D – and f – Block Elements: JEE Main Previous Year Questions (2021-2026)

 Page 1


JEE Main Previous Year Questions (2021-2026): 
d and f Block Elements  
 
(January 2026) 
 
Q1: Consider the following statements about manganate and permanganate ions. 
Identify the correct statements. 
A. The geometry of both manganate and permanganate ions is tetrahedral. 
B. The oxidation states of Mn in manganate and permanganate are +7 and +6, 
respectively. 
C. Oxidation of Mn(II) salt by peroxodisulphate gives manganate ion as the final 
product. 
D. Manganate ion is paramagnetic and permanganate ion is diamagnetic. 
E. Acidified permanganate ion reduces oxalate, nitrite and iodide ions. 
Choose the correct answer from the options given below: 
A: A and D Only 
B: A, B and C Only 
C: A, C and D Only 
D: A, D and E Only 
Answer: A 
Explanation: 
Manganate ion ? MnO 4?² 
Permanganate ion ? MnO 4? 
(A) In both ions, Mn is surrounded by four oxygen atoms. So, the shape of both MnO 4?² 
and MnO 4? is tetrahedral (often written as d³s hybridisation in this context). 
(B) For MnO 4?, let oxidation state of Mn be x. Then x + 4(-2) = -1 ? x = +7. 
For MnO 4?², x + 4(-2) = -2 ? x = +6. 
So, Mn is +7 in permanganate and +6 in manganate (the statement given is reversed). 
(C) When Mn²? salt is oxidised by peroxodisulphate S 2O 8² ?, the product formed is 
permanganate ion, not manganate: 
Mn²? + S 2O 8² ? ? MnO 4? (Permanganate ion) 
(D) In MnO 4?, Mn is in +7 state, which corresponds to 3d ° (no unpaired electrons), so it 
is diamagnetic. 
In MnO 4?², Mn is in +6 state, which corresponds to 3d¹ (one unpaired electron), so it is 
paramagnetic. 
(E) Acidified permanganate ion acts as a strong oxidising agent, so it oxidises (i.e., 
“reduces” itself while oxidising) oxalate, nitrite and iodide ions. 
 
Page 2


JEE Main Previous Year Questions (2021-2026): 
d and f Block Elements  
 
(January 2026) 
 
Q1: Consider the following statements about manganate and permanganate ions. 
Identify the correct statements. 
A. The geometry of both manganate and permanganate ions is tetrahedral. 
B. The oxidation states of Mn in manganate and permanganate are +7 and +6, 
respectively. 
C. Oxidation of Mn(II) salt by peroxodisulphate gives manganate ion as the final 
product. 
D. Manganate ion is paramagnetic and permanganate ion is diamagnetic. 
E. Acidified permanganate ion reduces oxalate, nitrite and iodide ions. 
Choose the correct answer from the options given below: 
A: A and D Only 
B: A, B and C Only 
C: A, C and D Only 
D: A, D and E Only 
Answer: A 
Explanation: 
Manganate ion ? MnO 4?² 
Permanganate ion ? MnO 4? 
(A) In both ions, Mn is surrounded by four oxygen atoms. So, the shape of both MnO 4?² 
and MnO 4? is tetrahedral (often written as d³s hybridisation in this context). 
(B) For MnO 4?, let oxidation state of Mn be x. Then x + 4(-2) = -1 ? x = +7. 
For MnO 4?², x + 4(-2) = -2 ? x = +6. 
So, Mn is +7 in permanganate and +6 in manganate (the statement given is reversed). 
(C) When Mn²? salt is oxidised by peroxodisulphate S 2O 8² ?, the product formed is 
permanganate ion, not manganate: 
Mn²? + S 2O 8² ? ? MnO 4? (Permanganate ion) 
(D) In MnO 4?, Mn is in +7 state, which corresponds to 3d ° (no unpaired electrons), so it 
is diamagnetic. 
In MnO 4?², Mn is in +6 state, which corresponds to 3d¹ (one unpaired electron), so it is 
paramagnetic. 
(E) Acidified permanganate ion acts as a strong oxidising agent, so it oxidises (i.e., 
“reduces” itself while oxidising) oxalate, nitrite and iodide ions. 
 
