s and p Block Elements — NEET Chemistry MCQs with answers

s and p Block Elements is a Chemistry chapter in the NEET (UG) syllabus. NEET720 has 133 reviewed practice questions on it, each with a quick answer and a step-by-step explanation. The 8 questions below are free and fixed, so you can bookmark this page; the full chapter, plus mistake tracking and spaced revision, is in the app.

Topics covered

Alkali Metals · Alkaline Earth Metals · Boron Family · Carbon Family · Nitrogen Family · Oxygen Family · Halogens · Noble Gases · s-Block and p-Block Elements

8 free s and p Block Elements practice questions with answers

Choose an answer in your head before opening it. Each explanation says why the correct option is right and, where relevant, why the tempting wrong option is wrong.

1. Question 1 · medium · Alkali Metals

   Lithium is the only alkali metal that reacts directly with atmospheric nitrogen to form a nitride (Li3N) under ordinary conditions. This is attributed to:

   • A.Lithium's small size and high charge density giving it enough lattice energy to stabilize the small, highly charged nitride ion
   • B.Lithium being the most reactive of all alkali metals toward every reagent
   • C.Nitrogen being uniquely reactive only with metals from period 2
   • D.Li3N being an unstable, transient species that decomposes immediately
   Show answer and explanation

   Answer: A. Lithium's small size and high charge density giving it enough lattice energy to stabilize the small, highly charged nitride ion

   Only lithium among the alkali metals has a high enough charge density (small ionic radius, concentrated positive charge) to release sufficient lattice energy upon combining with the small, highly charged N3− ion, making Li3N formation thermodynamically favourable, unlike for Na, K, Rb, Cs.

   Direct combination with N2 to form a nitride (M3N) requires forming the small, highly charged N3− ion, which has a very high lattice energy requirement to be stabilized. Lithium, being the smallest alkali metal cation with the highest charge density, can provide enough lattice energy in Li3N to make the overall process (6Li + N2 → 2Li3N) exothermic and favourable. The larger alkali metal ions (Na+, K+, Rb+, Cs+) cannot release sufficient lattice energy to stabilize the N3− ion, so they do not form stable nitrides directly with atmospheric nitrogen under ordinary conditions, distinguishing lithium's behaviour, which is also explained by its diagonal relationship with magnesium (Mg3N2 forms similarly).

   Common mistake: Assuming all alkali metals react with N2 similarly since they are in the same group

2. Question 2 · medium · Alkali Metals

   Among the alkali metal carbonates, Li2CO3 decomposes on heating to Li2O and CO2 at a relatively low temperature, while Na2CO3, K2CO3, Rb2CO3, and Cs2CO3 are thermally very stable and do not decompose even at red heat. This trend is explained by:

   • A.Smaller cations like Li+ polarize the large carbonate anion more strongly, weakening the C-O bonds and facilitating decomposition to the oxide
   • B.Larger cations have greater polarizing power, destabilizing the carbonate ion more
   • C.Lithium carbonate is more soluble in water, which makes it thermally unstable
   • D.All alkali metal carbonates decompose at similar temperatures around 300°C
   Show answer and explanation

   Answer: A. Smaller cations like Li+ polarize the large carbonate anion more strongly, weakening the C-O bonds and facilitating decomposition to the oxide

   Lithium's small ionic radius and high charge density give it strong polarizing power, which distorts and weakens the large carbonate ion's C-O bonds enough to promote decomposition into Li2O and CO2 on heating, unlike the larger, weakly polarizing Na+ through Cs+ ions.

   The thermal stability of an ionic carbonate depends on the polarizing power of the cation: a small, highly charged cation strongly polarizes the large, polarizable carbonate ion (CO3 2−), weakening its internal C-O bonds and facilitating the release of CO2, converting the carbonate to the corresponding oxide. Lithium, being the smallest alkali metal cation, has the greatest polarizing power among the group and shows the least thermally stable carbonate. As cation size increases down the group (Na+ < K+ < Rb+ < Cs+), polarizing power decreases sharply, so their carbonates become progressively more thermally stable, with Na2CO3 through Cs2CO3 not decomposing even when heated strongly (they melt without appreciable decomposition), unlike Li2CO3.

