P-Block Anomalies and Exceptions Quiz for JEE Main and NEET

Reducing power follows $Al > Ga > In > Tl$. $Al$ strongly prefers the +3 state, so $Al$ metal readily gives up three electrons (acts as a reductant). $Tl$ prefers +1 (inert pair effect), so $Tl^{3+}$ is actually a strong oxidising agent — the opposite behaviour. This is why $Al$ is used in thermite reactions but $Tl$ is not.

For a fixed halogen, Lewis acidity decreases down Group 13. $B$ is the smallest and most electron-deficient, so $BX_3$ accepts a lone pair most readily. $Al$, $Ga$ and $In$ have empty d-orbitals available for accepting electron pairs but their larger size dilutes the effect. (Within the boron halides themselves the order is reversed because of $p\pi$–$p\pi$ back bonding — that's a separate trend.)

Due to the inert pair effect, $Tl^{3+}$ is a strong oxidising agent that would oxidise $I^-$ to $I_2$. So $TlI_3$ cannot be a $Tl^{3+}$ salt. It is actually $Tl^+(I_3^-)$ — the triiodide salt of $Tl^+$. This is the same logic that makes $PbI_4$ non-existent. The 6s² lone pair on Tl prefers to stay inert, stabilising the +1 state.

Back bonding is most efficient when the orbitals overlapping are similar in size. In $BF_3$ the overlap is 2p (B) with 2p (F) — best size match, strongest back bonding. As we go to $BCl_3$ (2p-3p), $BBr_3$ (2p-4p), $BI_3$ (2p-5p), the size mismatch grows and back bonding weakens. This is why $BF_3$ is the weakest Lewis acid among boron halides — its electron deficiency is partially relieved by back bonding.

Between Al (period 3) and Ga (period 4) sits the first row of the d-block (Sc to Zn). The 3d electrons shield the outer electrons very poorly, so the effective nuclear charge on Ga's 4p electron is higher than expected. The shell is pulled inward, making Ga ($135$ pm) actually smaller than Al ($143$ pm). This is the d-block contraction.

Down Group 13 the M–OH bond becomes more ionic and $OH^-$ is released more easily. $B(OH)_3$ is actually acidic — boric acid donates a proton to water, since the small B atom polarises the O–H bond. $Al(OH)_3$ is amphoteric. $Ga(OH)_3$ is weakly basic. $Tl(OH)_3$ is the most basic — it behaves like a typical alkali. The pattern parallels the metallic-character increase down the group.

A positive $E°$ for $M^{3+}/M$ means the reduction $Tl^{3+} + 3e^- \rightarrow Tl$ is spontaneous — i.e. $Tl^{3+}$ readily accepts electrons. So $Tl^{3+}$ behaves as an oxidising agent. The other Group 13 metals all have negative $E°$ (Al most negative at $-1.66$ V), so they're reducing agents — $Al \rightarrow Al^{3+} + 3e^-$ is spontaneous. The smooth slide from very negative (Al) to positive (Tl) is the inert pair effect quantified: the 6s² electrons in Tl resist removal, making $Tl^+$ the stable oxidation state and $Tl^{3+}$ a strong oxidiser.

Between Sn (period 5) and Pb (period 6) lies the lanthanoid series (4f filling). The 4f electrons shield very poorly, so the 6p electron in Pb feels a higher effective nuclear charge than expected. This raises Pb's IE above Sn's — a clean fingerprint of lanthanoid contraction in the p-block. Compare with the same effect making $Tl$'s IE anomalously high in Group 13.

As you go down Group 14, the M–H bond length grows and bond strength falls steadily. Thermal stability follows bond strength, so $CH_4$ (strongest C–H bond) is most stable while $PbH_4$ is so unstable that it has barely been characterised. The order is monotonic: $CH_4 > SiH_4 > GeH_4 > SnH_4 > PbH_4$.

Catenation depends on M–M bond strength. The C–C single bond is one of the strongest homoatomic single bonds (~$348$ kJ/mol) because both atoms are small and 2p orbitals overlap optimally. As atomic size grows, M–M bond strength falls sharply: $C \gg Si > Ge \approx Sn$, and $Pb$ shows essentially no catenation. Note: carbon does not have d-orbitals — that's what stops $CCl_4$ from hydrolysing.

