Co-ordination Compounds: JEE Main Previous Year Questions (2021-2026)

 Page 1


JEE Main Previous Year Questions (2021-2026): 
Coordination Compounds  
 
(January 2026) 
 
 
Q1: The correct increasing order of spin-only magnetic moment values of the complex 
ions 
[MnBr 4]² ? (A), [Cu(H 2O) 6]² ? (B), [Ni(CN) 4]² ? (C) and [Ni(H 2O) 6]² ? (D) is: 
A: A = B < D < C 
B: C = D < B < A 
C: C < B < D < A 
D: A = B < C < D 
Answer: C 
Explanation: 
Mn² ? : 3d 5 
In [MnBr 4]² ?, Mn is in +2 oxidation state, so configuration is 3d 5. 
Br ? is a weak field ligand and the complex is tetrahedral, so it is high spin. 
Hence number of unpaired electrons (n) = 5. 
Cu² ? : 3d ? 
In [Cu(H 2O) 6]² ?, Cu is in +2 oxidation state, so configuration is 3d ?. 
This is an octahedral complex (six water ligands). In d ?, there is always one unpaired electron. 
 
Ni² ? : 3d 8 
In [Ni(CN) 4]² ?, Ni is in +2 oxidation state, so configuration is 3d 8. 
CN ? is a strong field ligand, so electrons pair up and the complex becomes square planar. 
In square planar d 8, all electrons are paired, so (n) = 0. 
Ni² ? : 3d 8 (tetrahedral) 
In [Ni(H 2O) 6]² ?, Ni is in +2 oxidation state, so configuration is 3d 8. 
With H 2O (weak field), the complex is high spin; the distribution gives two unpaired electrons. 
 
 
Q2: The correct statement among the following is: 
A: [Ni(CN) 4]² ? and [NiCl 4]² ? are diamagnetic and Ni(CO) 4 is paramagnetic. 
B: Ni(CO) 4 is diamagnetic and [NiCl 4]² ? and [Ni(CN) 4]² ? are paramagnetic. 
C: Ni(CO) 4 and [NiCl 4]² ? are diamagnetic and [Ni(CN) 4]² ? is paramagnetic. 
D: Ni(CO) 4 and [Ni(CN) 4]² ? are diamagnetic and [NiCl 4]² ? is paramagnetic. 
Answer: D 
Explanation: 
Option D is correct. 
Page 2


JEE Main Previous Year Questions (2021-2026): 
Coordination Compounds  
 
(January 2026) 
 
 
Q1: The correct increasing order of spin-only magnetic moment values of the complex 
ions 
[MnBr 4]² ? (A), [Cu(H 2O) 6]² ? (B), [Ni(CN) 4]² ? (C) and [Ni(H 2O) 6]² ? (D) is: 
A: A = B < D < C 
B: C = D < B < A 
C: C < B < D < A 
D: A = B < C < D 
Answer: C 
Explanation: 
Mn² ? : 3d 5 
In [MnBr 4]² ?, Mn is in +2 oxidation state, so configuration is 3d 5. 
Br ? is a weak field ligand and the complex is tetrahedral, so it is high spin. 
Hence number of unpaired electrons (n) = 5. 
Cu² ? : 3d ? 
In [Cu(H 2O) 6]² ?, Cu is in +2 oxidation state, so configuration is 3d ?. 
This is an octahedral complex (six water ligands). In d ?, there is always one unpaired electron. 
 
Ni² ? : 3d 8 
In [Ni(CN) 4]² ?, Ni is in +2 oxidation state, so configuration is 3d 8. 
CN ? is a strong field ligand, so electrons pair up and the complex becomes square planar. 
In square planar d 8, all electrons are paired, so (n) = 0. 
Ni² ? : 3d 8 (tetrahedral) 
In [Ni(H 2O) 6]² ?, Ni is in +2 oxidation state, so configuration is 3d 8. 
With H 2O (weak field), the complex is high spin; the distribution gives two unpaired electrons. 
 
