NEET Previous Year Questions (2014-2024): The p-Block Elements (Group 15, 16, 17 & 18) | Chemistry Class 11 PDF Download

2024

Q1: Among Group 16 elements, which one does NOT show -2 oxidation state?
(a) O
(b) Se
(c) CTe
(d) Po         (NEET 2024)
Ans: (d)
The elements in Group 16 of the periodic table, also known as the chalcogens, include oxygen (O), sulfur (S), selenium (Se), tellurium (Te), and polonium (Po). Generally, these elements are known to commonly exhibit a -2 oxidation state because they have six electrons in their outermost shell and can gain two electrons to complete their octet, thereby achieving a stable electronic configuration similar to the nearest noble gas.
Oxygen, being the most electronegative element in this group, predominantly shows a -2 oxidation state in most of its compounds, such asn water (H₂O) and oxides like SO₂.
Selenium and Tellurium also typically exhibit the -2 oxidation state. For instance, in compounds like H₂Se (hydrogen selenide) and H₂Te (hydrogen telluride), these elements are in the -2 state.
Polonium (Po), on the other hand, is different from the lighter chalcogens. Being a metalloid with more metallic character, it does not favor the -2 oxidation state nearly as strongly as the other elements in its group. Polonium most commonly exhibits +2 and +4 oxidation states in its compounds, like PoO₂ and PoCl₂. Due to the relativistic effects and its position in the periodic table, the -2 oxidation state is unstable and rare in polonium compounds.
Given this, the correct answer is Option D: Po. Polonium does not commonly exhibit the -2 oxidation state like the other Group 16 elements.

Q.2. The element expected to form largest ion to achieve the nearest noble gas configuration is    (NEET 2023)
(a) F
(b) N
(c) Na
(d) O
Ans: (b)

Q3: Match List-I with List-II.    (NEET 2023)

Choose the correct answer from the options given below:
(a) A-II, B-IV, C-I, D-III
(b) A-IV, B-I, C-II, D-III
(c) A-III, B-I, C-IV, D-II
(d) A-III, B-IV, C-I, D-II
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.

Q4:  Match List-I with List-II:     (NEET 2023)

Choose the correct answer from the options given below.
(a) A–I, B–III, C–II, D–IV
(b) A–III, B–IV, C–I, D–II
(c) A–I, B–III, C–IV, D–II
(d) A–III, B–IV, C–II, D–I

Ans: (b)

2022

Q1: Which of the following statement is not correct about diborane?    (NEET 2022 Phase 1)
(a) The four terminal Hydrogen atoms and the two Boron atoms lie in one plane.
(b) Both the Boron atoms are sp² hybridised.
(c) There are two 3-centre-2-eIectron bonds.
(d) The four terminal B-H bonds are two centre two electron bonds.
Ans: (b)
In diborane both the boran atoms are sp³ hybridized.

Q2: Given below are two statements
Statement I : The boiling points of the following hydrides of group 16 elements increases in the order -
H₂O < H₂S < H₂Se < H₂Te
Statement II : The boiling points of these hydrides increase with increase in molar mass.
In the light of the above statements, choose the most appropriate answer from the options given below :
(a) Both Statement I and Statement II are correct
(b) Both Statement I and Statement II are incorrect
(c) Statement I is correct but Statement II is incorrect
(d) Statement I is incorrect but Statement II is correct    (NEET 2022 Phase 1)
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.

Q3: Match List-I with List-II.  (NEET 2022 Phase 1)

Choose the correct answer from the options given below:
(a) (a) - (iv), (b) - (i), (c) - (ii), (d) - (iii)
(b) (a) - (iii), (b) - (i), (c) - (ii), (d) - (iv)
(c) (a) - (i), (b) - (ii), (c) - (iv), (d) - (iii)
(d) (a) - (ii), (b) - (iii), (c) - (iv), (d) - (i)
Ans: (a)

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

Q4: Match List-I with List-II :  (NEET 2022 Phase 2)

Choose the correct answer from the options given below :
(a) (a) - (i), (b) - (iii), (c) - (iv), (d) - (ii)
(b) (a) - (iv), (b) - (ii), (c) - (iii), (d) - (i)
(c) (a) - (ii), (b) - (iv), (c) - (iii), (d) - (i)
(d) (a) - (iii), (b) - (i), (c) - (iv), (d) - (ii)
Ans: (b)

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

 
Q5: Which of the following reactions is a part of the large scale industrial preparation of nitric acid?    (NEET 2022 Phase 2)

(a)

(b)

(c)

(d)

Ans: (c)

This is industrial method of preparation of nitric acid.

