JEE Main Previous Year Questions (2025): p-Block Elements

 Page 1


JEE Main Previous Year Questions 
(2025): p-Block Elements 
Q1: A group 15 element forms ?? ?? - ?? ?? bond with transition metals. It also forms 
hydride, which is a strongest base among the hydrides of other group members that 
form ???? - ???? bond. 
The atomic number of the element is ____ . 
JEE Main 2025 (Online) 28th January Evening Shift 
Ans: 15 
Solution: 
Phosphorus, an element from group 15 of the periodic table, can form a ???? - ???? bond with 
transition metals. Among the hydrides of group 15, phosphine ( PH
3
 ) is considered the 
strongest base, except for ammonia ( NH
3
 ). 
Q2: The maximum covalency of a non-metallic group 15 element ' ?? ' with weakest 
?? - ?? bond is : 
JEE Main 2025 (Online) 22nd January Evening Shift 
Options: 
A. 5 
B. 3 
C. 4 
D. 6 
Ans: C 
Solution: 
Among the group 15 elements ( N, P, As, Sb, Bi ), the lightest two ( N and P ) are commonly 
considered nonmetals. Of these, nitrogen ( N ) is known to have the weakest single bond to 
itself ( N - N ). 
The ?? - ?? single-bond enthalpy ( ˜ 160 \ kJ/mol ) is lower (weaker) than the ?? - ?? single-
bond enthalpy ( ˜ 200 (,kJ/mol). 
Therefore, the "nonmetallic group 15 element with the weakest E - E bond" is nitrogen. 
Maximum Covalency of Nitrogen 
Although nitrogen typically forms three covalent bonds in neutral compounds (e.g., NH
3
 ), it can 
expand to four bonds in certain cationic species such as ammonium NH
4
 
+
or NF
4
 
+
. In such 
species, nitrogen has a formal positive charge but is still forming four covalent bonds. 
Hence, the maximum covalency of nitrogen is ?? . 
Answer: 4 (Option C). 
Page 2


JEE Main Previous Year Questions 
(2025): p-Block Elements 
Q1: A group 15 element forms ?? ?? - ?? ?? bond with transition metals. It also forms 
hydride, which is a strongest base among the hydrides of other group members that 
form ???? - ???? bond. 
The atomic number of the element is ____ . 
JEE Main 2025 (Online) 28th January Evening Shift 
Ans: 15 
Solution: 
Phosphorus, an element from group 15 of the periodic table, can form a ???? - ???? bond with 
transition metals. Among the hydrides of group 15, phosphine ( PH
3
 ) is considered the 
strongest base, except for ammonia ( NH
3
 ). 
Q2: The maximum covalency of a non-metallic group 15 element ' ?? ' with weakest 
?? - ?? bond is : 
JEE Main 2025 (Online) 22nd January Evening Shift 
Options: 
A. 5 
B. 3 
C. 4 
D. 6 
Ans: C 
Solution: 
Among the group 15 elements ( N, P, As, Sb, Bi ), the lightest two ( N and P ) are commonly 
considered nonmetals. Of these, nitrogen ( N ) is known to have the weakest single bond to 
itself ( N - N ). 
The ?? - ?? single-bond enthalpy ( ˜ 160 \ kJ/mol ) is lower (weaker) than the ?? - ?? single-
bond enthalpy ( ˜ 200 (,kJ/mol). 
Therefore, the "nonmetallic group 15 element with the weakest E - E bond" is nitrogen. 
Maximum Covalency of Nitrogen 
Although nitrogen typically forms three covalent bonds in neutral compounds (e.g., NH
3
 ), it can 
expand to four bonds in certain cationic species such as ammonium NH
4
 
+
or NF
4
 
+
. In such 
species, nitrogen has a formal positive charge but is still forming four covalent bonds. 
Hence, the maximum covalency of nitrogen is ?? . 
Answer: 4 (Option C). 
 
