P-Block Elements: JEE Main Previous Year Questions (2021-2026)

 Page 1


JEE Main Previous Year Questions (2021-2026): 
p-Block Elements  
 
(January 2026) 
 
Q1: Consider the following reactions 
 
The oxidation states of Cu in Z and Q, respectively are: 
A: +2 and +1 
B: +1 and +1 
C: +1 and +2 
D: +2 and +2 
Answer: A 
Explanation: 
 
 
Q2: Given below are two statements : 
Statement I: The increasing order of boiling point of hydrogen halides is HCl < 
Page 2


JEE Main Previous Year Questions (2021-2026): 
p-Block Elements  
 
(January 2026) 
 
Q1: Consider the following reactions 
 
The oxidation states of Cu in Z and Q, respectively are: 
A: +2 and +1 
B: +1 and +1 
C: +1 and +2 
D: +2 and +2 
Answer: A 
Explanation: 
 
 
Q2: Given below are two statements : 
Statement I: The increasing order of boiling point of hydrogen halides is HCl < 
HBr < HI < HF. 
Statement II: The increasing order of melting point of hydrogen halides is HCl < 
HBr < HF < HI. 
In the light of the above statements, choose the correct answer from the options 
given below: 
A: Statement I is false but Statement II is true 
B: Both Statement I and Statement II are false 
C: Statement I is true but Statement II is false 
D: Both Statement I and Statement II are true 
Answer: D 
Explanation: 
Correct order of 
(i) Boiling point: HF > HI > HBr > HCl 
As we go from HCl to HI, the size and molar mass increase, so van der Waals forces 
increase and the boiling point increases. 
But HF has strong hydrogen bonding because fluorine is highly electronegative, so its 
boiling point becomes the highest. 
(ii) Melting point : HI > HF > HBr > HCl 
In general, from HCl to HI, increasing size and mass increase intermolecular forces, so 
melting point increases. 
However, HF shows hydrogen bonding, so its melting point is higher than HCl and HBr. 
Overall, HI has the highest melting point among hydrogen halides due to stronger 
dispersion forces in the solid state. 
 
Q3: Regarding the hydrides of group 15 elements EH
3
 (E = N, P, As, Sb), select the 
correct statement from the following: 
A. The stability of hydrides decreases down the group. 
B. The basicity of hydrides decreases down the group. 
C. The reducing character increases down the group. 
D. The boiling point increases down the group. 
Choose the correct answer from the options given below : 
A: A, B & C only 
B: A, B, C& D 
C: A & D only 
D: B & C only 
Answer: A 
Explanation: 
Hydrides of group 15: NH 3, PH 3, AsH 3, SbH 3 
A. Stability decreases down the group - Correct 
Down the group, the E – H bond strength decreases because atomic size increases, so 
Page 3


JEE Main Previous Year Questions (2021-2026): 
p-Block Elements  
 
(January 2026) 
 
Q1: Consider the following reactions 
 
The oxidation states of Cu in Z and Q, respectively are: 
A: +2 and +1 
B: +1 and +1 
C: +1 and +2 
D: +2 and +2 
Answer: A 
Explanation: 
 
 
Q2: Given below are two statements : 
Statement I: The increasing order of boiling point of hydrogen halides is HCl < 
HBr < HI < HF. 
Statement II: The increasing order of melting point of hydrogen halides is HCl < 
HBr < HF < HI. 
In the light of the above statements, choose the correct answer from the options 
given below: 
A: Statement I is false but Statement II is true 
B: Both Statement I and Statement II are false 
C: Statement I is true but Statement II is false 
D: Both Statement I and Statement II are true 
Answer: D 
Explanation: 
Correct order of 
(i) Boiling point: HF > HI > HBr > HCl 
As we go from HCl to HI, the size and molar mass increase, so van der Waals forces 
increase and the boiling point increases. 
But HF has strong hydrogen bonding because fluorine is highly electronegative, so its 
boiling point becomes the highest. 
(ii) Melting point : HI > HF > HBr > HCl 
In general, from HCl to HI, increasing size and mass increase intermolecular forces, so 
melting point increases. 
However, HF shows hydrogen bonding, so its melting point is higher than HCl and HBr. 
Overall, HI has the highest melting point among hydrogen halides due to stronger 
dispersion forces in the solid state. 
 