Q2: Given below are two statements: 
Statement I : The number of pairs, from the following, in which both the ions are 
coloured in aqueous solution is 3.  
[Sc
3+
,Ti
3+
],[Mn
2+
,Cr
2+
],[Cu
2+
,Zn
2+
] and [Ni
2+
,Ti
4+
] 
Statement II : Th
4+
 is the strongest reducing agent among Th
4+
,Ce
4+
,Gd
3+
 and Eu
2+
. 
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: Both Statement I and Statement II are true 
D: Statement I is true but Statement II is false 
Answer: B 
Explanation: 
Statement I 
An ion is coloured in aqueous solution if it has partially filled d-orbitals (so that d-d 
transitions are possible). Ions with d ° or d¹ ° configurations are colourless. 
Now check each pair: 
[Sc³ ?, Ti³ ?] 
Sc³ ? : Sc (Z = 21) is [Ar] 3d¹4s², so Sc³ ? = [Ar] 3d ° (colourless) 
Ti³ ? : Ti (Z = 22) is [Ar] 3d²4s², so Ti³ ? = [Ar] 3d¹ (coloured) 
? Both not coloured 
[Mn² ?, Cr² ?] 
Mn²? = [Ar] 3d 5 (coloured) 
Cr² ? = [Ar] 3d 4 (coloured) 
? Both coloured 
[Cu² ?, Zn² ?] 
Cu² ? = [Ar] 3d ? (coloured) 
Zn² ? = [Ar] 3d¹ ° (colourless) 
? Both not coloured 
[Ni² ?, Ti 4?] 
Ni² ? = [Ar] 3d 8 (coloured) 
Ti 4? = [Ar] 3d ° (colourless) 
? Both not coloured 
So, the number of pairs in which both ions are coloured is only 1, not 3. 
Hence, Statement I is false. 
Statement II 
A reducing agent is a species that gets oxidised (it donates electrons). 
Ce 4? is a well-known strong oxidising agent (it gets reduced to Ce³ ?). 
Th 4? is also in a high oxidation state and is not a strong reducing agent. 
Gd³ ? is a very stable common oxidation state for Gd; not strongly reducing. 
Page 3


JEE Main Previous Year Questions (2021-2026): 
d and f Block Elements  
 
(January 2026) 
 
Q1: Consider the following statements about manganate and permanganate ions. 
Identify the correct statements. 
A. The geometry of both manganate and permanganate ions is tetrahedral. 
B. The oxidation states of Mn in manganate and permanganate are +7 and +6, 
respectively. 
C. Oxidation of Mn(II) salt by peroxodisulphate gives manganate ion as the final 
product. 
D. Manganate ion is paramagnetic and permanganate ion is diamagnetic. 
E. Acidified permanganate ion reduces oxalate, nitrite and iodide ions. 
Choose the correct answer from the options given below: 
A: A and D Only 
B: A, B and C Only 
C: A, C and D Only 
D: A, D and E Only 
Answer: A 
Explanation: 
Manganate ion ? MnO 4?² 
Permanganate ion ? MnO 4? 
(A) In both ions, Mn is surrounded by four oxygen atoms. So, the shape of both MnO 4?² 
and MnO 4? is tetrahedral (often written as d³s hybridisation in this context). 
(B) For MnO 4?, let oxidation state of Mn be x. Then x + 4(-2) = -1 ? x = +7. 
For MnO 4?², x + 4(-2) = -2 ? x = +6. 
So, Mn is +7 in permanganate and +6 in manganate (the statement given is reversed). 
(C) When Mn²? salt is oxidised by peroxodisulphate S 2O 8² ?, the product formed is 
permanganate ion, not manganate: 
Mn²? + S 2O 8² ? ? MnO 4? (Permanganate ion) 
(D) In MnO 4?, Mn is in +7 state, which corresponds to 3d ° (no unpaired electrons), so it 
is diamagnetic. 
In MnO 4?², Mn is in +6 state, which corresponds to 3d¹ (one unpaired electron), so it is 
paramagnetic. 
(E) Acidified permanganate ion acts as a strong oxidising agent, so it oxidises (i.e., 
“reduces” itself while oxidising) oxalate, nitrite and iodide ions. 
 