Hydrolysis of $SiCl_4$ proceeds by lone-pair donation from water onto a vacant 3d orbital on Si, which makes the central atom temporarily 5-coordinate. Carbon is in period 2 and has no accessible d-orbitals, so the same mechanism is closed off. $CCl_4$ is therefore kinetically inert toward water. Same logic explains why $NF_3$ doesn't hydrolyse but $NCl_3$ does (Cl has 3d available, F doesn't).

For a given metal, M–F is the shortest and strongest M–X bond (F is small and highly electronegative), so $MF_4$ is most thermally stable. Stability falls as the halogen grows. The trend is reinforced by the redox angle: large iodides like $PbI_4$ don't exist at all because $Pb^{4+}$ oxidises $I^-$ to $I_2$, leaving $PbI_2$ behind.

Both diamond and silicon adopt the same tetrahedral lattice geometry, but bond lengths differ sharply: C–C is ~$1.54$ Å while Si–Si is ~$2.35$ Å. Diamond therefore packs ~$3.5$× more atoms per unit volume than silicon. From Ge onward, atomic mass grows much faster than volume, so density rises smoothly: $Ge < Sn < Pb$. The Si dip is the lone density anomaly in Group 14.

$NH_3$ has the highest melting point ($195.2$ K) because of strong intermolecular hydrogen bonding, which creates a tightly held solid lattice. Among the rest, mass-driven Van der Waals forces dominate, giving $SbH_3 > AsH_3 > PH_3$. The same hydrogen-bonding effect explains $NH_3$'s anomalously high boiling point.

Thermal stability falls down the group: $NH_3 > PH_3 > AsH_3 > SbH_3 > BiH_3$. As the central atom grows, the M–H bond lengthens and bond dissociation enthalpy drops sharply (~$389$ kJ/mol for $NH_3$ vs ~$255$ kJ/mol for $SbH_3$). $BiH_3$ is so unstable it decomposes near room temperature.

Reducing character is inversely correlated with thermal stability — a hydride that decomposes easily readily releases H to reduce other species. So the order is $NH_3 < PH_3 < AsH_3 < SbH_3 < BiH_3$. The weakest M–H bond ($BiH_3$) makes it the strongest reductant.

Basicity follows $NH_3 > PH_3 > AsH_3 > SbH_3 > BiH_3$. The small N atom holds its lone pair tightly localised at high electron density, so $NH_3$ donates the pair vigorously. As the central atom grows, the lone pair spreads out over a larger volume — lower charge density, weaker donor.

$NO_2^+$ is sp-hybridised (no lone pair on N) and linear, so bond angle is $180°$. $NO_2$ has a single odd electron on N (mild repulsion), giving a bent structure with angle $\sim 134°$. $NO_2^-$ has a full lone pair on N (stronger lp–bp repulsion), bending it further to $\sim 115°$. Lone-pair repulsion squeezes the bond angle harder than a single odd electron does.

$NF_3$ is a stable, colourless gas (used as an etchant in semiconductor manufacturing). $NCl_3$ is an explosive yellow oil. $NBr_3$ and $NI_3$ are far worse — $NI_3 \cdot NH_3$ famously detonates from the lightest contact. The reason: N is small, so as halogen size grows, steric strain around N rises sharply while N–X bond strength falls. F is the only halogen small and electronegative enough to give a strong, short, well-fitting N–F bond.

Two effects open the X–P–X angle as we go from F to I: (i) larger halogens crowd each other sterically, and (ii) decreasing electronegativity ($F > Cl > Br > I$) means bond pairs sit closer to P, raising bp–bp repulsion. In $PF_3$, highly electronegative F pulls bond pairs away from P, weakening repulsion and giving the smallest angle. Note this is the opposite pattern to Group 15 hydrides ($NH_3 > PH_3 > AsH_3$) where the central atom changes — there, the angle drops as the central atom grows.

Down Group 15, oxides shift from acidic to basic. $N_2O_3$ and $P_2O_3$ are acidic (give $HNO_2$ and $H_3PO_3$ with water). $As_2O_3$ and $Sb_2O_3$ are amphoteric (react with both acids and bases). $Bi_2O_3$ is basic — it dissolves only in acids, like a typical metal oxide. The pattern reflects the standard increase in metallic character down a group.

