Inorganic Chemistry Exceptions Quiz for JEE Main Revision

As you go down Group 1, the cation grows larger and the ionic M–H bond lengthens and weakens. A weaker bond decomposes at a lower temperature. $LiH$ has the smallest cation and the strongest bond, so it is the most thermally stable Group 1 hydride.

$BeO$ is amphoteric — it reacts with both acids and bases. The other Group 2 oxides become progressively more basic down the group as the cation grows and the M–O bond becomes more ionic. $Be(OH)_2$ is similarly amphoteric while $Ba(OH)_2$ is a strong base.

$Li_2CO_3 \rightarrow Li_2O + CO_2$ on heating. The very small $Li^+$ has high polarising power and destabilises the carbonate ion. All heavier alkali carbonates ($Na_2CO_3$, $K_2CO_3$, $Rb_2CO_3$, $Cs_2CO_3$) are thermally stable. This is a classic anomaly of lithium driven by its diagonal relationship with magnesium.

$SO_4^{2-}$ is a large anion, so lattice energy barely changes down the group, but hydration enthalpy of $M^{2+}$ falls sharply as the cation grows. $Be^{2+}$ has very high hydration enthalpy, making $BeSO_4$ highly soluble; $Ba^{2+}$ has low hydration enthalpy, so $BaSO_4$ is nearly insoluble (used in X-ray barium meals).

Atomic size generally increases down a group, but $Ga$ ($135$ pm) is smaller than $Al$ ($143$ pm). The reason is the poor shielding of the intervening 3d electrons (d-block contraction) which raises effective nuclear charge on the outer 4p electron and pulls the shell closer.

Group 13 ionization enthalpy follows the irregular order $B > Tl > Ga > Al > In$ (the so-called "W graph"). $Tl$ has anomalously high IE because of poor shielding by intervening 4f electrons (lanthanoid contraction), which raises the effective nuclear charge on the outer 6p electron.

The inert pair effect makes heavier Group 13 elements increasingly prefer the +1 state over the group oxidation state (+3). The 6s² electrons in $Tl$ are reluctant to participate in bonding, so $Tl^+$ is the most stable +1 ion of the group, while $B$ shows essentially no +1 chemistry.

Counter-intuitively, $BF_3$ is the weakest Lewis acid of the boron halides despite F being most electronegative. Strong $p\pi$–$p\pi$ back bonding between the small B (2p) and F (2p) orbitals partially satisfies boron's electron deficiency. As halogen size grows ($Cl \to I$), back-bonding weakens and Lewis acidity rises.

Catenation tendency follows $C \gg Si > Ge \approx Sn$, with $Pb$ showing essentially no catenation. The C–C bond is exceptionally strong (~$348$ kJ/mol). As atomic size grows down the group, M–M bond strength falls sharply, killing catenation.

Boiling point of Group 15 hydrides generally rises with mass (Van der Waals forces). $NH_3$ breaks the trend due to strong intermolecular hydrogen bonding from the highly electronegative N. The full BP order is $PH_3 < AsH_3 < NH_3 < SbH_3 < BiH_3$.

Bond angle order: $NH_3$ ($107.8°$) $> PH_3$ ($93.6°$) $> AsH_3$ ($91.8°$) $> SbH_3$ ($91.3°$). As the central atom grows larger and less electronegative, bonding electrons sit further from it and the bonds approach pure-p character (~$90°$). Only $NH_3$ retains substantial sp³ hybridisation, giving it a near-tetrahedral angle.

Due to the inert pair effect, $Pb^{4+}$ is a strong oxidising agent — it strongly prefers $Pb^{2+}$. $I^-$ is a reducing agent. The two react internally: $PbI_4 \rightarrow PbI_2 + I_2$, so $PbI_4$ cannot be isolated. The same effect makes $Tl^{3+}$ unstable and $TlI_3$ actually exist as $Tl^+(I_3)^-$.

The inert pair effect makes the +4 state progressively less stable down the group. $C^{4+}$ (effectively $CO_2$, $CCl_4$ etc.) is most stable; $Pb^{4+}$ is a strong oxidiser that readily reverts to $Pb^{2+}$, where the 6s² lone pair stays inert.

The 2p orbital in $O$ (and similarly in $F$) is small and already crowded with electrons. Adding another electron causes large inter-electronic repulsion that partially offsets the energy released. The roomier 3p orbital of $S$ (and $Cl$) accommodates the new electron more comfortably, so $S$ has the most negative $\Delta_{eg}H$ in Group 16.

Among $H_2S$, $H_2Se$ and $H_2Te$, BP rises with molecular mass (Van der Waals). $H_2O$ sits anomalously above all three because the high electronegativity of oxygen enables strong intermolecular hydrogen bonding, raising the BP from the predicted $\sim -90°C$ up to $100°C$.

Acidity is governed by H–X bond dissociation energy, which decreases down the group. The H–Te bond is the longest and weakest, releasing $H^+$ most easily. Despite being the most electronegative central atom, $H_2O$ is the weakest acid in the group.