Q2: Given below are two statements: 
Statement I : The number of pairs, from the following, in which both the ions are 
coloured in aqueous solution is 3.  
[Sc
3+
,Ti
3+
],[Mn
2+
,Cr
2+
],[Cu
2+
,Zn
2+
] and [Ni
2+
,Ti
4+
] 
Statement II : Th
4+
 is the strongest reducing agent among Th
4+
,Ce
4+
,Gd
3+
 and Eu
2+
. 
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: Both Statement I and Statement II are true 
D: Statement I is true but Statement II is false 
Answer: B 
Explanation: 
Statement I 
An ion is coloured in aqueous solution if it has partially filled d-orbitals (so that d-d 
transitions are possible). Ions with d ° or d¹ ° configurations are colourless. 
Now check each pair: 
[Sc³ ?, Ti³ ?] 
Sc³ ? : Sc (Z = 21) is [Ar] 3d¹4s², so Sc³ ? = [Ar] 3d ° (colourless) 
Ti³ ? : Ti (Z = 22) is [Ar] 3d²4s², so Ti³ ? = [Ar] 3d¹ (coloured) 
? Both not coloured 
[Mn² ?, Cr² ?] 
Mn²? = [Ar] 3d 5 (coloured) 
Cr² ? = [Ar] 3d 4 (coloured) 
? Both coloured 
[Cu² ?, Zn² ?] 
Cu² ? = [Ar] 3d ? (coloured) 
Zn² ? = [Ar] 3d¹ ° (colourless) 
? Both not coloured 
[Ni² ?, Ti 4?] 
Ni² ? = [Ar] 3d 8 (coloured) 
Ti 4? = [Ar] 3d ° (colourless) 
? Both not coloured 
So, the number of pairs in which both ions are coloured is only 1, not 3. 
Hence, Statement I is false. 
Statement II 
A reducing agent is a species that gets oxidised (it donates electrons). 
Ce 4? is a well-known strong oxidising agent (it gets reduced to Ce³ ?). 
Th 4? is also in a high oxidation state and is not a strong reducing agent. 
Gd³ ? is a very stable common oxidation state for Gd; not strongly reducing. 
Eu² ? readily gets oxidised to Eu³ ?, so it acts as a strong reducing agent among the 
given species. 
Therefore, the strongest reducing agent among Th 4?, Ce 4?, Gd³ ? and Eu² ? is Eu² ?, not 
Th 4?. 
Hence, Statement II is false. 
Correct Option 
Since both Statement I and Statement II are false, the correct choice is: 
Option B: Both Statement I and Statement II are false. 
 
Q3: Given below are two statements: 
Statement I : The first ionization enthalpy of Cr is lower than that of Mn. 
Statement II : The second and third ionization enthalpies of Cr are higher than 
those of Mn. 
In the light of the above statements, choose the correct answer from the options 
given below : 
A: Statement I is true but Statement II is false 
B: Both Statement I and Statement II are true 
C: Both Statement I and Statement II are false 
D: Statement I is false but Statement II is true 
Answer: A 
Explanation: 
First, write the electronic configurations of Cr and Mn. 
Cr = (Ar)3d 54s¹ 
Mn = (Ar)3d 54s² 
IE 1(Cr) < IE 1(Mn) 
IE 2(Cr) > IE 2(Mn) 
IE 3(Cr) < IE 3(Mn) 
For first ionization enthalpy (IE 1): In Cr, the electron is removed from 4s¹. After losing 
this one 4s electron, Cr becomes Cr ? with a stable half-filled 3d 5 configuration. So 
removing the first electron is comparatively easier, hence IE 1(Cr) < IE 1(Mn). 
For second ionization enthalpy (IE 2): After losing one electron, Cr ? already has the 
stable 3d 5 configuration. Removing the second electron would disturb this stable 
half-filled 3d arrangement, so it needs more energy. But in Mn, after the first electron 
loss, removing the next electron can lead to Mn² ? with stable 3d 5. Therefore, IE 2(Cr) > 
IE 2(Mn). 
For third ionization enthalpy (IE 3): Mn² ? is very stable due to 3d 5. Removing one more 
electron from this stable state is difficult, so IE 3 for Mn becomes high. In comparison, 
Cr² ? does not have this extra stability, so removing the third electron is relatively easier. 
Hence, IE 3(Cr) < IE 3(Mn). 
Page 4