Q6:  in the above reaction the product "X" is :
(a) NaB₃O₅
(b) H₃BO₃
(c) B₂O₃
(d) Na₂B₂O₅    (NEET 2022 Phase 2)
Ans: (c)

Product X is B₂O₃

2021

Q1: Statement I: Acid strength increases in the order given as HF << HCl << HBr << HI.
Statement II: As the size of the elements F, Cl, Br, I increase down the group, the bond strength of HF, HCI, HBr, and HI decreases and so the acid strength increases.
In the light of the above statements, choose the correct answer from the options given below.     (NEET 2021)
(a) Statement I is correct but Statement II is false.
(b) Statement I is incorrect but Statement II is true.
(c) Both Statement I and Statement II are true.
(d) Both Statement I and Statement II are false.
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

Q2: Noble gases are named because of their inertness towards reactivity. Identify an incorrect statement about them.
(a) Noble gases have large positive values of electron gain enthalpy
(b) Noble gases are sparingly soluble in water
(c) Noble gases have very high melting and boiling points
(d) Noble gases have weak dispersion forces    (NEET 2021)
Ans: (c)
Nobles gases have weak interatomic forces (van der Waals' forces). So, they have low melting and boiling points.

Q3: In which one of the following arrangements the given sequence is not strictly according to the properties indicated against it?     (NEET 2021)
(a) NH₃ < PH₃ < AsH₃ < SbH₃ : Increasing acidic character
(b) CO₂ < SiO₂ < SnO₂ < PbO₂ : Increasing oxidizing power
(c) HF < HCl < HBr < HI : Increasing acidic strength
(d) H₂O < H₂S < H₂Se < H₂Te : Increasing pK_a values
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

2020

Q1: Which of the following is not correct about carbon monoxide?     (NEET 2020)
(a) The carboxyhaemoglobin (haemoglobin bound to CO) is less stable than oxyhaemoglobin.
(b) It is produced due to incomplete combustion.
(d) It forms carboxyhaemoglobin.
(d) It reduces oxygen carrying ability of blood.
Ans: (a)
The carboxyhaemoglobin is about 300 times more stable than oxyhaemoglobin.

Q2: Identify the correct statements from the following:      (NEET 2020)
(a) CO₂(g) is used as refrigerant for ice-cream and frozen food.
(b) The structure of C₆₀ contains twelve six carbon rings and twenty five carbon rings.
(c) ZSM-5, a type of zeolite, is used to convert alcohols into gasoline.
(d) CO is colourless and odourless gas.
(a) (b) and (c) only
(b) (c) and (d) only
(c) (a), (b) and (c) only
(d) (a) and (c) only
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. 

Q3: Which of the following oxoacid of sulphur has –O–O– linkage?      (NEET 2020)
(a) H₂S₂O₈, peroxodisulphuric acid 
(b) H₂S₂O₇, pyrosulphuric acid 
(c) H₂SO₃, sulphurous acid 
(d) H₂SO₄, sulphuric acid
Ans: (a)

Q4: Match the following and identify the correct option.    (NEET 2020)

Which of the following is correct option?
(a)

(b)

(c)

(d)

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.

2019

Q1: Which of the following species is not stable?    (NEET 2019)
(a) [SiF₆]^2-
(b) [GeCl₆]^2-
(c) [Sn(OH)₆]^2-
(d) [SiCl₆]^2-

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.

Q2: Which is the correct thermal stability order for H₂E (E = O, S, Se, Te and Po)?    (NEET 2019)
(a) H₂S < H₂O < H₂Se < H₂Te < H₂Po
(b) H₂O < H₂S < H2Se < H₂Te < H₂Po
(c) H₂Po < H₂Te < H₂Se < H₂S < H₂O
(d) H₂Se < H₂Te < H₂Po < H₂O < H₂S

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

Q3: Identify the incorrect statement related to PCl5 from the following :   (NEET 2019)
(a) Axial P–Cl bonds are longer than equatorial P–Cl bonds
(b) PCl₅ molecule is non-reactive
(c) Three equatorial P–Cl bonds make an angle of 120º with each other
(d) Two axial P–Cl bonds make an angle of 180º with each other
Ans: (b)

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

Q4: Which of the following is incorrect statement?
(a) GeX₄(X = F, Cl, Br, I) is more stable than GeX₂
(b) SnF₄ is ionic in nature
(c) PbF₄ is covalent in nature
(d) SiCl₄ is easily hydrolysed     (NEET 2019)
Ans: (c)
PbF4 is an ionic compound.

