GATE Main-Group Chemistry — Structure and Bonding, Solved by Electron Counting

Main-group inorganic chemistry looks like a memory subject and is not. Almost every GATE question on p-block structure can be answered by counting electrons correctly and then applying one of four rules: the steric-number rule for shapes, Wade's rules for cluster compounds, the back-bonding argument for anomalous reactivity, and Bent's rule for which position a substituent takes. Learn the counting and the "facts" mostly generate themselves. This guide works through each rule with the species that actually appear in papers.

Rule 1 — steric number decides the shape

The steric number of a central atom is the number of sigma-bonded neighbours plus the number of lone pairs on it. Get the lone-pair count right and the shape follows.

Add one electron for each negative charge on the ion and subtract one for each positive charge. A double bond to oxygen is one sigma bond for shape purposes, but it consumes two of the central atom's electrons in the simple Lewis count used below.

Worked example 1 — three shapes from one method

Rule 2 — electron-deficient and hypervalent bonding

Two bonding types outside the ordinary two-centre two-electron picture carry a lot of marks.

Three-centre two-electron (3c–2e) bonds occur where there are too few electrons for conventional bonds. Diborane, B₂H₆, is the standard case: 2 × 3 + 6 × 1 = 12 valence electrons. Four terminal B–H bonds use 8, leaving 4 electrons for two B–H–B bridges — one pair per bridge, spread over three atoms. The molecule is not planar; the bridging hydrogens sit above and below the B₂H₄ plane. Al₂(CH₃)₆ has the same bridging arrangement with methyl groups.

Three-centre four-electron (3c–4e) bonds explain so-called hypervalent species such as XeF₂, I₃⁻ and the axial bonds of PF₅ without needing d orbitals. A p orbital on the central atom overlaps with one orbital on each of two collinear neighbours, giving a bonding, a non-bonding and an antibonding combination; four electrons fill the first two. This is the modern description, and it is why the axial bonds in a trigonal bipyramid are longer and weaker than the equatorial ones. If an option says "expansion of the octet requires the participation of 3d orbitals", treat it with suspicion — high-level calculations show d-orbital involvement in second-row hypervalent compounds is small.

Rule 3 — Wade's rules for boranes and clusters

Wade's rules turn cluster classification into arithmetic. Count the skeletal electron pairs and compare with the number of vertices n.

Worked example 2 — classifying B₅H₉ and [B₆H₆]²⁻

Rule 4 — back-bonding explains the "wrong" trends

Whenever a main-group trend looks backwards, back-bonding is usually the reason.

• Lewis acidity of boron halides: BF₃ < BCl₃ < BBr₃ < BI₃. Electro­negativity alone would predict the opposite. Fluorine's 2p orbital matches boron's empty 2p almost perfectly, so pπ–pπ back-donation from F to B is strong and partly fills boron's vacant orbital. Down the halogen group the orbital sizes mismatch, back-donation weakens, and the boron becomes hungrier for a lone pair.
• N(SiH₃)₃ is planar while N(CH₃)₃ is pyramidal. In trisilylamine the nitrogen lone pair is delocalised onto silicon, which flattens the nitrogen; in trimethylamine the lone pair stays put and the molecule keeps its pyramid. As a consequence N(SiH₃)₃ is a much weaker base than N(CH₃)₃.
• Silicon does not form stable pπ–pπ double bonds easily the way carbon does, because the larger 3p orbitals overlap sideways poorly. That is why CO₂ is a discrete linear molecule while SiO₂ is a three-dimensional network of corner-shared SiO₄ tetrahedra.

Two smaller rules worth memorising

Bent's rule: a central atom directs hybrid orbitals of greater p character towards more electronegative substituents. In a trigonal bipyramid the axial positions have more p character than the equatorial ones, so the more electronegative substituents go axial. In PF₃Cl₂ the two axial sites are taken by fluorines and the two chlorines sit equatorial with the third fluorine.

Drago's rule: in hydrides of heavier p-block elements with low electronegativity, hybridisation is negligible and the central atom bonds through nearly pure p orbitals, so the bond angle drops towards 90°.

Do not explain the NH₃ → SbH₃ trend by lone-pair repulsion alone; the near-90° angles in the heavier hydrides are the signature of unhybridised p bonding.

Isoelectronic families — free marks if you know them

Species with the same number of atoms and the same valence electron count have the same shape and similar bonding.

Silicates, the inert pair and diagonal relationships

Silicate classification is asked as a straight matching question. Every silicate is built from corner-sharing SiO₄ tetrahedra, and the number of shared corners fixes the formula:

The inert pair effect is the growing preference for an oxidation state two below the group valence as you descend groups 13 to 15: Tl(I) is more stable than Tl(III), Pb(II) than Pb(IV), Bi(III) than Bi(V). The practical consequence tested in exams is oxidising power — PbO₂ and NaBiO₃ are strong oxidants precisely because the higher state is unstable.

Diagonal relationships — Li with Mg, Be with Al, B with Si — arise because the increase in charge going right across a period roughly cancels the increase in size going down a group, leaving similar charge-to-radius ratios. This is why Li forms a nitride like Mg, why BeCl₂ and AlCl₃ are both covalent and polymeric or dimeric, and why boron and silicon both form covalent network hydrides and oxides rather than ionic ones.

Common mistakes that cost marks

Where this appears in GATE Chemistry

Main-group structure and bonding sits in the inorganic portion of the GATE Chemistry (CY) syllabus and overlaps with group theory and solid-state chemistry. For the exact syllabus wording, question count and marking scheme in the year you are appearing, read the current official GATE information brochure rather than any summary.