JEE Main Previous Year Questions (2021-2026): 
d and f Block Elements  
 
(January 2026) 
 
Q1: Consider the following statements about manganate and permanganate ions. 
Identify the correct statements. 
A. The geometry of both manganate and permanganate ions is tetrahedral. 
B. The oxidation states of Mn in manganate and permanganate are +7 and +6, 
respectively. 
C. Oxidation of Mn(II) salt by peroxodisulphate gives manganate ion as the final 
product. 
D. Manganate ion is paramagnetic and permanganate ion is diamagnetic. 
E. Acidified permanganate ion reduces oxalate, nitrite and iodide ions. 
Choose the correct answer from the options given below: 
A: A and D Only 
B: A, B and C Only 
C: A, C and D Only 
D: A, D and E Only 
Answer: A 
Explanation: 
Manganate ion ? MnO 4?² 
Permanganate ion ? MnO 4? 
(A) In both ions, Mn is surrounded by four oxygen atoms. So, the shape of both MnO 4?² 
and MnO 4? is tetrahedral (often written as d³s hybridisation in this context). 
(B) For MnO 4?, let oxidation state of Mn be x. Then x + 4(-2) = -1 ? x = +7. 
For MnO 4?², x + 4(-2) = -2 ? x = +6. 
So, Mn is +7 in permanganate and +6 in manganate (the statement given is reversed). 
(C) When Mn²? salt is oxidised by peroxodisulphate S 2O 8² ?, the product formed is 
permanganate ion, not manganate: 
Mn²? + S 2O 8² ? ? MnO 4? (Permanganate ion) 
(D) In MnO 4?, Mn is in +7 state, which corresponds to 3d ° (no unpaired electrons), so it 
is diamagnetic. 
In MnO 4?², Mn is in +6 state, which corresponds to 3d¹ (one unpaired electron), so it is 
paramagnetic. 
(E) Acidified permanganate ion acts as a strong oxidising agent, so it oxidises (i.e., 
“reduces” itself while oxidising) oxalate, nitrite and iodide ions. 
 
