NEET The p-Block Elements Previous Year Questions with Solutions

All Group 15 elements form volatile covalent hydrides of the general formula EH₃, where E = N, P, As, Sb, Bi:

• • Arsine (AsH₃) is a well-known, highly toxic gas formed by the reduction of arsenic compounds.
• • Therefore, Statement I is unequivocally CORRECT.

Group 15 elements form two primary series of oxides: trioxides (E₂O₃) and pentoxides (E₂O₅). The oxidation state of the element in the pentoxide is +5:

• • Antimony pentoxide (Sb₂O₅) is synthesized by oxidizing antimony or antimony trioxide with concentrated nitric acid.
• • The statement asserts that "Antimony cannot form antimony pentoxide", which is scientifically FALSE.
• • (It is bismuth, Bi, that does not form stable pentoxide due to the dominant inert pair effect).

Statement I is correct and Statement II is incorrect, which corresponds to Option (3).

• • 100% aligned with the official NTA Final Answer Key.

Xenon has 8 valence electrons (5s² 5p⁶):

• • A matches II (A-II).
• • XeF₂: 2 single bonds (2σ) + 3 lone pairs = 5 electron pairs ⟹ sp³d hybridization. The three lone pairs occupy equatorial positions, giving a Linear geometry (B-I).

In XeOF₄, xenon forms 1 double bond to oxygen (1σ + 1π) and 4 single bonds to fluorine (4σ):

• • The four fluorine atoms form the square base, oxygen occupies one axial position, and the lone pair occupies the opposite axial position.
• • C matches IV (C-IV).

In XeF₆, xenon forms 6 single bonds to fluorine (6σ) and retains 1 lone pair:

• • The lone pair distorts the regular octahedral symmetry, resulting in a distorted octahedral geometry.
• • D matches III (D-III).
• • Complete match: A-II, B-I, C-IV, D-III, matching Option (1).

The measured boiling points of Group 16 hydrides are: - Water (H2O): 373 K - Hydrogen telluride (H2Te): 269 K - Hydrogen selenide (H2Se): 232 K - Hydrogen sulphide (H2S): 213 K The observed sequence is strictly: H2O > H2Te > H2Se > H2S. Statement I is correct.

For non-hydrogen bonding hydrides (H2S, H2Se, H2Te), boiling points increase regularly down the group as molecular mass and electron cloud polarizability increase, strengthening London dispersion forces. Based purely on molecular mass (18 g/mol), H2O should possess the weakest dispersion forces and lowest boiling point. However, each H2O molecule forms up to four strong intermolecular hydrogen bonds in a tetrahedral network, requiring substantial thermal energy to disrupt. Statement II is correct.

Group 16 elements have the valence shell electronic configuration ns² np⁴. They can achieve a noble gas octet by gaining two electrons to form the -2 oxidation state. However, the stability of the -2 oxidation state decreases sharply down the group as electronegativity drops and metallic character rises:

Polonium (Po, Z = 84) is a radioactive, distinctly metallic element. As stated verbatim in NCERT, because of its pronounced metallic character and electropositive nature, polonium hardly shows the -2 oxidation state; its predominant and stable oxidation states are +2 and +4:

Oxygen is predominantly -2 (in H2O, metal oxides). Sulfur forms sulphides (S²⁻, -2). Selenium forms selenides (Se²⁻, -2). Tellurium forms tellurides (Te²⁻, -2, e.g. H2Te, Na2Te). All of O, S, Se, and Te exhibit -2, whereas Polonium does not.

As we descend Group 13 from boron to thallium, the valence electron configuration is ns² np¹. In heavier members (especially thallium, Z = 81), the intervening 4f and 5d electrons provide poor screening of the nuclear charge. Consequently, the effective nuclear charge experienced by the valence 6s electrons increases sharply, holding them tightly and preventing their participation in chemical bonding. This reluctance of the s-electron pair to participate in bonding is known as the inert pair effect.

Because the inert pair effect reaches its maximum at thallium, the +1 oxidation state is far more stable than the +3 oxidation state:

Tl³⁺ has a high reduction potential (E° = +1.26 V) and acts as a powerful oxidizing agent. Iodide (I⁻) is a strong reducing agent. Tl³⁺ readily oxidizes I⁻ to I₂, rendering covalent TlI₃ thermodynamically unstable:

Therefore, TlI is much more stable than TlI₃ (TlI > TlI₃). In contrast, TlCl is more stable than TlCl₃ (making Option 2 false); InI₃ is more stable than InI (making Option 3 state InI₃ > InI, but between TlI > TlI₃ it is the quintessential example tested); AlCl₃ is far more stable than AlCl (making Option 4 false).

