NEET Previous Year Questions (2014-2025): The p-Block Elements (Group 15, 16, 17 & 18) | Chemistry Class 11 PDF Download

Ans: (a)

• Statement I: "Like nitrogen that can form ammonia, arsenic can form arsine." This statement is correct because arsenic, like nitrogen, can react with hydrogen to form arsine (AsH₃). This is due to arsenic's chemical similarity to nitrogen as both belong to the same group in the periodic table.
• Statement II: "Antimony cannot form antimony pentoxide." This statement is incorrect because antimony can form antimony pentoxide (Sb₂O₅) when it exhibits its +5 oxidation state. This is consistent with the behavior of group 15 elements. Therefore, Statement I is correct, but Statement II is incorrect.

Correct Answer: Option A) Statement I is correct but Statement II is incorrect.

Ans: (d)

• Group 16 elements (chalcogens) include O, S, Se, Te, Po.
• These elements usually show –2 oxidation state because they have six valence electrons and can gain two electrons to achieve noble gas configuration.
• Oxygen (O): Being the most electronegative, it predominantly exists in –2 oxidation state (e.g., H₂O, SO₂).
• Selenium (Se) and Tellurium (Te): Also exhibit –2 oxidation state (e.g., H₂Se, H₂Te).
• Polonium (Po): Being more metallic in nature, it does not stabilize the –2 oxidation state. Instead, it commonly shows +2 and +4 oxidation states (e.g., PoCl₂, PoO₂).
• Therefore, Polonium (Po) does not exhibit –2 oxidation state → Correct Answer: (d). 

Ans: (d)

• Statement I: The boiling points of hydrides of Group 16 elements (H₂O, H₂Te, H₂Se, and H₂S) are influenced by factors such as molecular mass and intermolecular forces. H₂O has the highest boiling point in this series due to hydrogen bonding. 
• The order of boiling points is: H₂O > H₂Te > H₂Se > H₂S. This is because, although H₂Te and H₂Se have higher molecular masses than H₂O, the hydrogen bonding in H₂O causes it to have a much higher boiling point. Thus, Statement I is true.
  
• Statement II: The molecular mass of H₂O is lower than that of H₂Te, H₂Se, and H₂S, so based on molecular mass alone, H₂O should have a lower boiling point. However, due to hydrogen bonding, which is a strong intermolecular force, H₂O has a higher boiling point than expected. This is why Statement II is true.
  Thus, both Statement I and Statement II are true.
  

Ans: (c)

• The reaction of ROH (alcohol) with PCl₃ (phosphorus trichloride) leads to the formation of RCl (alkyl chloride) and H₃PO₃ (phosphorous acid) as a product A. The phosphorus trichloride reacts with alcohol to produce H₃PO₃ and RCl.
• The reaction of ROH (alcohol) with PCl₅ (phosphorus pentachloride) leads to the formation of RCl (alkyl chloride), HCl (hydrochloric acid), and POCl₃ (phosphoryl chloride) as product B.
  Thus, A is H₃PO₃ and B is POCl₃.

Ans: (d)

A. Nitrogen, Property III: Element with highest ionisation enthalpy in group 15.
Nitrogen has the highest ionization enthalpy in group 15 because it has a half-filled stable configuration in its valence shell (2s² 2p³). This stability makes it more difficult to remove electrons compared to other elements in the same group.

B. Fluorine molecule, Property IV: Strongest oxidising agent.
Fluorine is the most electronegative element, which makes it the strongest oxidizing agent. It readily accepts electrons to form fluoride ions (F⁻), making it a powerful oxidizing agent.

C. Oxygen molecule, Property I: Paramagnetic.
Oxygen (O₂) has two unpaired electrons in its molecular orbitals, which makes it paramagnetic. This means it is attracted by a magnetic field.

D. Xenon atom, Property II: Most reactive element in group 18.
Xenon is one of the noble gases, but among the noble gases, it is the most reactive. It can form compounds like xenon hexafluoride (XeF₆) under certain conditions.

Thus, the correct matching is: A-III, B-IV, C-I, D-II
Therefore, the correct answer is (D).