Q2: Given below are two statements: 
Statement I : The number of pairs, from the following, in which both the ions are 
coloured in aqueous solution is 3.  
[Sc
3+
,Ti
3+
],[Mn
2+
,Cr
2+
],[Cu
2+
,Zn
2+
] and [Ni
2+
,Ti
4+
] 
Statement II : Th
4+
 is the strongest reducing agent among Th
4+
,Ce
4+
,Gd
3+
 and Eu
2+
. 
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: Both Statement I and Statement II are true 
D: Statement I is true but Statement II is false 
Answer: B 
Explanation: 
Statement I 
An ion is coloured in aqueous solution if it has partially filled d-orbitals (so that d-d 
transitions are possible). Ions with d ° or d¹ ° configurations are colourless. 
Now check each pair: 
[Sc³ ?, Ti³ ?] 
Sc³ ? : Sc (Z = 21) is [Ar] 3d¹4s², so Sc³ ? = [Ar] 3d ° (colourless) 
Ti³ ? : Ti (Z = 22) is [Ar] 3d²4s², so Ti³ ? = [Ar] 3d¹ (coloured) 
? Both not coloured 
[Mn² ?, Cr² ?] 
Mn²? = [Ar] 3d 5 (coloured) 
Cr² ? = [Ar] 3d 4 (coloured) 
? Both coloured 
[Cu² ?, Zn² ?] 
Cu² ? = [Ar] 3d ? (coloured) 
Zn² ? = [Ar] 3d¹ ° (colourless) 
? Both not coloured 
[Ni² ?, Ti 4?] 
Ni² ? = [Ar] 3d 8 (coloured) 
Ti 4? = [Ar] 3d ° (colourless) 
? Both not coloured 
So, the number of pairs in which both ions are coloured is only 1, not 3. 
Hence, Statement I is false. 
Statement II 
A reducing agent is a species that gets oxidised (it donates electrons). 
Ce 4? is a well-known strong oxidising agent (it gets reduced to Ce³ ?). 
Th 4? is also in a high oxidation state and is not a strong reducing agent. 
Gd³ ? is a very stable common oxidation state for Gd; not strongly reducing. 
Eu² ? readily gets oxidised to Eu³ ?, so it acts as a strong reducing agent among the 
given species. 
Therefore, the strongest reducing agent among Th 4?, Ce 4?, Gd³ ? and Eu² ? is Eu² ?, not 
Th 4?. 
Hence, Statement II is false. 
Correct Option 
Since both Statement I and Statement II are false, the correct choice is: 
Option B: Both Statement I and Statement II are false. 
 
Q3: Given below are two statements: 
Statement I : The first ionization enthalpy of Cr is lower than that of Mn. 
Statement II : The second and third ionization enthalpies of Cr are higher than 
those of Mn. 
In the light of the above statements, choose the correct answer from the options 
given below : 
A: Statement I is true but Statement II is false 
B: Both Statement I and Statement II are true 
C: Both Statement I and Statement II are false 
D: Statement I is false but Statement II is true 
Answer: A 
Explanation: 
First, write the electronic configurations of Cr and Mn. 
Cr = (Ar)3d 54s¹ 
Mn = (Ar)3d 54s² 
IE 1(Cr) < IE 1(Mn) 
IE 2(Cr) > IE 2(Mn) 
IE 3(Cr) < IE 3(Mn) 
For first ionization enthalpy (IE 1): In Cr, the electron is removed from 4s¹. After losing 
this one 4s electron, Cr becomes Cr ? with a stable half-filled 3d 5 configuration. So 
removing the first electron is comparatively easier, hence IE 1(Cr) < IE 1(Mn). 
For second ionization enthalpy (IE 2): After losing one electron, Cr ? already has the 
stable 3d 5 configuration. Removing the second electron would disturb this stable 
half-filled 3d arrangement, so it needs more energy. But in Mn, after the first electron 
loss, removing the next electron can lead to Mn² ? with stable 3d 5. Therefore, IE 2(Cr) > 
IE 2(Mn). 
For third ionization enthalpy (IE 3): Mn² ? is very stable due to 3d 5. Removing one more 
electron from this stable state is difficult, so IE 3 for Mn becomes high. In comparison, 
Cr² ? does not have this extra stability, so removing the third electron is relatively easier. 
Hence, IE 3(Cr) < IE 3(Mn). 
 
Q4: On heating a mixture of common salt and K 2Cr 2O 7 in equal amount along with 
concentrated H 2SO 4 in a test tube, a gas is evolved. Formula of the gas evolved 
and oxidation state of the central metal atom in the gas respectively are: 
A: Cr 2O 2Cl 2 and +3 
B: Cr 2O 2Cl 2 and +6 
C: CrO 2Cl 2 and +6 
D: CrO 2Cl 2 and +5 
Answer: C 
Explanation:  
Gas evolved is chromyl chloride, formed in the chromyl chloride test. 
Step 1: Reaction (NCERT) 
On heating K 2Cr 2O 7 with a chloride salt (common salt NaCl) and conc. H 2SO 4, red fumes 
of chromyl chloride are obtained: 
K 2Cr 2O 7 + 4NaCl + 6H 2SO 4 ? 2CrO 2Cl 2 + 2KHSO 4 + 4NaHSO 4 + 3H 2O 
So, the gas is CrO 2Cl 2. 
Step 2: Oxidation state of Cr in CrO 2Cl 2 
Let oxidation state of Cr be x. 
Oxygen: 2 × (-2) = -4 
Chlorine: 2 × (-1) = -2 
Sum = x - 4 - 2 = 0 
x - 6 = 0 ? x = +6 
Correct option 
Option C: CrO 2Cl 2 and +6 
 