Q5: Match the Xenon compounds in Column-I with its structure in Column-II and assign the correct code:    (NEET 2019)


(a) (1)
(b) (2)
(c) (3)
(d) (4)

Ans: (b)

2018

Q1: Which of the following statements is not true for halogens ?    (NEET 2018)
(a) All form monobasic oxyacids.
(b) All are oxidizing agents.
(c) All but fluorine show positive oxidation states.
(d) Chlorine has the highest electron-gain enthalpy.
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.

Q2: Which one of the following elements is unable to form MF₆^3– ion?    (NEET 2018)
(a) Ga
(b) Al
(c)  B
(d) In
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–. 

Q.23. The correct order of atomic radii in group 13 elements is    (2018)
(a) B < Al < In < Ga < Tl
(b) B < Al < Ga < In < Tl
(c) B < Ga < Al < Tl < In
(d) B < Ga < Al < In < Tl

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).

2017

Q2: Match the interhalogen compounds of column-I with the geometry in column II and assign the correct code.    (NEET 2017)

(a) (1)
(b) (2)
(c) (3)
(d) (4)
Ans: (a)

Q3: In which pair of ions both the species contain S–S bond?    (NEET 2017)
(a)
(b)
(c)
(d)

HClO < HClO₂ < HClO₃ < HClO₄

Q4: Match the compound given in column I with the hybridization and shape given in column II and mark the correct option.     (NEET 2016 Phase 1)
Column-I                Column-II
(a) XeF₆            (i) distorted octahedral
(b) XeO₃           (ii) square planar
(c) XeOF₄         (iii) pyramidal
(d) XeF₄            (iv) square pyramidal
Code :
(a) (b) (c) (d)
(a) (iv) (i) (ii) (iii)
(b) (i) (iii) (iv) (ii)
(c) (i) (ii) (iv) (iii)
(d) (iv) (iii) (i) (ii)
Ans: (b)
(i) (iii) (iv) (ii)

Q5: Which is the correct statement for the given acids?      (NEET 2016 Phase 1)
(a) Phosphinic acid is a diprotic acid while phosphonic acid is a monoprotic acid.
(b) Phosphinic acid is a monoprotic acid while phosphonic acid is a diprotic acid.
(c) Both are diprotic acids.
(d) Both are triprotic acids.
Ans: (b)

Q6: Boric acid is an acid because its molecule     (NEET 2016 Phase 2)
(a) contains replaceable H+ ion
(b) gives up a proton
(c) accepts OH− from water releasing proton
(d) combines with proton from water molecule.
Ans: (c)
Boric acid behaves as a Lewis acid, by accepting a pair of electrons from OH^– ion of water thereby releasing a proton.
H-OH+B(OH)₃ $\to$ [B(OH)₄]^– + H⁺ 

Q1: Which of the statements given below is incorrect?      (NEET / AIPMT 2015)
(a) O₃ molecule is bent.
(b) ONF is isoelectronic with O₂N
(c) OF₂ is an oxides of fluorine.
(d) Cl₂O₇ is an anhydride of perchloric acid.
Ans: (c)
OF₂ (oxygen difluoride) is a fluoride of oxygen because fluorine is more electronegative than oxygen.

Q2: Strong reducing behavior of H3PO2 is due to      (NEET / AIPMT 2015)
(a) high electron gain enthalpy of phosphorus
(b) high oxidation state of phosphorus
(c) presence of two −OH groups and one P−H bond
(d) presence of one −OH group and two P−H bonds.
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.

Q3: The variation of the boiling points of the hydrogen halides is in the order HF > HI > HBr > HCl.     (NEET / AIPMT 2015)
What explains the higher boiling point of hydrogen fluoride?
(a) There is strong hydrogen bonding between HF molecules.
(b) The bond energy of HF molecules is greater than in other hydrogen halides.
(c) The effect of nuclear shielding is much reduced in fluorine which polarises the HF molecule.
(d) The electronegativity of fluorine is much higher than for other elements in the group.
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.

Q4: The stability of +1 oxidation state among Al, Ga, In and Tl increases in the sequence     (NEET / AIPMT 2015)
(a) Al < Ga < In < Tl
(b) Tl < In < Ga < Al
(c) In < Tl < Ga < Al
(d) Ga < In < Al < Tl
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