Lewis basicity depends on N's lone-pair availability. In $NF_3$, the highly electronegative F atoms pull electron density away from N, leaving the lone pair impoverished — $NF_3$ is a very poor Lewis base. As halogen electronegativity drops ($F > Cl > Br > I$), more density stays on N and the lone pair becomes increasingly available. $NI_3$ is the strongest donor in the series. The same reasoning explains why $NH_3$ has a much larger dipole moment than $NF_3$.

Acidic character of $NO_x$ rises with the oxidation state of nitrogen. $N_2O$ (N is +1) and $NO$ (+2) are essentially neutral — they don't form acids in water. $N_2O_3$ (+3) is the anhydride of $HNO_2$. $N_2O_4$ (+4) disproportionates to $HNO_2 + HNO_3$. $N_2O_5$ (+5) is the anhydride of nitric acid — it dissolves vigorously in water to give $HNO_3$, the standard strong mineral acid. Higher OS on N → more delocalisation in the conjugate base → stronger acid.

Between Sb (period 5) and Bi (period 6) sit the 14 lanthanoids, where the 4f shell fills. The 4f electrons shield extremely poorly, so Bi's outer 6p electrons feel a higher effective nuclear charge than expected. This raises $IE_2$ and $IE_3$ for Bi above the Sb value, breaking the down-the-group trend. Same fingerprint as $Pb > Sn$ in Group 14 IE and $Tl$'s anomalously high IE in Group 13 — all three are signatures of lanthanoid contraction in the heavy p-block.

In $H_2O$ the small, highly electronegative O pulls bond pairs close, increasing bp–bp repulsion and pushing the angle outward to $104°$ (sp³-like). In $H_2S$, the S–H bonds are longer and the bond pairs sit far from S, so repulsion is small — the bonds use almost pure p-orbitals (Drago's rule), giving the natural $90°$ p–p angle. The same trend continues in $H_2Se$ ($91°$) and $H_2Te$ ($90°$).

In O–O, the two small oxygen atoms have multiple lone pairs uncomfortably close, leading to strong electron-pair repulsion that weakens the single bond ($\sim 142$ kJ/mol). S is larger, so S–S lone pairs sit further apart and the bond is much stronger ($\sim 226$ kJ/mol). Strong S–S bonds enable extensive catenation; weak O–O bonds limit oxygen mostly to $O=O$ (which uses a strong $\pi$ system instead).

The 2p orbital of O is small and already crowded with electrons. Adding another electron forces it into a tight, electron-rich region, so the energy released (electron gain enthalpy) is partially offset by inter-electronic repulsion. Sulphur's 3p orbital is larger and roomier, so $S$ accommodates the extra electron more comfortably and has a more negative $\Delta_{eg}H$. The same reasoning explains $F < Cl$ for halogen EA.

For Group 16 hydrides without hydrogen bonding, BP rises with molecular mass because Van der Waals forces dominate. $H_2Te$ is the heaviest, so it has the highest BP. The full order excluding water is $H_2S < H_2Se < H_2Te$. $H_2O$ sits anomalously above all three because of strong intermolecular hydrogen bonding.

Acid strength order: $HF < HCl < HBr < HI$. Acidity is governed by H–X bond dissociation enthalpy, which falls down the group as H–X bond length increases. $H–I$ is the weakest H–X bond, so it releases $H^+$ most readily. Counter-intuitively, $HF$ is the weakest acid in this series despite F being the most electronegative — its strong H–F bond holds the proton tightly.

Oxidising power follows $F_2 > Cl_2 > Br_2 > I_2$. $F_2$ is the strongest oxidant because it combines (i) low bond dissociation enthalpy of $F_2$ (lone-pair–lone-pair repulsion makes F–F weak) and (ii) very high hydration enthalpy of $F^-$ (small ion, strong solvation). Both factors release a lot of energy when $F_2$ accepts electrons in solution.

For the same oxidation state (+1 here), acid strength depends on the electronegativity of the halogen. More electronegative central atoms pull electron density away from the O–H bond, weakening it and making $H^+$ release easier. So $HClO > HBrO > HIO$. (Note: this is the opposite trend from the HX series, where bond dissociation enthalpy dominates.)