Coke is a carbonaceous solid residue derived from destructive distillation of low-ash, low-sulfur bituminous coal. Because of its high carbon content and thermal stability, it is predominantly utilized in extraction metallurgy (e.g., blast furnace reduction of iron ore) as a reducing agent:

In diamond, each carbon atom undergoes sp³ hybridization and is linked to four neighboring carbon atoms by single covalent bonds (length 154 pm) directed tetrahedrally, producing a rigid, three-dimensional covalent lattice:

Fullerenes are cage-like hollow spheroidal molecules consisting of interconnected hexagonal and pentagonal rings (such as C₆₀ buckminsterfullerene, featuring 20 six-membered and 12 five-membered rings in a soccer-ball shape):

Graphite possesses a two-dimensional layered sheet structure where sp²-hybridized carbon atoms form planar hexagonal rings. The individual layers are separated by 340 pm and held together only by weak van der Waals forces, allowing them to glide over one another. This cleavability makes graphite slippery and suitable as a dry lubricant at elevated temperatures:

Known as Marshall’s acid, each sulphur is in oxidation state +6 and is bonded to one -OH, two =O, and joined by a peroxo linkage:

Sulphur is surrounded by four oxygen atoms in a tetrahedral geometry:

Also known as oleum, formed by condensing two molecules of sulphuric acid with elimination of water:

Sulphur has one lone pair, with oxidation state +4:

Noble gases (Group 18) are monoatomic gases with stable closed-shell electronic configurations ($ns^2 np^6$, except He which has $1s^2$). Because they possess neither permanent electric dipole moments nor covalent bonds between atoms, the only attractive forces holding them together in condensed states are exceedingly weak London dispersion forces (induced dipole - induced dipole interactions).

Owing to the weakness of dispersion forces, very little thermal energy is required to overcome these interactions. Consequently, noble gases have extremely low melting and boiling points. Helium has the lowest boiling point of any known element at $4.2\text{ K}$ ($-268.95^\circ\text{C}$). Therefore, Statement (2) asserting that noble gases have "very high melting and boiling points" is factually and scientifically incorrect.

- Statement 1: Noble gases are sparingly soluble in water due to weak dipole-induced dipole forces with polar water molecules (True). - Statement 3: Interatomic forces are indeed weak dispersion forces (True). - Statement 4: Noble gases possess stable octets; accepting an electron into the next higher quantum shell requires substantial energy input, resulting in large positive electron gain enthalpy ($\Delta_{\text{eg}}H > 0$) (True).

Hydrogen halides in aqueous solution dissociate to produce hydronium ions and halide ions: $HX + H_2O \rightleftharpoons H_3O^+ + X^-$. Their experimental acid dissociation constants ($K_a$) and $pK_a$ values demonstrate that $HF$ is a weak acid ($pK_a \approx 3.2$), whereas $HCl$ ($pK_a \approx -7$), $HBr$ ($pK_a \approx -9$), and $HI$ ($pK_a \approx -10$) are strong acids:

As we descend Group 17 from Fluorine to Iodine, the atomic radius increases significantly ($F: 72\text{ pm} \to I: 133\text{ pm}$). The overlap between the small $1s$ orbital of hydrogen and the increasingly diffuse valence $p$ orbitals of halogen weakens dramatically. The $H-X$ bond length increases and the bond dissociation enthalpy decreases sharply:

Because the $H-I$ bond is the longest and weakest, it cleaves most readily to furnish $H^+$ ions in solution. Hence, the decrease in bond strength is the primary physical driving force behind the increase in acid strength. Statement II is scientifically true and correctly explains Statement I.

For Group 16 hydrides ($H_2O, H_2S, H_2Se, H_2Te$), the size of the central element $E$ increases down the group ($O < S < Se < Te$), leading to longer $E-H$ bonds and decreasing $E-H$ bond dissociation enthalpies: $$H_2O (463\text{ kJ/mol}) > H_2S (347\text{ kJ/mol}) > H_2Se (276\text{ kJ/mol}) > H_2Te (238\text{ kJ/mol})$$ Consequently, ease of proton release increases down the group, so acidic strength increases: $$\text{Acidic Strength: } H_2O < H_2S < H_2Se < H_2Te \implies K_a(H_2O) < K_a(H_2S) < K_a(H_2Se) < K_a(H_2Te)$$

Because $pK_a = -\log_{10} K_a$, an increase in acid dissociation constant $K_a$ strictly corresponds to a decrease in $pK_a$: - $pK_a(H_2O) \approx 14.0$ (or $15.74$ auto-ionization) - $pK_a(H_2S) \approx 7.0$ - $pK_a(H_2Se) \approx 3.89$ - $pK_a(H_2Te) \approx 2.6$ Thus, the sequence $H_2O < H_2S < H_2Se < H_2Te$ is the order of DECREASING pKa, making the label "Increasing $pK_a$ values" false.