Q3: The incorrect statement among the following is 
JEE Main 2025 (Online) 23rd January Morning Shift 
Options: 
A. PH
3
 shows lower proton affinity than NH
3
. 
B. SO
2
 can act as an oxidizing agent, but not as a reducing agent. 
C. NO
2
 can dimerise easily. 
D. PF
3
 exists but NF
5
 does not. 
Ans: B 
Solution: 
PH
3
 shows lower proton affinity than NH
3
. 
Ammonia ( NH
3
 ) has a more available lone pair on nitrogen than PH
3
 on phosphorus due to 
smaller size and higher electronegativity. Thus, NH
3
 is a stronger base and has a higher proton 
affinity than PH
3
. 
This statement is correct. 
SO
2
 can act as an oxidizing agent, but not as a reducing agent. 
Sulfur dioxide ( SO
2
 ) is quite versatile in redox chemistry. In many reactions, it actually behaves 
as a reducing agent by being oxidized (e.g., to sulfate, where sulfur goes from +4 to +6 oxidation 
state). While under some conditions it may act as an oxidizing agent, it is well known and widely 
used for its reducing properties. 
Hence, the claim that it "cannot act as a reducing agent" is incorrect. 
NO
2
 can dimerise easily. 
Nitrogen dioxide ( NO
2
 ) is known to dimerise to form dinitrogen tetroxide ( N
2
O
4
 ), especially at 
lower temperatures. 
This statement is correct. 
PF
3
 exists but NF
5
 does not. 
Phosphorus trifluoride ( PF
3
 ) is a known stable compound. In contrast, a compound like 
nitrogen pentafluoride ( NF
5
 ) is not observed, largely due to the limitations of nitrogen's size 
and bonding capabilities in forming such a structure. 
This statement is correct. 
Based on the analysis, the incorrect statement is: 
Option B. 
Q4: The large difference between the melting and boiling points of oxygen and 
sulphur may be explained on the basis of 
JEE Main 2025 (Online) 24th January Morning Shift 
Options: 
A. Atomicity 
B. Electron gain enthalpy 
C. Atomic size 
D. Electronegativity 
Page 3


JEE Main Previous Year Questions 
(2025): p-Block Elements 
Q1: A group 15 element forms ?? ?? - ?? ?? bond with transition metals. It also forms 
hydride, which is a strongest base among the hydrides of other group members that 
form ???? - ???? bond. 
The atomic number of the element is ____ . 
JEE Main 2025 (Online) 28th January Evening Shift 
Ans: 15 
Solution: 
Phosphorus, an element from group 15 of the periodic table, can form a ???? - ???? bond with 
transition metals. Among the hydrides of group 15, phosphine ( PH
3
 ) is considered the 
strongest base, except for ammonia ( NH
3
 ). 
Q2: The maximum covalency of a non-metallic group 15 element ' ?? ' with weakest 
?? - ?? bond is : 
JEE Main 2025 (Online) 22nd January Evening Shift 
Options: 
A. 5 
B. 3 
C. 4 
D. 6 
Ans: C 
Solution: 
Among the group 15 elements ( N, P, As, Sb, Bi ), the lightest two ( N and P ) are commonly 
considered nonmetals. Of these, nitrogen ( N ) is known to have the weakest single bond to 
itself ( N - N ). 
The ?? - ?? single-bond enthalpy ( ˜ 160 \ kJ/mol ) is lower (weaker) than the ?? - ?? single-
bond enthalpy ( ˜ 200 (,kJ/mol). 
Therefore, the "nonmetallic group 15 element with the weakest E - E bond" is nitrogen. 
Maximum Covalency of Nitrogen 
Although nitrogen typically forms three covalent bonds in neutral compounds (e.g., NH
3
 ), it can 
expand to four bonds in certain cationic species such as ammonium NH
4
 
+
or NF
4
 
+
. In such 
species, nitrogen has a formal positive charge but is still forming four covalent bonds. 
Hence, the maximum covalency of nitrogen is ?? . 
Answer: 4 (Option C). 
 