 
Q2: The correct statement among the following is: 
A: [Ni(CN) 4]² ? and [NiCl 4]² ? are diamagnetic and Ni(CO) 4 is paramagnetic. 
B: Ni(CO) 4 is diamagnetic and [NiCl 4]² ? and [Ni(CN) 4]² ? are paramagnetic. 
C: Ni(CO) 4 and [NiCl 4]² ? are diamagnetic and [Ni(CN) 4]² ? is paramagnetic. 
D: Ni(CO) 4 and [Ni(CN) 4]² ? are diamagnetic and [NiCl 4]² ? is paramagnetic. 
Answer: D 
Explanation: 
Option D is correct. 
1. Ni(CO) 4 
Oxidation state of Ni: CO is a neutral ligand, so Ni is in 0 state. 
Electronic configuration: Ni(0) ? 3d¹ °4s ° 
Since it is d¹ °, all electrons are paired. 
Hence, Ni(CO) 4 is diamagnetic. 
2. [Ni(CN) 4]² ? 
Oxidation state of Ni: 
x + 4(-1) = -2 ? x = +2 
So, Ni² ? is 3d 8. 
CN ? is a strong field ligand, causes pairing of electrons. 
For Ni² ? (d 8) with strong field ligands, complex becomes square planar (inner orbital), and all 
electrons get paired. 
Hence, [Ni(CN) 4]² ? is diamagnetic. 
3. [NiCl 4]² ? 
Oxidation state of Ni: 
x + 4(-1) = -2 ? x = +2 
So, again Ni² ? is 3d 8. 
Cl ? is a weak field ligand, so it does not cause pairing. 
Such complexes are generally tetrahedral (outer orbital) and remain high spin, leaving 2 
unpaired electrons. 
Hence, [NiCl 4]² ? is paramagnetic. 
Therefore: 
Ni(CO) 4 ? diamagnetic 
[Ni(CN) 4]² ? ? diamagnetic 
[NiCl 4]² ? ? paramagnetic 
So, the correct statement is Option D. 
 
Q3: The wavelength of light absorbed for the following complexes are in the order 
 
A: III < IV < I < II < V 
B: III < I < IV < V < II 
C: III < I < IV < II < V 
D: III < I < II < IV < V 
Answer: C 
Explanation: 
Wavelength of light absorbed increases as C.F.S.E of complex decreases. 
[Co(CN) 6]³ ? has maximum CFSE 
[CoF 6]³ ? has least CFSE 
Ligand field strength ? : C.F.S.E ? 
Correct wavelength order. 
V > II > IV > I > III 
Page 3


JEE Main Previous Year Questions (2021-2026): 
Coordination Compounds  
 
(January 2026) 
 
 
Q1: The correct increasing order of spin-only magnetic moment values of the complex 
ions 
[MnBr 4]² ? (A), [Cu(H 2O) 6]² ? (B), [Ni(CN) 4]² ? (C) and [Ni(H 2O) 6]² ? (D) is: 
A: A = B < D < C 
B: C = D < B < A 
C: C < B < D < A 
D: A = B < C < D 
Answer: C 
Explanation: 
Mn² ? : 3d 5 
In [MnBr 4]² ?, Mn is in +2 oxidation state, so configuration is 3d 5. 
Br ? is a weak field ligand and the complex is tetrahedral, so it is high spin. 
Hence number of unpaired electrons (n) = 5. 
Cu² ? : 3d ? 
In [Cu(H 2O) 6]² ?, Cu is in +2 oxidation state, so configuration is 3d ?. 
This is an octahedral complex (six water ligands). In d ?, there is always one unpaired electron. 
 
Ni² ? : 3d 8 
In [Ni(CN) 4]² ?, Ni is in +2 oxidation state, so configuration is 3d 8. 
CN ? is a strong field ligand, so electrons pair up and the complex becomes square planar. 
In square planar d 8, all electrons are paired, so (n) = 0. 
Ni² ? : 3d 8 (tetrahedral) 
In [Ni(H 2O) 6]² ?, Ni is in +2 oxidation state, so configuration is 3d 8. 
With H 2O (weak field), the complex is high spin; the distribution gives two unpaired electrons. 
 