Q5: Given below are some of the statements about Mn and Mn 2O 7. Identify the 
correct statements. 
A. Mn forms the oxide Mn 2O 7, in which Mn is in its highest oxidation state. 
B. Oxygen stabilizes the Mn in higher oxidation states by forming multiple bonds 
with Mn. 
C. Mn 2O 7 is an ionic oxide. 
D. The structure of Mn 2O 7 consists of one bridged oxygen. 
Choose the correct answer from the options given below: 
A: A, B, C and D 
B: A, B and D Only 
C: A, B and C Only 
D: A, C and D Only 
Answer: B 
Explanation: 
A. Mn forms the oxide Mn 2O 7, in which Mn is in its highest oxidation state. 
Page 5


JEE Main Previous Year Questions (2021-2026): 
d and f Block Elements  
 
(January 2026) 
 
Q1: Consider the following statements about manganate and permanganate ions. 
Identify the correct statements. 
A. The geometry of both manganate and permanganate ions is tetrahedral. 
B. The oxidation states of Mn in manganate and permanganate are +7 and +6, 
respectively. 
C. Oxidation of Mn(II) salt by peroxodisulphate gives manganate ion as the final 
product. 
D. Manganate ion is paramagnetic and permanganate ion is diamagnetic. 
E. Acidified permanganate ion reduces oxalate, nitrite and iodide ions. 
Choose the correct answer from the options given below: 
A: A and D Only 
B: A, B and C Only 
C: A, C and D Only 
D: A, D and E Only 
Answer: A 
Explanation: 
Manganate ion ? MnO 4?² 
Permanganate ion ? MnO 4? 
(A) In both ions, Mn is surrounded by four oxygen atoms. So, the shape of both MnO 4?² 
and MnO 4? is tetrahedral (often written as d³s hybridisation in this context). 
(B) For MnO 4?, let oxidation state of Mn be x. Then x + 4(-2) = -1 ? x = +7. 
For MnO 4?², x + 4(-2) = -2 ? x = +6. 
So, Mn is +7 in permanganate and +6 in manganate (the statement given is reversed). 
(C) When Mn²? salt is oxidised by peroxodisulphate S 2O 8² ?, the product formed is 
permanganate ion, not manganate: 
Mn²? + S 2O 8² ? ? MnO 4? (Permanganate ion) 
(D) In MnO 4?, Mn is in +7 state, which corresponds to 3d ° (no unpaired electrons), so it 
is diamagnetic. 
In MnO 4?², Mn is in +6 state, which corresponds to 3d¹ (one unpaired electron), so it is 
paramagnetic. 
(E) Acidified permanganate ion acts as a strong oxidising agent, so it oxidises (i.e., 
“reduces” itself while oxidising) oxalate, nitrite and iodide ions. 
 