Q3: The incorrect statement among the following is 
JEE Main 2025 (Online) 23rd January Morning Shift 
Options: 
A. PH
3
 shows lower proton affinity than NH
3
. 
B. SO
2
 can act as an oxidizing agent, but not as a reducing agent. 
C. NO
2
 can dimerise easily. 
D. PF
3
 exists but NF
5
 does not. 
Ans: B 
Solution: 
PH
3
 shows lower proton affinity than NH
3
. 
Ammonia ( NH
3
 ) has a more available lone pair on nitrogen than PH
3
 on phosphorus due to 
smaller size and higher electronegativity. Thus, NH
3
 is a stronger base and has a higher proton 
affinity than PH
3
. 
This statement is correct. 
SO
2
 can act as an oxidizing agent, but not as a reducing agent. 
Sulfur dioxide ( SO
2
 ) is quite versatile in redox chemistry. In many reactions, it actually behaves 
as a reducing agent by being oxidized (e.g., to sulfate, where sulfur goes from +4 to +6 oxidation 
state). While under some conditions it may act as an oxidizing agent, it is well known and widely 
used for its reducing properties. 
Hence, the claim that it "cannot act as a reducing agent" is incorrect. 
NO
2
 can dimerise easily. 
Nitrogen dioxide ( NO
2
 ) is known to dimerise to form dinitrogen tetroxide ( N
2
O
4
 ), especially at 
lower temperatures. 
This statement is correct. 
PF
3
 exists but NF
5
 does not. 
Phosphorus trifluoride ( PF
3
 ) is a known stable compound. In contrast, a compound like 
nitrogen pentafluoride ( NF
5
 ) is not observed, largely due to the limitations of nitrogen's size 
and bonding capabilities in forming such a structure. 
This statement is correct. 
Based on the analysis, the incorrect statement is: 
Option B. 
Q4: The large difference between the melting and boiling points of oxygen and 
sulphur may be explained on the basis of 
JEE Main 2025 (Online) 24th January Morning Shift 
Options: 
A. Atomicity 
B. Electron gain enthalpy 
C. Atomic size 
D. Electronegativity 
Ans: A 
Solution: 
The significant difference in the melting and boiling points of oxygen and sulfur can be explained 
by considering atomicity. 
Oxygen exists as O
2
 (Atomicity = 2 ). 
Sulfur exists as S
8
 (Atomicity = 8). 
Due to sulfur's higher atomicity (forming S8 molecules), its melting and boiling points are 
considerably higher than those of oxygen. 
Q5: The nature of oxide (?????? ?? ) and hydride (?????? ?? ) formed by Te , respectively are : 
JEE Main 2025 (Online) 2nd April Evening Shift 
Options: 
A. Reducing and basic 
B. Reducing and acidic 
C. Oxidising and acidic 
D. Oxidising and basic 
Ans: C 
Solution: 
Tellurium dioxide ( TeO
2
 ) acts as an oxidizing agent. This is because it can accept electrons and 
be reduced from its +4 oxidation state to a lower oxidation state. 
Tellurium hydride ( TeH
2
 ) is considered acidic. This is due to its relatively low bond dissociation 
energy, which makes it prone to breaking, releasing protons ( H
+
). 
Q6: Given below are two statements : 
Statement I : The N - N single bond is weaker and longer than that of P - P single bond. 
Statement II: Compounds of group 15 elements in +3 oxidation states readily undergo 
disproportionation reactions. 
In the light of the above statements, choose the correct answer from the options 
given below 
JEE Main 2025 (Online) 3rd April Morning Shift 
Options: 
A. Statement I is false but Statement II is true 
B. Both Statement I and Statement II are true 
C. Both Statement I and Statement II are false 
D. Statement I is true but Statement II is false 
Ans: C 
Solution: 
Page 4


JEE Main Previous Year Questions 
(2025): p-Block Elements 
Q1: A group 15 element forms ?? ?? - ?? ?? bond with transition metals. It also forms 
hydride, which is a strongest base among the hydrides of other group members that 
form ???? - ???? bond. 
The atomic number of the element is ____ . 
JEE Main 2025 (Online) 28th January Evening Shift 
Ans: 15 
Solution: 
Phosphorus, an element from group 15 of the periodic table, can form a ???? - ???? bond with 
transition metals. Among the hydrides of group 15, phosphine ( PH
3
 ) is considered the 
strongest base, except for ammonia ( NH
3
 ). 
Q2: The maximum covalency of a non-metallic group 15 element ' ?? ' with weakest 
?? - ?? bond is : 
JEE Main 2025 (Online) 22nd January Evening Shift 
Options: 
A. 5 
B. 3 
C. 4 
D. 6 
Ans: C 
Solution: 
Among the group 15 elements ( N, P, As, Sb, Bi ), the lightest two ( N and P ) are commonly 
considered nonmetals. Of these, nitrogen ( N ) is known to have the weakest single bond to 
itself ( N - N ). 
The ?? - ?? single-bond enthalpy ( ˜ 160 \ kJ/mol ) is lower (weaker) than the ?? - ?? single-
bond enthalpy ( ˜ 200 (,kJ/mol). 
Therefore, the "nonmetallic group 15 element with the weakest E - E bond" is nitrogen. 
Maximum Covalency of Nitrogen 
Although nitrogen typically forms three covalent bonds in neutral compounds (e.g., NH
3
 ), it can 
expand to four bonds in certain cationic species such as ammonium NH
4
 
+
or NF
4
 
+
. In such 
species, nitrogen has a formal positive charge but is still forming four covalent bonds. 
Hence, the maximum covalency of nitrogen is ?? . 
Answer: 4 (Option C). 
 