   Common mistake: Assuming larger cations are more destabilizing, reversing the actual polarizing power trend

3. Question 3 · medium · Alkaline Earth Metals

   Among group 2 hydroxides, solubility in water INCREASES down the group (Mg(OH)2 sparingly soluble, Ba(OH)2 fairly soluble), but among group 2 sulfates, solubility DECREASES down the group (MgSO4 very soluble, BaSO4 essentially insoluble). This apparently opposite trend is explained by:

   • A.For hydroxides, lattice energy falls off faster than hydration enthalpy down the group; for sulfates, the large anion keeps lattice energy relatively constant while hydration enthalpy falls off, reversing the net solubility trend
   • B.Both hydroxide and sulfate solubility trends are actually identical and increase down the group
   • C.Sulfate ions are always insoluble with any metal cation, regardless of size
   • D.Hydroxide solubility trends are irrelevant to lattice/hydration energy considerations and depend only on the metal's reactivity
   Show answer and explanation

   Answer: A. For hydroxides, lattice energy falls off faster than hydration enthalpy down the group; for sulfates, the large anion keeps lattice energy relatively constant while hydration enthalpy falls off, reversing the net solubility trend

   Solubility depends on the relative magnitudes of lattice energy (which resists dissolution) and hydration enthalpy (which favours it); for the small hydroxide ion, lattice energy drops faster than hydration enthalpy as cation size increases, favouring dissolution, while for the large sulfate ion, lattice energy stays relatively constant (dominated by anion size) while hydration enthalpy of the cation falls, disfavouring dissolution down the group.

   For a salt MX, solubility broadly correlates with ΔHsolution = lattice energy − hydration enthalpy (more negative/favourable when hydration enthalpy dominates). For hydroxides (small OH− anion), lattice energy is sensitive to cation size and falls off substantially as the cation grows down the group (Mg2+ to Ba2+), while hydration enthalpy also falls but less steeply; the net effect favours increasing solubility down the group. For sulfates (large SO4 2− anion), lattice energy is dominated by the large anion and changes relatively little as the cation size increases (since anion size difference dominates the sum of radii), while cationic hydration enthalpy decreases substantially down the group (larger cations hydrate less exothermically); since hydration enthalpy falls faster than lattice energy in this case, solubility decreases down the group, explaining why BaSO4 is famously insoluble (used as a radio-opaque 'barium meal' in X-ray imaging) while MgSO4 (Epsom salt) is highly soluble.

   Common mistake: Assuming all group 2 salts follow the same solubility trend direction, without considering how anion size affects the lattice energy vs hydration enthalpy balance differently

4. Question 4 · medium · Alkaline Earth Metals

   Beryllium, unlike other group 2 elements, forms predominantly covalent compounds (e.g., BeCl2 is covalent and readily hydrolyzed, sublimes, and dissolves in organic solvents) rather than ionic ones. This is primarily due to:

   • A.Beryllium's very small size and high charge density giving it exceptionally strong polarizing power toward anions
   • B.Beryllium having a full valence octet unlike other group 2 elements
   • C.Beryllium being radioactive and hence forming unstable ionic lattices
   • D.Beryllium having the lowest electronegativity in group 2
   Show answer and explanation

   Answer: A. Beryllium's very small size and high charge density giving it exceptionally strong polarizing power toward anions

   Beryllium's extremely small ionic radius and 2+ charge give it very high charge density (polarizing power per Fajans' rules), causing it to distort anion electron clouds strongly enough to give its compounds substantial covalent character, unlike the more purely ionic compounds of Mg, Ca, Sr, and Ba.

   Beryllium is the smallest element in group 2 (ionic radius Be2+ ≈ 31 pm) and carries a 2+ charge, giving it an unusually high charge density among the alkaline earth metals. By Fajans' rules, high cationic charge density leads to strong polarization of the accompanying anion's electron cloud, distorting it enough to give substantial covalent character to the bond. This explains beryllium's many anomalous properties compared to Mg-Ba: BeCl2 is covalent, volatile, fumes in moist air, dissolves in organic solvents, and exists as a chain polymer in the solid state (each Be bridging two Cl atoms) rather than as a simple ionic lattice; Be(OH)2 is amphoteric (unlike the basic hydroxides of heavier group 2 members); and beryllium shows a strong diagonal relationship with aluminium, another small, highly charge-dense, covalent-bond-forming element.

