009 asked
Contains N–N bond. Also has N₂O and N₂O₄. JEE 2009 directly asked which nitrogen oxides contain N–N bonds → N₂O, N₂O₃, N₂O₄.
Boron Compounds (Frequently Asked)
H₃BO₃
Orthoboric acid
B OH OH OH empty p-orbital
Trigonal planar · sp² · Lewis acid
Lewis acid (NOT Brønsted) H-bonding layers
B is sp² with empty p-orbital → Lewis acid. Accepts OH⁻ from water. NOT self-ionising. Layered structure with H-bonds between layers (JEE 2014).
B₂H₆
Diborane — 3-centre 2-electron (3c2e) bonds
H H B B H H H H terminal 3c2e bridge terminal
2 bridging H (3c2e) · 4 terminal H · B is sp³
Banana bonds Electron deficient JEE favourite
4 terminal B–H (normal 2c2e) + 2 bridging B–H–B (3c2e, banana bonds). B is sp³. Each bridge H has partial bonds to both B atoms. Hydrolysis gives B(OH)₃ + H₂.