Q3: The incorrect statement among the following is 
JEE Main 2025 (Online) 23rd January Morning Shift 
Options: 
A. PH
3
 shows lower proton affinity than NH
3
. 
B. SO
2
 can act as an oxidizing agent, but not as a reducing agent. 
C. NO
2
 can dimerise easily. 
D. PF
3
 exists but NF
5
 does not. 
Ans: B 
Solution: 
PH
3
 shows lower proton affinity than NH
3
. 
Ammonia ( NH
3
 ) has a more available lone pair on nitrogen than PH
3
 on phosphorus due to 
smaller size and higher electronegativity. Thus, NH
3
 is a stronger base and has a higher proton 
affinity than PH
3
. 
This statement is correct. 
SO
2
 can act as an oxidizing agent, but not as a reducing agent. 
Sulfur dioxide ( SO
2
 ) is quite versatile in redox chemistry. In many reactions, it actually behaves 
as a reducing agent by being oxidized (e.g., to sulfate, where sulfur goes from +4 to +6 oxidation 
state). While under some conditions it may act as an oxidizing agent, it is well known and widely 
used for its reducing properties. 
Hence, the claim that it "cannot act as a reducing agent" is incorrect. 
NO
2
 can dimerise easily. 
Nitrogen dioxide ( NO
2
 ) is known to dimerise to form dinitrogen tetroxide ( N
2
O
4
 ), especially at 
lower temperatures. 
This statement is correct. 
PF
3
 exists but NF
5
 does not. 
Phosphorus trifluoride ( PF
3
 ) is a known stable compound. In contrast, a compound like 
nitrogen pentafluoride ( NF
5
 ) is not observed, largely due to the limitations of nitrogen's size 
and bonding capabilities in forming such a structure. 
This statement is correct. 
Based on the analysis, the incorrect statement is: 
Option B. 
Q4: The large difference between the melting and boiling points of oxygen and 
sulphur may be explained on the basis of 
JEE Main 2025 (Online) 24th January Morning Shift 
Options: 
A. Atomicity 
B. Electron gain enthalpy 
C. Atomic size 
D. Electronegativity 
Ans: A 
Solution: 
The significant difference in the melting and boiling points of oxygen and sulfur can be explained 
by considering atomicity. 
Oxygen exists as O
2
 (Atomicity = 2 ). 
Sulfur exists as S
8
 (Atomicity = 8). 
Due to sulfur's higher atomicity (forming S8 molecules), its melting and boiling points are 
considerably higher than those of oxygen. 
Q5: The nature of oxide (?????? ?? ) and hydride (?????? ?? ) formed by Te , respectively are : 
JEE Main 2025 (Online) 2nd April Evening Shift 
Options: 
A. Reducing and basic 
B. Reducing and acidic 
C. Oxidising and acidic 
D. Oxidising and basic 
Ans: C 
Solution: 
Tellurium dioxide ( TeO
2
 ) acts as an oxidizing agent. This is because it can accept electrons and 
be reduced from its +4 oxidation state to a lower oxidation state. 
Tellurium hydride ( TeH
2
 ) is considered acidic. This is due to its relatively low bond dissociation 
energy, which makes it prone to breaking, releasing protons ( H
+
). 
Q6: Given below are two statements : 
Statement I : The N - N single bond is weaker and longer than that of P - P single bond. 
Statement II: Compounds of group 15 elements in +3 oxidation states readily undergo 
disproportionation reactions. 
In the light of the above statements, choose the correct answer from the options 
given below 
JEE Main 2025 (Online) 3rd April Morning Shift 
Options: 
A. Statement I is false but Statement II is true 
B. Both Statement I and Statement II are true 
C. Both Statement I and Statement II are false 
D. Statement I is true but Statement II is false 
Ans: C 
Solution: 
 
single bond weaker than 
 
 
 