   Common mistake: Assuming all group 2 metals form purely ionic compounds without accounting for beryllium's exceptionally small size and high polarizing power

5. Question 5 · medium · Boron Family

   Down group 13, the stability of the +1 oxidation state increases relative to +3 (e.g., Tl+ is more stable than Tl3+, while for boron and aluminium the +3 state dominates). This trend is explained by the:

   • A.Inert pair effect, where the ns2 electron pair becomes increasingly reluctant to participate in bonding for heavier p-block elements
   • B.Increasing atomic radius making all three valence electrons equally easy to remove down the group
   • C.Decreasing nuclear charge down the group
   • D.The increasing tendency of heavier elements to form covalent rather than ionic bonds
   Show answer and explanation

   Answer: A. Inert pair effect, where the ns2 electron pair becomes increasingly reluctant to participate in bonding for heavier p-block elements

   The inert pair effect arises because the ns2 electron pair (especially in heavier p-block elements) is poorly shielded by intervening filled d and f orbitals and is stabilized by relativistic contraction, making it increasingly reluctant to participate in bonding; hence heavier group 13 elements like thallium preferentially retain this pair, favouring the +1 over +3 oxidation state.

   The inert pair effect describes the increasing reluctance of the outermost ns2 electron pair to take part in bond formation as one descends a p-block group, particularly noticeable from period 4 onward. This arises because the ns orbital penetrates closer to the nucleus and is poorly shielded by the intervening filled (n-1)d10 (and for heavier elements, 4f14) subshells, along with relativistic effects that further stabilize (contract) the s-orbital in heavier elements. As a result, this pair becomes energetically more stable to retain (as a lone pair) rather than to use in bonding, favouring lower oxidation states. In group 13: B and Al almost exclusively show +3 (little inert pair effect at these light elements), Ga and In show both +1 and +3 (with +3 still generally more common/stable), and Tl strongly favours +1 (Tl+ salts are common and stable, while Tl3+ compounds are strong oxidizing agents that readily revert to Tl+).

   Common mistake: Confusing the inert pair effect with simple ionization enthalpy trends without recognizing its specific basis in poor d/f-orbital shielding and relativistic contraction

6. Question 6 · hard · Boron Family

   Diborane, B2H6, has a distinctive structure containing two bridging hydrogen atoms connecting the two boron atoms via 3-centre-2-electron (banana) bonds. Regarding this structure, which statement is correct?

   • A.Each B-H-B bridge bond involves 2 electrons shared among 3 atoms (one B, one H, one B), distinct from a normal 2-centre-2-electron bond
   • B.Diborane contains 8 normal 2-centre-2-electron B-H bonds, with no unusual bonding at all
   • C.The bridging hydrogens in B2H6 each contribute 2 electrons to their respective bridge bond
   • D.B2H6 has no bridging bonds; all six hydrogens are equivalent terminal atoms
   Show answer and explanation

   Answer: A. Each B-H-B bridge bond involves 2 electrons shared among 3 atoms (one B, one H, one B), distinct from a normal 2-centre-2-electron bond

   Diborane is electron-deficient (only 12 valence electrons total, insufficient for 8 normal 2-electron bonds), so it adopts a unique structure with 4 terminal normal B-H bonds and 2 bridging B-H-B bonds, each of which is a 3-centre-2-electron bond where 2 electrons are delocalized over 3 nuclei (B, H, B).

   B2H6 has a total of 2(3) + 6(1) = 12 valence electrons. If it had a conventional structure with 8 normal B-H bonds (as might be naively expected from 2 BH3 units), 16 electrons would be needed, but only 12 are available — making diborane electron-deficient. Structurally, diborane has 4 terminal hydrogens (2 on each boron) forming normal 2-centre-2-electron (2c-2e) B-H bonds using 8 electrons, and 2 bridging hydrogens positioned between the two borons, each involved in a 3-centre-2-electron (3c-2e) bond, where a single pair of electrons is delocalized over the B-H-B triangle (one B, one bridging H, the other B), using the remaining 4 electrons. This banana-bond bridging arrangement is a hallmark of electron-deficient boron hydride chemistry and gives diborane its characteristic non-classical bonding structure.