Q3: Regarding the hydrides of group 15 elements EH
3
 (E = N, P, As, Sb), select the 
correct statement from the following: 
A. The stability of hydrides decreases down the group. 
B. The basicity of hydrides decreases down the group. 
C. The reducing character increases down the group. 
D. The boiling point increases down the group. 
Choose the correct answer from the options given below : 
A: A, B & C only 
B: A, B, C& D 
C: A & D only 
D: B & C only 
Answer: A 
Explanation: 
Hydrides of group 15: NH 3, PH 3, AsH 3, SbH 3 
A. Stability decreases down the group - Correct 
Down the group, the E – H bond strength decreases because atomic size increases, so 
overlap becomes poorer. Hence hydrides become less stable: 
NH 3 > PH 3 > AsH 3 > SbH 3 
B. Basicity decreases down the group - Correct 
Basicity depends on the availability of the lone pair on E. Down the group, the lone pair 
becomes more diffused and less available for donation, so basicity decreases: 
NH 3 > PH 3 > AsH 3 > SbH 3 
C. Reducing character increases down the group- Correct 
Reducing nature increases as the E - H bond becomes weaker down the group, so the 
hydride can donate hydrogen (or get oxidised) more easily: 
NH 3 < PH 3 < AsH 3 < SbH 3 
D. Boiling point increases down the group - Not correct as a general statement 
Boiling points generally increase down the group due to increasing molar mass, but NH 3 
is anomalous because of hydrogen bonding, which raises its boiling point above what is 
expected. 
So the trend is not strictly increasing from NH 3 to SbH 3. 
Correct option: Option A (A, B & C only) 
 
Q4: "X" is an oxoanion of the lightest element of group 7 (in the periodic table). 
The metal is in +6 oxidation state in "X". The color of the potassium salt of X is 
A: Purple 
B: Orange 
C: Yellow 
D: green 
Answer: D 
Explanation: 
The lightest element in group 7 is manganese (Mn). 
An oxoanion where Mn is in the +6 oxidation state is the manganate ion, . 
Its potassium salt is potassium manganate, K
2
MnO
4
, which is green. 
Correct option: D) green 
 
Q5: Choose the INCORRECT statement 
A: Carbon exhibits negative oxidation states along with +4 and +2 . 
B: Carbon cannot exceed its covalency more than four. 
C: Among the isotopes of carbon, 
13
C is a radioactive isotope. 
D: CO
2
 is the most acidic oxide among the dioxides of group of 14 elements. 
Answer: C 
Explanation: 
Incorrect statement: Option C 
13
C is stable (non-radioactive). 
Page 4


JEE Main Previous Year Questions (2021-2026): 
p-Block Elements  
 
(January 2026) 
 
Q1: Consider the following reactions 
 
The oxidation states of Cu in Z and Q, respectively are: 
A: +2 and +1 
B: +1 and +1 
C: +1 and +2 
D: +2 and +2 
Answer: A 
Explanation: 
 
 
Q2: Given below are two statements : 
Statement I: The increasing order of boiling point of hydrogen halides is HCl < 
HBr < HI < HF. 
Statement II: The increasing order of melting point of hydrogen halides is HCl < 
HBr < HF < HI. 
In the light of the above statements, choose the correct answer from the options 
given below: 
A: Statement I is false but Statement II is true 
B: Both Statement I and Statement II are false 
C: Statement I is true but Statement II is false 
D: Both Statement I and Statement II are true 
Answer: D 
Explanation: 
Correct order of 
(i) Boiling point: HF > HI > HBr > HCl 
As we go from HCl to HI, the size and molar mass increase, so van der Waals forces 
increase and the boiling point increases. 
But HF has strong hydrogen bonding because fluorine is highly electronegative, so its 
boiling point becomes the highest. 
(ii) Melting point : HI > HF > HBr > HCl 
In general, from HCl to HI, increasing size and mass increase intermolecular forces, so 
melting point increases. 
However, HF shows hydrogen bonding, so its melting point is higher than HCl and HBr. 
Overall, HI has the highest melting point among hydrogen halides due to stronger 
dispersion forces in the solid state. 
 