due to more lp - lp repulsion. 
Bond length ? ?? ?? -?? > ?? N-N
( size ?, B.L. ?) 
In group 15 elements only N \ & P show disproportionation in +3 oxidation state, As, ???? &Bi 
have almost inert for disproportionation in +3 oxidation state. 
So both statements are false. 
Q7: Given below are the pairs of group 13 elements showing their relation in terms of 
atomic radius. (?? < ???? ), (???? < ???? ), (???? < ???? ) and ( ???? < ???? )Identify the elements 
present in the incorrect pair and in that pair find out the element ( ?? ) that has higher 
ionic radius ( ?? ?? +
 ) than the other one. The atomic number of the element (?? ) is 
JEE Main 2025 (Online) 4th April Morning Shift 
Options: 
A. 49 
B. 81 
C. 31 
D. 13 
Ans: C 
Solution: 
Size order 
Al > Ga
Al
3+
< Ga
3+
 
Page 5


JEE Main Previous Year Questions 
(2025): p-Block Elements 
Q1: A group 15 element forms ?? ?? - ?? ?? bond with transition metals. It also forms 
hydride, which is a strongest base among the hydrides of other group members that 
form ???? - ???? bond. 
The atomic number of the element is ____ . 
JEE Main 2025 (Online) 28th January Evening Shift 
Ans: 15 
Solution: 
Phosphorus, an element from group 15 of the periodic table, can form a ???? - ???? bond with 
transition metals. Among the hydrides of group 15, phosphine ( PH
3
 ) is considered the 
strongest base, except for ammonia ( NH
3
 ). 
Q2: The maximum covalency of a non-metallic group 15 element ' ?? ' with weakest 
?? - ?? bond is : 
JEE Main 2025 (Online) 22nd January Evening Shift 
Options: 
A. 5 
B. 3 
C. 4 
D. 6 
Ans: C 
Solution: 
Among the group 15 elements ( N, P, As, Sb, Bi ), the lightest two ( N and P ) are commonly 
considered nonmetals. Of these, nitrogen ( N ) is known to have the weakest single bond to 
itself ( N - N ). 
The ?? - ?? single-bond enthalpy ( ˜ 160 \ kJ/mol ) is lower (weaker) than the ?? - ?? single-
bond enthalpy ( ˜ 200 (,kJ/mol). 
Therefore, the "nonmetallic group 15 element with the weakest E - E bond" is nitrogen. 
Maximum Covalency of Nitrogen 
Although nitrogen typically forms three covalent bonds in neutral compounds (e.g., NH
3
 ), it can 
expand to four bonds in certain cationic species such as ammonium NH
4
 
+
or NF
4
 
+
. In such 
species, nitrogen has a formal positive charge but is still forming four covalent bonds. 
Hence, the maximum covalency of nitrogen is ?? . 
Answer: 4 (Option C). 
 