   Common mistake: Assuming diborane has 8 conventional bonds like a simple dimer of BH3, ignoring its electron deficiency and unique bridge bonding

7. Question 7 · easy · Carbon Family

   Carbon shows the strongest tendency toward catenation (self-linking to form chains and rings) among all elements, far exceeding that of silicon, germanium, or any other element in group 14. This is primarily because:

   • A.The C-C bond is exceptionally strong due to carbon's small size allowing effective orbital overlap, and carbon forms strong pi bonds enabling multiple bonding within chains
   • B.Carbon has more valence electrons than silicon
   • C.Silicon-silicon bonds are stronger than carbon-carbon bonds
   • D.Carbon atoms are larger than silicon atoms, allowing more stable chain formation
   Show answer and explanation

   Answer: A. The C-C bond is exceptionally strong due to carbon's small size allowing effective orbital overlap, and carbon forms strong pi bonds enabling multiple bonding within chains

   Carbon's small atomic size allows for highly effective orbital overlap, giving unusually strong and short C-C single bonds (bond energy ~348 kJ/mol) compared to Si-Si (~226 kJ/mol) or Ge-Ge bonds, and carbon uniquely forms strong, stable pi bonds, enabling extensive catenation with both single and multiple bonds, unlike heavier group 14 elements.

   Catenation (the ability of atoms of the same element to bond to each other in chains or rings) is strongly favoured by carbon due to several factors: its small atomic size permits close approach and maximal orbital overlap, giving a strong C-C single bond; carbon can also effectively form pi bonds via sideways overlap of 2p orbitals (unlike heavier group 14 elements, whose larger, more diffuse p-orbitals overlap poorly sideways), allowing carbon to form stable double and triple bonds within chains; and the C-C bond energy substantially exceeds the corresponding Si-Si, Ge-Ge, or Sn-Sn bond energies, which decrease down the group as atomic size increases and orbital overlap becomes less effective. This combination of strong sigma AND pi bonding capability, absent to the same degree in heavier congeners, is the foundation of the vast diversity of organic (carbon) chemistry.

   Common mistake: Assuming catenation tendency should increase down the group like atomic size, rather than recognizing it actually decreases due to weakening bond strength

8. Question 8 · medium · Carbon Family

   Graphite conducts electricity well, while diamond is an electrical insulator, even though both are pure allotropes of the same element, carbon. This difference is explained by:

   • A.In graphite, each carbon is sp2 hybridized with one delocalized electron per atom free to move within the planar layers, while in diamond, each carbon is sp3 hybridized with all four electrons localized in strong sigma bonds
   • B.Graphite contains metallic impurities that conduct electricity, while pure diamond has none
   • C.Diamond has more carbon atoms per unit cell than graphite, making it an insulator
   • D.Graphite is denser than diamond, and higher density always implies better conductivity
   Show answer and explanation

   Answer: A. In graphite, each carbon is sp2 hybridized with one delocalized electron per atom free to move within the planar layers, while in diamond, each carbon is sp3 hybridized with all four electrons localized in strong sigma bonds

   In graphite's layered structure, each carbon is sp2 hybridized, using 3 electrons for sigma bonds within the hexagonal planar layer and leaving one electron per carbon delocalized in a pi system that extends across the layer, enabling electrical conduction; diamond's sp3 carbons use all 4 valence electrons in strong, fully localized sigma bonds, leaving no free electrons for conduction.

   Graphite has a layered structure where each carbon atom is sp2 hybridized, forming three strong sigma bonds to neighboring carbons within a planar hexagonal sheet, using 3 of its 4 valence electrons. The fourth electron occupies an unhybridized p-orbital perpendicular to the plane, and these p-orbitals overlap sideways across the entire sheet to form a delocalized pi electron system, similar to a 2D 'electron sea,' allowing electrons to move freely within the plane and conduct electricity along the layers. Diamond, in contrast, has a 3D tetrahedral network where each carbon is sp3 hybridized, using all 4 valence electrons to form four strong, localized sigma bonds to neighboring carbons; with no delocalized or free electrons available, diamond cannot conduct electricity and behaves as an electrical insulator (and is in fact one of the hardest known materials due to this extensive 3D covalent network).

   Common mistake: Attributing conductivity differences to impurities or density rather than the fundamental difference in hybridization and electron delocalization

Questions about s and p Block Elements for NEET

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