Q3: Regarding the hydrides of group 15 elements EH
3
 (E = N, P, As, Sb), select the 
correct statement from the following: 
A. The stability of hydrides decreases down the group. 
B. The basicity of hydrides decreases down the group. 
C. The reducing character increases down the group. 
D. The boiling point increases down the group. 
Choose the correct answer from the options given below : 
A: A, B & C only 
B: A, B, C& D 
C: A & D only 
D: B & C only 
Answer: A 
Explanation: 
Hydrides of group 15: NH 3, PH 3, AsH 3, SbH 3 
A. Stability decreases down the group - Correct 
Down the group, the E – H bond strength decreases because atomic size increases, so 
overlap becomes poorer. Hence hydrides become less stable: 
NH 3 > PH 3 > AsH 3 > SbH 3 
B. Basicity decreases down the group - Correct 
Basicity depends on the availability of the lone pair on E. Down the group, the lone pair 
becomes more diffused and less available for donation, so basicity decreases: 
NH 3 > PH 3 > AsH 3 > SbH 3 
C. Reducing character increases down the group- Correct 
Reducing nature increases as the E - H bond becomes weaker down the group, so the 
hydride can donate hydrogen (or get oxidised) more easily: 
NH 3 < PH 3 < AsH 3 < SbH 3 
D. Boiling point increases down the group - Not correct as a general statement 
Boiling points generally increase down the group due to increasing molar mass, but NH 3 
is anomalous because of hydrogen bonding, which raises its boiling point above what is 
expected. 
So the trend is not strictly increasing from NH 3 to SbH 3. 
Correct option: Option A (A, B & C only) 
 
Q4: "X" is an oxoanion of the lightest element of group 7 (in the periodic table). 
The metal is in +6 oxidation state in "X". The color of the potassium salt of X is 
A: Purple 
B: Orange 
C: Yellow 
D: green 
Answer: D 
Explanation: 
The lightest element in group 7 is manganese (Mn). 
An oxoanion where Mn is in the +6 oxidation state is the manganate ion, . 
Its potassium salt is potassium manganate, K
2
MnO
4
, which is green. 
Correct option: D) green 
 
Q5: Choose the INCORRECT statement 
A: Carbon exhibits negative oxidation states along with +4 and +2 . 
B: Carbon cannot exceed its covalency more than four. 
C: Among the isotopes of carbon, 
13
C is a radioactive isotope. 
D: CO
2
 is the most acidic oxide among the dioxides of group of 14 elements. 
Answer: C 
Explanation: 
Incorrect statement: Option C 
13
C is stable (non-radioactive). 
The common radioactive isotope of carbon is 
14
C (also 
11
C is radioactive, but 
short-lived). 
The other statements (A, B, D) are correct in standard chemical context. 
 
Q6: It is noticed that Pb²? is more stable than Pb4? but Sn²? is less stable than Sn4?. 
Observe the following reactions. 
PbO2 + Pb ? 2PbO; ??G°(1) 
SnO2 + Sn ? 2SnO; ??G°(2) 
Identify the correct set from the following 
A. ??G°(1) > 0; ??G°(2) < 0 
B. ??G°(1) < 0; ??G°(2) > 0 
C. ??G°(1) < 0; ??G°(2) < 0 
D: ??G°(1) > 0; ??G°(2) > 0 
Answer: B 
Explanation: 
PbO2 + Pb ? 2PbO 
In this reaction, lead changes from +4 (in PbO2) and 0 (in Pb) to +2 (in PbO). 
Since Pb²? is more stable than Pb4?, the reaction prefers to form Pb²?. 
So, this reaction is spontaneous, hence ?G°(1) is negative. 
SnO2 + Sn ? 2SnO 
In this reaction, tin changes from +4 (in SnO2) and 0 (in Sn) to +2 (in SnO). 
But Sn²? is less stable than Sn4?, so forming Sn²? is not preferred. 
Therefore, this reaction is non-spontaneous, hence ?G°(2) is positive. 
 