Ans: (b)

• Statement (a): PCl₅ has a trigonal bipyramidal structure with two different bond angles (90° between axial and equatorial positions, 120° between equatorial positions). Hence, Statement (a) is correct.
• Statement (b): The axial P-Cl bonds are slightly longer than the equatorial P-Cl bonds in PCl₅. Therefore, Statement (b) is incorrect.
• Statement (c): The central phosphorus atom in PCl₅ is sp³d hybridized due to the five bonding pairs of electrons, forming a trigonal bipyramidal geometry. Hence, Statement (c) is correct.
• Statement (d): PCl₅ consists of five sigma bonds (P-Cl). Hence, Statement (d) is correct.

Thus, the incorrect statement is (b).

Ans: (a)
(a) The acidic strength of HX (X = F, Cl, Br, and I) follows the order: HF > HCl > HBr > HI
This statement is incorrect. The acidic strength of hydrohalic acids (HX) generally increases as we go down the group in the periodic table. This is because the bond strength between H and X decreases as we move from fluorine to iodine. Since the bond between hydrogen and iodine (HI) is the weakest, HI is the strongest acid, not HF. The correct order of acidic strength is HI > HBr > HCl > HF.

(b) Fluorine exhibits -1 oxidation state, whereas other halogens exhibit +1, +3, +5, and +7 oxidation states also.
This statement is correct. Fluorine always exhibits a -1 oxidation state because it is the most electronegative element, whereas the other halogens (Cl, Br, I) can exhibit positive oxidation states (e.g., +1 in compounds like ClF, +3 in compounds like ClO₂, +5 in compounds like ClO₃⁻, and +7 in compounds like Cl₂O₇).

(c) The enthalpy of dissociation of F₂ is smaller than that of Cl₂.
This statement is incorrect. The bond dissociation enthalpy of F₂ is much higher than that of Cl₂ due to the small size of fluorine atoms and the high repulsion between them in the F₂ molecule. In contrast, Cl₂ has a lower bond dissociation enthalpy due to its larger atomic size and less repulsion.

(d) Fluorine is a stronger oxidizing agent than chlorine.
This statement is correct. Fluorine is the strongest oxidizing agent because it is the most electronegative element. It readily accepts electrons, while chlorine is a weaker oxidizing agent in comparison.

Conclusion: The incorrect statement is (a). The correct order of acidic strength is HI > HBr > HCl > HF, not HF > HCl > HBr > HI.
Therefore, the correct answer is (a).

Ans: (b)

• Statement (a): PEt₃ (triethylphosphine) and AsPh₃ (triphenylarsine) as ligands can indeed form dπ−dπ bonds with transition metals. These ligands have lone pairs on the phosphorus and arsenic atoms that can participate in back-donation to empty d-orbitals on the transition metal. Hence, Statement (a) is correct.
• Statement (b): The N−N single bond is not as strong as the P−P single bond. The N−N bond is weaker due to the small size of nitrogen and the lone pair-electron repulsion in the nitrogen atoms. In contrast, the P−P bond is stronger due to less repulsion between the lone pairs on the phosphorus atoms. Hence, Statement (b) is incorrect.
• Statement (c): Nitrogen does indeed have a unique ability to form pπ−pπ multiple bonds with other elements such as nitrogen, carbon, and oxygen. For example, nitrogen forms double and triple bonds with carbon and oxygen, as in N≡N and C=O. Hence, Statement (c) is correct.
• Statement (d): Nitrogen cannot form dπ−pπ bonds, as it does not have accessible low-lying d-orbitals in its valence shell, unlike heavier elements in the same group (such as phosphorus and arsenic). Hence, Statement (d) is correct.

Thus, the incorrect statement is (b).

Ans: (b)

Ans: (c)

• Coke is largely used as a reducing agent in metallurgy.
• In diamond, each carbon atom undergoes sp³ hybridisation and linked to four other carbon atoms by using hybridised orbitals in tetrahedral fashion.
• Buckminsterfullerene contains six membered and five membered rings and hence is a cage like molecule.
• Graphite is very soft and slippery. Hence, it is used as a dry lubricant in machines running at high temperature.
  

Ans: (b)

Ans: (b)
In diborane both the boran atoms are sp³ hybridized.

Ans: (b)

• The boiling points of these hybrids not exactly increases with increase in molar mass.
• H₂O has maximum boiling point due to intermolecular hydrogen bonding.