Q2: Given below are two statements: 
Statement I : The number of pairs, from the following, in which both the ions are 
coloured in aqueous solution is 3.  
[Sc
3+
,Ti
3+
],[Mn
2+
,Cr
2+
],[Cu
2+
,Zn
2+
] and [Ni
2+
,Ti
4+
] 
Statement II : Th
4+
 is the strongest reducing agent among Th
4+
,Ce
4+
,Gd
3+
 and Eu
2+
. 
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: Both Statement I and Statement II are true 
D: Statement I is true but Statement II is false 
Answer: B 
Explanation: 
Statement I 
An ion is coloured in aqueous solution if it has partially filled d-orbitals (so that d-d 
transitions are possible). Ions with d ° or d¹ ° configurations are colourless. 
Now check each pair: 
[Sc³ ?, Ti³ ?] 
Sc³ ? : Sc (Z = 21) is [Ar] 3d¹4s², so Sc³ ? = [Ar] 3d ° (colourless) 
Ti³ ? : Ti (Z = 22) is [Ar] 3d²4s², so Ti³ ? = [Ar] 3d¹ (coloured) 
? Both not coloured 
[Mn² ?, Cr² ?] 
Mn²? = [Ar] 3d 5 (coloured) 
Cr² ? = [Ar] 3d 4 (coloured) 
? Both coloured 
[Cu² ?, Zn² ?] 
Cu² ? = [Ar] 3d ? (coloured) 
Zn² ? = [Ar] 3d¹ ° (colourless) 
? Both not coloured 
[Ni² ?, Ti 4?] 
Ni² ? = [Ar] 3d 8 (coloured) 
Ti 4? = [Ar] 3d ° (colourless) 
? Both not coloured 
So, the number of pairs in which both ions are coloured is only 1, not 3. 
Hence, Statement I is false. 
Statement II 
A reducing agent is a species that gets oxidised (it donates electrons). 
Ce 4? is a well-known strong oxidising agent (it gets reduced to Ce³ ?). 
Th 4? is also in a high oxidation state and is not a strong reducing agent. 
Gd³ ? is a very stable common oxidation state for Gd; not strongly reducing. 
Eu² ? readily gets oxidised to Eu³ ?, so it acts as a strong reducing agent among the 
given species. 
Therefore, the strongest reducing agent among Th 4?, Ce 4?, Gd³ ? and Eu² ? is Eu² ?, not 
Th 4?. 
Hence, Statement II is false. 
Correct Option 
Since both Statement I and Statement II are false, the correct choice is: 
Option B: Both Statement I and Statement II are false. 
 
Q3: Given below are two statements: 
Statement I : The first ionization enthalpy of Cr is lower than that of Mn. 
Statement II : The second and third ionization enthalpies of Cr are higher than 
those of Mn. 
In the light of the above statements, choose the correct answer from the options 
given below : 
A: Statement I is true but Statement II is false 
B: Both Statement I and Statement II are true 
C: Both Statement I and Statement II are false 
D: Statement I is false but Statement II is true 
Answer: A 
Explanation: 
First, write the electronic configurations of Cr and Mn. 
Cr = (Ar)3d 54s¹ 
Mn = (Ar)3d 54s² 
IE 1(Cr) < IE 1(Mn) 
IE 2(Cr) > IE 2(Mn) 
IE 3(Cr) < IE 3(Mn) 
For first ionization enthalpy (IE 1): In Cr, the electron is removed from 4s¹. After losing 
this one 4s electron, Cr becomes Cr ? with a stable half-filled 3d 5 configuration. So 
removing the first electron is comparatively easier, hence IE 1(Cr) < IE 1(Mn). 
For second ionization enthalpy (IE 2): After losing one electron, Cr ? already has the 
stable 3d 5 configuration. Removing the second electron would disturb this stable 
half-filled 3d arrangement, so it needs more energy. But in Mn, after the first electron 
loss, removing the next electron can lead to Mn² ? with stable 3d 5. Therefore, IE 2(Cr) > 
IE 2(Mn). 
For third ionization enthalpy (IE 3): Mn² ? is very stable due to 3d 5. Removing one more 
electron from this stable state is difficult, so IE 3 for Mn becomes high. In comparison, 
Cr² ? does not have this extra stability, so removing the third electron is relatively easier. 
Hence, IE 3(Cr) < IE 3(Mn). 
 