Q7: The correct statements from the following are: 
A. Ionic radii of trivalent cations of group 13 elements decreases down the group. 
B. Electronegativity of group 13 elements decreases down the group. 
C. Among the group 13 elements, Boron has highest ?rst ionisation enthalpy. 
D. The trichloride and triiodide of group 13 elements are covalent in nature. 
Choose the correct answer from the options given below : 
A: B and D Only 
B: C and D Only 
C: A and D Only 
D: A and C Only 
Answer: B 
Explanation: 
(A) B³? < Al³? < Ga³? < In³? < Tl³?: ionic size. This order means that the ionic radius 
increases as we go down the group (from B³? to Tl³?). So the ionic radii do not decrease 
down the group. Therefore, statement (A) is incorrect. 
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JEE Main Previous Year Questions (2021-2026): 
p-Block Elements  
 
(January 2026) 
 
Q1: Consider the following reactions 
 
The oxidation states of Cu in Z and Q, respectively are: 
A: +2 and +1 
B: +1 and +1 
C: +1 and +2 
D: +2 and +2 
Answer: A 
Explanation: 
 
 
Q2: Given below are two statements : 
Statement I: The increasing order of boiling point of hydrogen halides is HCl < 
HBr < HI < HF. 
Statement II: The increasing order of melting point of hydrogen halides is HCl < 
HBr < HF < HI. 
In the light of the above statements, choose the correct answer from the options 
given below: 
A: Statement I is false but Statement II is true 
B: Both Statement I and Statement II are false 
C: Statement I is true but Statement II is false 
D: Both Statement I and Statement II are true 
Answer: D 
Explanation: 
Correct order of 
(i) Boiling point: HF > HI > HBr > HCl 
As we go from HCl to HI, the size and molar mass increase, so van der Waals forces 
increase and the boiling point increases. 
But HF has strong hydrogen bonding because fluorine is highly electronegative, so its 
boiling point becomes the highest. 
(ii) Melting point : HI > HF > HBr > HCl 
In general, from HCl to HI, increasing size and mass increase intermolecular forces, so 
melting point increases. 
However, HF shows hydrogen bonding, so its melting point is higher than HCl and HBr. 
Overall, HI has the highest melting point among hydrogen halides due to stronger 
dispersion forces in the solid state. 
 
Q3: Regarding the hydrides of group 15 elements EH
3
 (E = N, P, As, Sb), select the 
correct statement from the following: 
A. The stability of hydrides decreases down the group. 
B. The basicity of hydrides decreases down the group. 
C. The reducing character increases down the group. 
D. The boiling point increases down the group. 
Choose the correct answer from the options given below : 
A: A, B & C only 
B: A, B, C& D 
C: A & D only 
D: B & C only 
Answer: A 
Explanation: 
Hydrides of group 15: NH 3, PH 3, AsH 3, SbH 3 
A. Stability decreases down the group - Correct 
Down the group, the E – H bond strength decreases because atomic size increases, so 
overlap becomes poorer. Hence hydrides become less stable: 
NH 3 > PH 3 > AsH 3 > SbH 3 
B. Basicity decreases down the group - Correct 
Basicity depends on the availability of the lone pair on E. Down the group, the lone pair 
becomes more diffused and less available for donation, so basicity decreases: 
NH 3 > PH 3 > AsH 3 > SbH 3 
C. Reducing character increases down the group- Correct 
Reducing nature increases as the E - H bond becomes weaker down the group, so the 
hydride can donate hydrogen (or get oxidised) more easily: 
NH 3 < PH 3 < AsH 3 < SbH 3 
D. Boiling point increases down the group - Not correct as a general statement 
Boiling points generally increase down the group due to increasing molar mass, but NH 3 
is anomalous because of hydrogen bonding, which raises its boiling point above what is 
expected. 
So the trend is not strictly increasing from NH 3 to SbH 3. 
Correct option: Option A (A, B & C only) 
 
Q4: "X" is an oxoanion of the lightest element of group 7 (in the periodic table). 
The metal is in +6 oxidation state in "X". The color of the potassium salt of X is 
A: Purple 
B: Orange 
C: Yellow 
D: green 
Answer: D 
Explanation: 
The lightest element in group 7 is manganese (Mn). 
An oxoanion where Mn is in the +6 oxidation state is the manganate ion, . 
Its potassium salt is potassium manganate, K
2
MnO
4
, which is green. 
Correct option: D) green 
 
Q5: Choose the INCORRECT statement 
A: Carbon exhibits negative oxidation states along with +4 and +2 . 
B: Carbon cannot exceed its covalency more than four. 
C: Among the isotopes of carbon, 
13
C is a radioactive isotope. 
D: CO
2
 is the most acidic oxide among the dioxides of group of 14 elements. 
Answer: C 
Explanation: 
Incorrect statement: Option C 
13
C is stable (non-radioactive). 
The common radioactive isotope of carbon is 
14
C (also 
11
C is radioactive, but 
short-lived). 
The other statements (A, B, D) are correct in standard chemical context. 
 
