CSIR-NET Main-Group Chemistry Trends
Main-group (s- and p-block) trend questions in CSIR-NET Part C ask you to explain why a property changes down a group or across a period, not just to recall the trend direction. Three ideas cover most of what is tested: the inert pair effect on oxidation-state stability, the sharp drop in catenation ability down group 14, and the debated explanation for expanded-octet bonding in heavier p-block halides.
The inert pair effect
Going down groups 13–16, the lower oxidation state (group number minus 2) becomes progressively more stable relative to the group's maximum oxidation state. This is most dramatic in period 6: Tl(I) is more stable than Tl(III); Pb(II) is more stable than Pb(IV); Bi(III) is more stable than Bi(V). The accepted cause is poor shielding of the ns² electron pair by the intervening (n−1)d and, for period 6, 4f electrons, combined with a relativistic contraction of the 6s orbital that makes those two electrons less energetically available for bonding — effectively "inert."
Worked example 1 — predicting the more stable oxidation state
Q. Between TlCl and TlCl₃, and between AlCl and AlCl₃, which member of each pair is thermodynamically more stable, and why does the answer differ between the two elements?
Thallium (period 6, group 13) shows a strong inert pair effect, so TlCl (Tl(I)) is the more stable chloride — Tl(III) compounds are comparatively strong oxidising agents. Aluminium (period 3) shows essentially no inert pair effect, since its 3s² electrons are well shielded only by core electrons with no intervening d-block contraction — so AlCl₃ (Al(III)), the group's maximum oxidation state, is overwhelmingly the stable, characteristic chloride. The same group, opposite answer, purely because of where each element sits in the periodic table.
Catenation — the sharp fall down group 14
Catenation (an element bonding to itself in chains) is dominated in the p-block by carbon, and falls away quickly down group 14. The underlying reason is simply bond strength: the element-element single-bond enthalpy drops steeply from carbon to its heavier congeners (approximate literature values, which vary slightly by source: C–C ≈ 348 kJ/mol, Si–Si ≈ 226 kJ/mol, Ge–Ge ≈ 188 kJ/mol, Sn–Sn ≈ 151 kJ/mol).
Worked example 2 — ranking catenation tendency
Q. Rank C, Si, Ge and Sn by their tendency to form long catenated chains, and justify the order using bond enthalpies.
Order: C >> Si > Ge > Sn. Using the approximate values above, the C–C bond (≈348 kJ/mol) is more than twice as strong as the Sn–Sn bond (≈151 kJ/mol). Long chains built from weaker element-element bonds are both harder to form and more thermodynamically prone to cleavage (often by disproportionation or reaction with the element-oxygen bond, which is comparatively much stronger for the heavier elements) — this single trend in bond enthalpy is why organic chemistry (carbon catenation) is a vast field while stable long-chain silanes or germanes are comparatively rare and reactive.
Expanded octets — a genuinely debated explanation
Nitrogen never forms NCl₅ (or any five-coordinate halide) because it has only 2s and 2p valence orbitals available — a maximum coordination number of 4. Phosphorus, however, readily forms PCl₅. The traditional explanation taught in many courses is that heavier p-block elements can use empty 3d (or higher) orbitals to expand beyond an octet (dsp³ hybridisation for PCl₅). This d-orbital picture is now considered largely incorrect by much of the modern computational chemistry literature — high-level calculations show negligible d-orbital occupation in these molecules. The favoured modern explanations instead invoke ionic resonance structures, negative hyperconjugation, or three-centre four-electron bonding models that do not require d-orbital participation at all. Both explanations still appear across different textbooks and question sources, so state which one you are using rather than presenting either as the single settled answer.
Worked example 3 — why bismuth pentahalides are so unstable
Q. BiCl₅ is far less stable than PCl₅, even though both are group 15 pentahalides. Explain using periodic trends.
Bismuth sits at the bottom of group 15, where the inert pair effect is strongest — its 6s² electron pair resists participating in bonding, so the +5 oxidation state (which requires using all five valence electrons, including both 6s electrons) is strongly disfavoured relative to the +3 state. BiCl₅ is therefore a powerful oxidising agent and extremely difficult to isolate, while phosphorus (period 3, negligible inert pair effect) forms a perfectly stable PCl₅. This is the inert pair effect and expanded-octet bonding interacting directly — the two trends together, not either alone, explain the observation.
Common mistakes that cost marks
- Confusing the inert pair effect with the lanthanide contraction. They are related but distinct phenomena — lanthanide contraction is the anomalous decrease in atomic/ionic radius across the lanthanides, which is one contributing cause of poor shielding, not the inert pair effect itself.
- Presenting the d-orbital explanation for expanded octets as the only correct one. As above, this is a genuine point of disagreement between older and modern treatments — state which framework you are using.
- Predicting catenation purely from electronegativity, ignoring bond enthalpy. The dominant factor is the element-element single-bond strength, not electronegativity differences.
- Assuming the inert pair effect is equally strong across all of periods 4, 5 and 6. It is weak in period 4 (e.g. Ga, Ge), moderate in period 5, and strongest in period 6 (Tl, Pb, Bi) — the effect itself has a trend.
Trend summary table
| Trend | Direction down a p-block group | Underlying cause |
|---|---|---|
| Stability of (group − 2) oxidation state | Increases | Poor shielding + relativistic contraction of ns² |
| Catenation tendency | Decreases sharply | Falling element-element bond enthalpy |
| Maximum coordination number | Increases from period 2 to period 3 onward | Debated: classical d-orbital vs. modern hyperconjugation/ionic models |
Checking oxidation numbers as part of a trend question? The Oxidation Number calculator assigns oxidation states to every atom in a formula instantly.
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