- Option (1): $HF < HCl < HBr < HI$ correctly follows decreasing $H-X$ bond dissociation enthalpy, increasing acidic strength. - Option (3): $NH_3 < PH_3 < AsH_3 < SbH_3$ correctly reflects decreasing $E-H$ bond stability, increasing acidic character. - Option (4): In Group 14, due to the inert pair effect, the stability of the $+2$ state increases relative to $+4$ down the column ($Pb^{2+} > Pb^{4+}$). Thus, $PbO_2$ readily accepts electrons to reduce to $Pb^{2+}$, functioning as a powerful oxidizing agent.

SOLID carbon dioxide (dry ice) sublimes directly at $$-78.5^\circ\text{C}$$ without melting into liquid, making it ideal as a refrigerant for ice cream and frozen food. Statement (a) specifically claims "$$\text{CO}_2\text{(g)}$$" (gaseous carbon dioxide) is used as the refrigerant, which is technically incorrect.

Buckminsterfullerene ($$C_{60}$$) has a soccer-ball shape containing 60 carbon vertices. It consists of exactly TWENTY 6-membered hexagonal rings and TWELVE 5-membered pentagonal rings. Statement (b) deliberately inverts these numbers ("twelve six-carbon rings and twenty five-carbon rings"), making it blatantly false.

ZSM-5 (Zeolite Socony Mobil-5) is a microporous aluminosilicate shape-selective catalyst used in the petroleum industry to dehydrate alcohols directly into synthetic gasoline/petrol hydrocarbons ($$\text{Alcohols} \xrightarrow{\text{ZSM-5}} \text{Gasoline}$$). This statement is completely accurate.

Carbon monoxide ($$\text{CO}$$) is a colorless, odorless, and virtually water-insoluble toxic gas formed by incomplete combustion of carbon. This statement is factually true.

Only statements (c) and (d) are correct, corresponding to Option 4.

Sulphur forms several oxoacids where sulphur exhibits tetrahedral coordination with oxygen: 1. Sulphurous acid ($$\text{H}_2\text{SO}_3$$): Contains one $$\text{S}=\text{O}$$, two $$\text{S}-\text{OH}$$, and one lone pair (oxidation state +4). 2. Sulphuric acid ($$\text{H}_2\text{SO}_4$$): Contains two $$\text{S}=\text{O}$$ and two $$\text{S}-\text{OH}$$ bonds (oxidation state +6).

Pyrosulphuric acid ($$\text{H}_2\text{S}_2\text{O}_7$$) contains two $$\text{SO}_3\text{H}$$ groups linked through a single oxygen bridge: $$\text{HO}-\text{SO}_2-\text{O}-\text{SO}_2-\text{OH}$$. It has an $$\text{S}-\text{O}-\text{S}$$ linkage, NOT a peroxide linkage.

Peroxodisulphuric acid (Marshall's acid, $$\text{H}_2\text{S}_2\text{O}_8$$) is formed by joining two sulphuric acid moieties through a peroxy (peroxide) bridge: $$\text{HO}-\text{SO}_2-\text{O}-\text{O}-\text{SO}_2-\text{OH}$$ The two central oxygen atoms are covalently linked by an $$\text{-O-O-}$$ single bond, each having an oxidation state of $$-1$$, allowing both sulphur atoms to retain the maximum +6 oxidation state.

Therefore, $$\text{H}_2\text{S}_2\text{O}_8$$ uniquely possesses the $$\text{-O-O-}$$ peroxide linkage.

On a small scale, ammonia is prepared from urea. Urea reacts with water to form ammonium carbonate ($$A$$): $$\text{NH}_2\text{CONH}_2 + 2\text{H}_2\text{O} \to (\text{NH}_4)_2\text{CO}_3 \text{ (Compound A)}$$

Ammonium carbonate is unstable and readily decomposes into ammonia, water, and carbon dioxide: $$(\text{NH}_4)_2\text{CO}_3 \rightleftharpoons 2\text{NH}_3\text{ (Gas B)} + \text{H}_2\text{O} + \text{CO}_2$$ Thus, component $$B$$ is ammonia ($$\text{NH}_3$$).

Due to the presence of a lone pair of electrons on the nitrogen atom, ammonia acts as a Lewis base and donates electron pairs to transition metal cations to form coordination complexes. When ammonia solution is added to pale blue aqueous copper sulphate, it initially precipitates pale blue $$\text{Cu(OH)}_2$$, which dissolves in excess ammonia to form an intense DEEP BLUE solution ($$C$$): $$\text{Cu}^{2+} \text{(aq, pale blue)} + 4\text{NH}_3 \text{(aq)} \rightleftharpoons [\text{Cu}(\text{NH}_3)_4]^{2+} \text{(aq, deep blue)}$$

The deep blue species $$C$$ is the tetraamminecopper(II) complex ion, $$[\text{Cu}(\text{NH}_3)_4]^{2+}$$.