Q6: It is noticed that Pb²? is more stable than Pb4? but Sn²? is less stable than Sn4?. 
Observe the following reactions. 
PbO2 + Pb ? 2PbO; ??G°(1) 
SnO2 + Sn ? 2SnO; ??G°(2) 
Identify the correct set from the following 
A. ??G°(1) > 0; ??G°(2) < 0 
B. ??G°(1) < 0; ??G°(2) > 0 
C. ??G°(1) < 0; ??G°(2) < 0 
D: ??G°(1) > 0; ??G°(2) > 0 
Answer: B 
Explanation: 
PbO2 + Pb ? 2PbO 
In this reaction, lead changes from +4 (in PbO2) and 0 (in Pb) to +2 (in PbO). 
Since Pb²? is more stable than Pb4?, the reaction prefers to form Pb²?. 
So, this reaction is spontaneous, hence ?G°(1) is negative. 
SnO2 + Sn ? 2SnO 
In this reaction, tin changes from +4 (in SnO2) and 0 (in Sn) to +2 (in SnO). 
But Sn²? is less stable than Sn4?, so forming Sn²? is not preferred. 
Therefore, this reaction is non-spontaneous, hence ?G°(2) is positive. 
 
Q7: The correct statements from the following are: 
A. Ionic radii of trivalent cations of group 13 elements decreases down the group. 
B. Electronegativity of group 13 elements decreases down the group. 
C. Among the group 13 elements, Boron has highest ?rst ionisation enthalpy. 
D. The trichloride and triiodide of group 13 elements are covalent in nature. 
Choose the correct answer from the options given below : 
A: B and D Only 
B: C and D Only 
C: A and D Only 
D: A and C Only 
Answer: B 
Explanation: 
(A) B³? < Al³? < Ga³? < In³? < Tl³?: ionic size. This order means that the ionic radius 
increases as we go down the group (from B³? to Tl³?). So the ionic radii do not decrease 
down the group. Therefore, statement (A) is incorrect. 
(B) B > Tl > In > Ga > Al: EN. This is the actual order of electronegativity in group 13. It 
does not simply decrease from top to bottom; instead, there are irregular variations. So 
the statement that electronegativity decreases down the group is not correct. Therefore, 
statement (B) is incorrect. 
(C) B > Tl > Ga > Al > In: IE. From this order of ?rst ionisation enthalpy, we can see that 
boron has the highest ?rst ionisation enthalpy among the group 13 elements. So 
statement (C) is correct. 
(D) Trichlorides and triiodides of group 13
th
 elements are covalent in nature. The MCl3 
and MI3 (where M is a group 13 element) have high polarising power due to the small, 
highly charged M³? ion, so they form covalent bonds with halide ions. Hence, statement 
(D) is also correct. 
 
Q8: Given below are two statements : 
Statement I: Elements 'X' and 'Y' are the most and least electronegative elements, 
respectively among N, As, Sb and P . The nature of the oxides X2O3 and Y2O3 is acidic 
and amphoteric, respectively. 
Statement II: BCl3 is covalent in nature and gets hydrolysed in water. It produces 
[B(OH)4]? and [B(H2O)6]³? in aqueous medium. 
In the light of the above statements, choose the correct answer from the options given 
below : 
A: Statement I is false but Statement II is true 
B: Statement I is true but Statement II is false 
C: Both Statement I and Statement II are false 
D: Both Statement I and Statement II are true 
Answer: B 
Explanation:  
 
X = N  X2O3 = N2O3 (Acidic) 
Y = Sb  Y2O3 = Sb2O3 (Amphoteric) 
Statement-I is true 
BCl3 + 3H2O ? B(OH)3 + 3HCl 
Statement-II is false