Order: $Cl > F > Br > I$. Fluorine's tiny 2p orbital is already crowded, so adding an extra electron causes large inter-electronic repulsion. Chlorine's 3p orbital is roomier and releases more energy on electron addition. $Cl$ has the most negative $\Delta_{eg}H$ of any element in the periodic table.

Order: $Cl_2 > Br_2 > F_2 > I_2$. Even though fluorine atoms are smallest (which usually predicts a strong bond), the two fluorines' three lone pairs each are pressed close together, causing large lone-pair–lone-pair repulsion. This weakens the F–F bond well below the Cl–Cl and Br–Br bonds — a classic favourite of JEE / NEET examiners.

BP order: $HCl < HBr < HI < HF$. The other three follow the Van der Waals (mass) trend, but $HF$ is anomalously high because of strong intermolecular hydrogen bonding ($F$ is small, highly electronegative, and has three lone pairs). $HF$ is liquid at room temperature; the others are gases.

Acid strength rises with the oxidation state of the central atom (+1 in $HClO$ to +7 in $HClO_4$). More terminal O atoms allow greater delocalisation of the negative charge in the conjugate base by resonance. A more stable conjugate base $\Rightarrow$ a stronger acid. $HClO_4$ is one of the strongest known oxoacids.

Between the 4d and 5d series sit the 14 lanthanoids, where the 4f shell is filled. The 4f electrons shield the outer shell very poorly, so effective nuclear charge rises sharply across the lanthanoids and pulls the next 5d shell closer in. The result: 5d elements are nearly identical in size to their 4d counterparts ($Zr \approx Hf$, $Nb \approx Ta$, $Mo \approx W$). This is the lanthanoid contraction.

Melting points across the 3d series rise as more unpaired d electrons participate in metallic bonding, peaking near $d^5$ (Cr). $Mn$ ($3d^5\, 4s^2$) holds its $d^5$ electrons tightly in an extra-stable half-filled configuration — they participate less in delocalised metallic bonding, so $Mn$'s metallic bonding is unusually weak and its melting point dips below both $Cr$ and $Fe$.

Spin-only magnetic moment is $\mu = \sqrt{n(n+2)}$ B.M., where $n$ is the number of unpaired electrons. $Mn^{2+}$ has a $3d^5$ configuration with five unpaired electrons (the maximum possible in d-orbitals), giving $\mu = 5.92$ B.M. — the highest among common 3d ions. $Sc^{3+}$ ($d^0$) and $Zn^{2+}$ ($d^{10}$) are diamagnetic.

$Mn^{2+}$ is $3d^5$ (stable half-filled). Removing one more electron disrupts this stability, costing extra energy. In contrast, $Fe^{2+}$ is $3d^6$ — removing one electron gives the stable $3d^5$ configuration, which is energetically favourable. Hence $IE_3(Mn) > IE_3(Fe)$, even though $Fe$ comes after $Mn$ in the period.

Hydration enthalpy rises with charge density (charge / size). $Li^+$ is the smallest alkali cation, so it attracts water molecules most strongly. Order: $Li^+ > Na^+ > K^+ > Rb^+ > Cs^+$. A direct consequence: $Li^+_{(aq)}$ is actually the largest alkali ion in solution because it carries the thickest hydration shell.

By Fajans' rule, covalent character of an ionic compound rises as the anion grows larger and more polarisable. $I^-$ is the largest, most polarisable halide, so $LiI$ is the most covalent. $LiF$, with the small, weakly-polarisable $F^-$, is the most ionic. Order of covalent character: $LiI > LiBr > LiCl > LiF$.

Both $NH_3$ and $NF_3$ are pyramidal, so the lone pair on N contributes a dipole along the molecular axis. In $NH_3$, the N–H bond dipoles point toward N (since N is more electronegative than H), reinforcing the lone-pair dipole. In $NF_3$, the N–F bond dipoles point away from N (toward F) and oppose the lone-pair dipole, partially cancelling it. Net: $\mu(NH_3) = 1.47$ D vs $\mu(NF_3) = 0.24$ D.

Hydrolysis of $NCl_3$ proceeds via lone-pair donation from water onto a vacant 3d orbital of Cl. Fluorine (period 2) has no accessible d-orbitals, so the same mechanism is unavailable to $NF_3$ — it remains kinetically inert toward water. Notice the parallel with $CCl_4$ (no d-orbitals on C) which also resists hydrolysis, while $SiCl_4$ (vacant 3d on Si) hydrolyses readily.

All six species have 10 electrons (the Ne core), so size is determined entirely by nuclear charge. More protons pull the same shell tighter. $Al^{3+}$ has 13 protons — the highest in the set — so it is the smallest. The order of radii is $N^{3-} > O^{2-} > F^- > Na^+ > Mg^{2+} > Al^{3+}$.

$NO$ has 11 valence electrons — an odd number — so one electron is unpaired in a $\pi^*$ molecular orbital, making it paramagnetic. $NO_2$ is similarly paramagnetic (17 valence electrons). Both dimerise on cooling ($N_2O_2$, $N_2O_4$) and become diamagnetic in the solid. $N_2O$, $N_2O_3$ and $N_2O_5$ all have even electron counts and are diamagnetic.