Q4: On heating a mixture of common salt and K 2Cr 2O 7 in equal amount along with 
concentrated H 2SO 4 in a test tube, a gas is evolved. Formula of the gas evolved 
and oxidation state of the central metal atom in the gas respectively are: 
A: Cr 2O 2Cl 2 and +3 
B: Cr 2O 2Cl 2 and +6 
C: CrO 2Cl 2 and +6 
D: CrO 2Cl 2 and +5 
Answer: C 
Explanation:  
Gas evolved is chromyl chloride, formed in the chromyl chloride test. 
Step 1: Reaction (NCERT) 
On heating K 2Cr 2O 7 with a chloride salt (common salt NaCl) and conc. H 2SO 4, red fumes 
of chromyl chloride are obtained: 
K 2Cr 2O 7 + 4NaCl + 6H 2SO 4 ? 2CrO 2Cl 2 + 2KHSO 4 + 4NaHSO 4 + 3H 2O 
So, the gas is CrO 2Cl 2. 
Step 2: Oxidation state of Cr in CrO 2Cl 2 
Let oxidation state of Cr be x. 
Oxygen: 2 × (-2) = -4 
Chlorine: 2 × (-1) = -2 
Sum = x - 4 - 2 = 0 
x - 6 = 0 ? x = +6 
Correct option 
Option C: CrO 2Cl 2 and +6 
 
Q5: Given below are some of the statements about Mn and Mn 2O 7. Identify the 
correct statements. 
A. Mn forms the oxide Mn 2O 7, in which Mn is in its highest oxidation state. 
B. Oxygen stabilizes the Mn in higher oxidation states by forming multiple bonds 
with Mn. 
C. Mn 2O 7 is an ionic oxide. 
D. The structure of Mn 2O 7 consists of one bridged oxygen. 
Choose the correct answer from the options given below: 
A: A, B, C and D 
B: A, B and D Only 
C: A, B and C Only 
D: A, C and D Only 
Answer: B 
Explanation: 
A. Mn forms the oxide Mn 2O 7, in which Mn is in its highest oxidation state. 
To find the oxidation state of Mn in Mn 2O 7, we set the sum of oxidation states to zero. 
Let Mn's oxidation state be 'x'. Oxygen's oxidation state is typically -2. 
2(x) + 7(-2) = 0 
2x - 14 = 0 
2x = 14 
x = +7 
Manganese is a Group 7 element with an electronic configuration of [Ar]3d 54s². The 
highest possible oxidation state for manganese is +7, which involves the loss of all 7 
valence electrons. 
Therefore, Mn is indeed in its highest oxidation state (+7) in Mn 2O 7. 
This statement is correct. 
B. Oxygen stabilizes the Mn in higher oxidation states by forming multiple bonds 
with Mn. 
In high oxidation states, transition metals strongly attract electrons, increasing the 
covalent character of their bonds with oxygen. This is particularly true for elements like 
Mn in a +7 oxidation state. 
The formation of multiple bonds (e.g., Mn=O) allows for better delocalization of electron 
density, which helps to stabilize the high positive charge on the central metal atom. This 
is a common phenomenon for high oxidation state metal oxides (e.g., CrO 3, V 2O 5). 
This statement is correct. 
C. Mn 2O 7 is an ionic oxide. 
Ionic oxides are typically formed when the metal has a low oxidation state and a 
significant electronegativity difference with oxygen leads to electron transfer (e.g., 
MnO). 
However, as the oxidation state of a metal increases, its ability to polarize electron 
clouds increases, and the covalent character of its bonds with oxygen becomes 
prominent. 
Mn 2O 7 is a dark green, viscous liquid with a melting point of -20 °C. These properties are 
characteristic of a molecular covalent compound, not an ionic solid which would have a 
high melting point. It is highly covalent due to the high electronegativity of oxygen and 
the high charge density of Mn 7?. 
Therefore, Mn 2O 7 is a covalent oxide, not an ionic oxide. 
This statement is incorrect. 
D. The structure of Mn 2O 7 consists of one bridged oxygen. 
The structure of Mn 2O 7 can be represented as O 3Mn - O - MnO 3. It consists of two 
MnO 4 tetrahedra sharing a common oxygen atom. 
The oxygen atom that links the two manganese atoms is a bridging oxygen. Each Mn 
atom has three terminal oxygen atoms and one bridging oxygen atom. 
This statement is correct. 
Based on the analysis, statements A, B, and D are correct, while statement C is