Melting points (and hence physical states) follow the regular Van der Waals trend for halogens: $F_2$ (gas, MP $-220°C$) $< Cl_2$ (gas, MP $-101°C$) $< Br_2$ (liquid, MP $-7°C$) $< I_2$ (solid, MP $114°C$). The order is monotonic — no anomalies here, since hydrogen bonding is absent in $X_2$ molecules.

In $OF_2$, the high electronegativity of F pulls bonding electrons away from O, reducing bp–bp repulsion at the central oxygen and giving an angle of $\sim 103°$. In $Cl_2O$, the much larger Cl atoms sit closer to each other and repel sterically, opening the angle to $\sim 110°$. Both effects exist, but in $Cl_2O$ steric crowding wins.

MP order: $HI(-50°C) > HF(-83.6°C) > HBr(-86°C) > HCl(-114°C)$. Counter-intuitively, $HI$ tops the MP list because Van der Waals forces (which scale with size/mass) dominate in the solid lattice — and HI is the heaviest. $HF$ comes second due to H-bonding. This is different from BP order ($HF > HI > HBr > HCl$), where H-bonding networks matter more in the liquid state and lift HF above HI. Both anomalies are JEE/NEET favourites.

Oxidising power is the inverse of acid strength here. $HClO$ has Cl in the $+1$ state with a single, easily-released O atom — kinetically reactive, generates nascent oxygen and hypochlorite readily. $HClO_4$ has Cl in $+7$ but is so resonance-stabilised that it's kinetically reluctant to oxidise. So while $HClO_4 > HClO_3 > HClO_2 > HClO$ for acid strength, oxidising power follows $HClO > HClO_2 > HClO_3 > HClO_4$. Students who memorise only the acidity order get this question wrong — a perennial JEE/NEET trap.

Noble gases have completely filled $ns^2 np^6$ valence shells. Any added electron must go into the next $(n+1)s$ orbital, which is much higher in energy. This costs energy rather than releasing it — hence positive $\Delta_{eg}H$. This is the reason noble gases are extremely reluctant to form anions and remain monatomic.

Noble gases are held together only by weak London dispersion forces, which scale with atomic size and electron count. Helium is the smallest and least polarisable, so its dispersion forces are minimal — He has the lowest BP of any substance ($4.2$ K). The full order is $He < Ne < Ar < Kr < Xe < Rn$.

$Al_2O_3$ dissolves in HCl (acting as a base) and in NaOH (acting as an acid, forming $NaAlO_2$) — it is the textbook amphoteric oxide. Other amphoteric oxides include $BeO$, $ZnO$, $Ga_2O_3$, $SnO/SnO_2$, $PbO/PbO_2$. $CO$, $NO$ and $N_2O$ are neutral. $MgO$ is purely basic.

In a single N–N bond, the two small N atoms have multiple lone pairs uncomfortably close, causing significant electron-pair repulsion that weakens the bond. P is larger, so P–P lone pairs sit further apart and the bond is stronger. This parallels the O–O vs S–S story. As a result, $P_4$ and polyphosphate $P–O–P$ chains are common, but N catenates only minimally (mostly via $\pi$ systems like $N_2$, azides).

Inert pair effect: the 6s² electrons on Pb are reluctant to participate in bonding because of poor shielding by intervening 4f and 5d electrons (relativistic + lanthanoid contraction). So $Pb^{2+}$ (with the 6s² pair retained) is much more stable than $Pb^{4+}$. Whenever $Pb^{4+}$ is forced into a compound, it strongly oxidises whatever is nearby to revert to $Pb^{2+}$ — that is why $PbO_2$ is a powerful oxidiser used in lead-acid batteries.

By Fajans' rule, ionic character decreases as the anion grows larger and more polarisable. $F^-$ is the smallest, least polarisable halide, so $NaF$ is the most ionic. $I^-$ is large and easily polarised, so $NaI$ has significant covalent character. Order of ionic character: $NaF > NaCl > NaBr > NaI$. (For lithium halides the trend is the same, and $LiI$ is essentially molecular.)