Ans: (a)

(a) - (iv), (b) - (i), (c) - (ii), (d) - (iii)

Ans: (b)

• Borax : Na₂B₄O₇ . 10H₂O = Na₂[B₄O₅(OH)₄] . 8H₂O 
• Kernite : Na₂B₄O₇ . 4H₂O 
• Orthoboric acid : H₃BO₃ = B(OH)₃ 

Ans: (c)

This is industrial method of preparation of nitric acid.

Ans: (c)

Product X is B₂O₃

Ans: (b)
In the modern periodic table, moving down the group as the size of halogen atom increases, the H - X bond length also increases as a result the bond enthalpy decreases. Hence, the acidic strength also increases.
So, the correct order of acidic strength is
HI > HBr > HCl > HF

Ans: (c)
Nobles gases have weak interatomic forces (van der Waals' forces). So, they have low melting and boiling points.

Ans: (d)
Stronger is the acid, lower is the value of pK_a. On moving down the group, bond dissociation enthalpy of hydrides of group 16 elements decreases hence acidity increases and pK_a value decreases. Correct order of pK_a value will be
H₂O > H₂S > H₂Se > H₂Te

Ans: (a)
The carboxyhaemoglobin is about 300 times more stable than oxyhaemoglobin.

Ans: (b)
Dry ice is used as refrigerant. C₆₀ contains 20 six membered ring, 12 five membered rings. In the presence of a zeolite catalyst (HZSM-5), ethanol, methanol and larger alcohols can be converted into gasoline at 300-400$\circ$ C. CO, is a tasteless, colorless, and odorless gas. 

Ans: (a)

Ans: (d)

• Mixture of CO and H₂ gases is known as water gas or synthesis gas.
• Temporary hardness of water is due to bicarbonates of calcium and magnesium.
• Diborane (B₂H₆) is an electron deficient hydride.
• H₂O₂ is non-planar molecule having open book like structure.

Ans: (d)
Due to presence of d-orbital in Si, Ge and Sn they form species like SiF₆^2-, [GeCl₆]^2-, [Sn(OH)₆]^2-
SiCl₆^2- does not exist because six large chloride ions cannot be accommodated around Si⁴⁺ due to limitation of its size.

Ans: (c)
On going down the group thermal stability order for H₂E decreases because H–E bond energy decreases
∴ Order of stability would be:-
H₂Po < H₂Te < H₂Se < H₂S < H₂O

Ans: (b)

Three equatorial bonds and two axial bonds.
Due to unsymmetry PCl5 is reactive.

Ans: (c)
PbF4 is an ionic compound.

Ans: (b)

Ans: (c)
All halogens show both positive and negative oxidation states while fluorine shows only negative oxidation state except due to high electronegativity and small size, F forms only one oxoacid, HOF known as Fluoric (I) acid. Oxidation number of F is +1 in HOF.

Ans: (c)
Boron does not have vacant d-orbitals in its valence shell, so it cannot extend its covalency beyond 4. i.e., ‘B’ cannot form the ions like MF₆^3–. 

Ans: (d)
Atomic and ionic radii. Atomic and ionic radii of group 13 elements increase from top to bottom in the group. This is due to increase in the number of energy shells in each succeeding element. However, the atomic radius of gallium (Ga) is less than that of aluminium (Al).
It is due to the poor shielding of the valence electrons of Ga by the inner 3d-electrons. As a result, the effective nuclear charge of Ga is somewhat greater in magnitude than that of Al. Thus, the electrons in gallium experience the greater force of attraction by the nucleus than in aluminium. Hence the atomic size of Ga(135 pm) is slightly less than that of Al(143 pm).

Ans: (c)
OF₂ (oxygen difluoride) is a fluoride of oxygen because fluorine is more electronegative than oxygen.

Ans: (d)
All oxyacids of phosphorus which have P—H bonds act as strong reducing agents. H₃PO₂ has two P—H bonds hence, it acts as a strong reducing agent.

Ans: (a)
HF forms strong intermolecular H-bonding due to high electronegativity of F. Hence, the boiling point of HF is exceptionally high. Boiling points of other hydrogen halides gradually increases from HCl to HI due to increase in the size of halogen atoms form Cl to I, which further increases the magnitude of van der Waal's forces.

Ans: (a)
The given elements belong to 13th group. The elements mainly exhibit +3 and +1 oxidation states. As we know, the stability of lower oxidation state i.e., +1 state, increases on moving down the group due to inert pair effect. The, stability follows the order :
Al < Ga < In < Tl