Q3: The incorrect statement among the following is 
JEE Main 2025 (Online) 23rd January Morning Shift 
Options: 
A. PH
3
 shows lower proton affinity than NH
3
. 
B. SO
2
 can act as an oxidizing agent, but not as a reducing agent. 
C. NO
2
 can dimerise easily. 
D. PF
3
 exists but NF
5
 does not. 
Ans: B 
Solution: 
PH
3
 shows lower proton affinity than NH
3
. 
Ammonia ( NH
3
 ) has a more available lone pair on nitrogen than PH
3
 on phosphorus due to 
smaller size and higher electronegativity. Thus, NH
3
 is a stronger base and has a higher proton 
affinity than PH
3
. 
This statement is correct. 
SO
2
 can act as an oxidizing agent, but not as a reducing agent. 
Sulfur dioxide ( SO
2
 ) is quite versatile in redox chemistry. In many reactions, it actually behaves 
as a reducing agent by being oxidized (e.g., to sulfate, where sulfur goes from +4 to +6 oxidation 
state). While under some conditions it may act as an oxidizing agent, it is well known and widely 
used for its reducing properties. 
Hence, the claim that it "cannot act as a reducing agent" is incorrect. 
NO
2
 can dimerise easily. 
Nitrogen dioxide ( NO
2
 ) is known to dimerise to form dinitrogen tetroxide ( N
2
O
4
 ), especially at 
lower temperatures. 
This statement is correct. 
PF
3
 exists but NF
5
 does not. 
Phosphorus trifluoride ( PF
3
 ) is a known stable compound. In contrast, a compound like 
nitrogen pentafluoride ( NF
5
 ) is not observed, largely due to the limitations of nitrogen's size 
and bonding capabilities in forming such a structure. 
This statement is correct. 
Based on the analysis, the incorrect statement is: 
Option B. 
Q4: The large difference between the melting and boiling points of oxygen and 
sulphur may be explained on the basis of 
JEE Main 2025 (Online) 24th January Morning Shift 
Options: 
A. Atomicity 
B. Electron gain enthalpy 
C. Atomic size 
D. Electronegativity 
Ans: A 
Solution: 
The significant difference in the melting and boiling points of oxygen and sulfur can be explained 
by considering atomicity. 
Oxygen exists as O
2
 (Atomicity = 2 ). 
Sulfur exists as S
8
 (Atomicity = 8). 
Due to sulfur's higher atomicity (forming S8 molecules), its melting and boiling points are 
considerably higher than those of oxygen. 
Q5: The nature of oxide (?????? ?? ) and hydride (?????? ?? ) formed by Te , respectively are : 
JEE Main 2025 (Online) 2nd April Evening Shift 
Options: 
A. Reducing and basic 
B. Reducing and acidic 
C. Oxidising and acidic 
D. Oxidising and basic 
Ans: C 
Solution: 
Tellurium dioxide ( TeO
2
 ) acts as an oxidizing agent. This is because it can accept electrons and 
be reduced from its +4 oxidation state to a lower oxidation state. 
Tellurium hydride ( TeH
2
 ) is considered acidic. This is due to its relatively low bond dissociation 
energy, which makes it prone to breaking, releasing protons ( H
+
). 
Q6: Given below are two statements : 
Statement I : The N - N single bond is weaker and longer than that of P - P single bond. 
Statement II: Compounds of group 15 elements in +3 oxidation states readily undergo 
disproportionation reactions. 
In the light of the above statements, choose the correct answer from the options 
given below 
JEE Main 2025 (Online) 3rd April Morning Shift 
Options: 
A. Statement I is false but Statement II is true 
B. Both Statement I and Statement II are true 
C. Both Statement I and Statement II are false 
D. Statement I is true but Statement II is false 
Ans: C 
Solution: 
 
single bond weaker than 
 
 
 
due to more lp - lp repulsion. 
Bond length ? ?? ?? -?? > ?? N-N
( size ?, B.L. ?) 
In group 15 elements only N \ & P show disproportionation in +3 oxidation state, As, ???? &Bi 
have almost inert for disproportionation in +3 oxidation state. 
So both statements are false. 
Q7: Given below are the pairs of group 13 elements showing their relation in terms of 
atomic radius. (?? < ???? ), (???? < ???? ), (???? < ???? ) and ( ???? < ???? )Identify the elements 
present in the incorrect pair and in that pair find out the element ( ?? ) that has higher 
ionic radius ( ?? ?? +
 ) than the other one. The atomic number of the element (?? ) is 
JEE Main 2025 (Online) 4th April Morning Shift 
Options: 
A. 49 
B. 81 
C. 31 
D. 13 
Ans: C 
Solution: 
Size order 
Al > Ga
Al
3+
< Ga
3+
 
Atomic number of Ga is 31 
Q8: Given below are two statements: 
Statement I: Nitrogen forms oxides with +?? to +?? oxidation states due to the 
formation of ?? ?? - ?? ?? bond with oxygen. 
Statement II: Nitrogen does not form halides with +?? oxidation state due to the 
absence of d-orbital in it. 
In the light of given statements, choose the correct answer from the options given 
below. 
JEE Main 2025 (Online) 4th April Morning Shift 
Options: 
A. Both Statement I and Statement II are true 
B. Statement I is true but Statement II is false 
C. Statement I is false but Statement II are true 
D. Both Statement I and Statement II are False 
Ans: A 
Solution: 
In oxide of nitrogen it can achieve +5 oxidation state because it can form p?? - p?? bond with 
oxygen e.g. N
2
O
5
 
 
Nitrogen cannot form halide in +5 oxidation state because it does not contain d-orbital. 
e.g.  NX
5
 does not exist 
X = halide 
Q9: Given below are two statements : 
Statement (I): The first ionisation enthalpy of group 14 elements is higher than the 
corresponding elements of group 13.