 
Q2: The correct statement among the following is: 
A: [Ni(CN) 4]² ? and [NiCl 4]² ? are diamagnetic and Ni(CO) 4 is paramagnetic. 
B: Ni(CO) 4 is diamagnetic and [NiCl 4]² ? and [Ni(CN) 4]² ? are paramagnetic. 
C: Ni(CO) 4 and [NiCl 4]² ? are diamagnetic and [Ni(CN) 4]² ? is paramagnetic. 
D: Ni(CO) 4 and [Ni(CN) 4]² ? are diamagnetic and [NiCl 4]² ? is paramagnetic. 
Answer: D 
Explanation: 
Option D is correct. 
1. Ni(CO) 4 
Oxidation state of Ni: CO is a neutral ligand, so Ni is in 0 state. 
Electronic configuration: Ni(0) ? 3d¹ °4s ° 
Since it is d¹ °, all electrons are paired. 
Hence, Ni(CO) 4 is diamagnetic. 
2. [Ni(CN) 4]² ? 
Oxidation state of Ni: 
x + 4(-1) = -2 ? x = +2 
So, Ni² ? is 3d 8. 
CN ? is a strong field ligand, causes pairing of electrons. 
For Ni² ? (d 8) with strong field ligands, complex becomes square planar (inner orbital), and all 
electrons get paired. 
Hence, [Ni(CN) 4]² ? is diamagnetic. 
3. [NiCl 4]² ? 
Oxidation state of Ni: 
x + 4(-1) = -2 ? x = +2 
So, again Ni² ? is 3d 8. 
Cl ? is a weak field ligand, so it does not cause pairing. 
Such complexes are generally tetrahedral (outer orbital) and remain high spin, leaving 2 
unpaired electrons. 
Hence, [NiCl 4]² ? is paramagnetic. 
Therefore: 
Ni(CO) 4 ? diamagnetic 
[Ni(CN) 4]² ? ? diamagnetic 
[NiCl 4]² ? ? paramagnetic 
So, the correct statement is Option D. 
 
Q3: The wavelength of light absorbed for the following complexes are in the order 
 
A: III < IV < I < II < V 
B: III < I < IV < V < II 
C: III < I < IV < II < V 
D: III < I < II < IV < V 
Answer: C 
Explanation: 
Wavelength of light absorbed increases as C.F.S.E of complex decreases. 
[Co(CN) 6]³ ? has maximum CFSE 
[CoF 6]³ ? has least CFSE 
Ligand field strength ? : C.F.S.E ? 
Correct wavelength order. 
V > II > IV > I > III 
 
Q4: Given below are two statements: 
Statement I: Hybridisation, shape and spin only magnetic moment of K 3[Co(CO 3) 3] is 
sp³d², octahedral and 4.9 BM respectively. 
Statement II: Geometry, hybridisation and spin only magnetic moment values (BM) of the 
ions [Ni(CN) 4]² ?, [MnBr 4]² ? and [CoF 6]³ ? respectively are square planar, tetrahedral, 
octahedral; dsp², sp³, sp³d² and 0, 5.9, 4.9. 
In the light of the above statements, choose the correct answer from the options given 
below: 
A: Both Statement I and Statement II are false 
B: Statement I is true but Statement II is false 
C: Both Statement I and Statement II are true 
D: Statement I is false but Statement II is true 
Answer: C 
Explanation: 
To decide hybridisation, geometry and magnetic moment, we mainly look at (i) the type of 
ligands (strong field or weak field) and (ii) the number of unpaired electrons. The spin-only 
magnetic moment depends on the number of unpaired electrons n. 
The given results for each complex are summarised below (hybridisation ? geometry ? 
number of unpaired electrons ? magnetic moment in B.M.). 
In K 3[Co(CO 3) 3]  
? sp³d² hybridized, octahedral 
? 4 unpaired electron 
? 4.9 B.M. 
[Ni(CN) 4]² ?  
? dsp² hybridized, square planar 
? 0 unpaired electron 
? 0 B.M. 
[MnBr 4]² ?  
? sp³ hybridized, tetrahedral 
? 5 unpaired electron 
? 5.9 B.M. 
[CoF 6]³ ?  
? sp³d² hybridized, octahedral 
? 4 unpaired electron 
? 4.9 B.M. 
So, the hybridisation/geometry and magnetic moments mentioned in the statements match 
these standard results (based on unpaired electrons in each case). 
 
Q5: Given below are two statements: 
Statement I: The number of paramagnetic species among [CoF 6]³ ?, [TiF 6]² ?, V 2O 5 and 
[Fe(CN) 6]³ ? is 3. 
Statement II: 
K 4[Fe(CN) 6] < K 3[Fe(CN) 6] < [Fe(H 2O) 6]SO 4 · H 2O < [Fe(H 2O) 6]Cl 3 is the correct order in terms 
Page 4


JEE Main Previous Year Questions (2021-2026): 
Coordination Compounds  
 
(January 2026) 
 
 
Q1: The correct increasing order of spin-only magnetic moment values of the complex 
ions 
[MnBr 4]² ? (A), [Cu(H 2O) 6]² ? (B), [Ni(CN) 4]² ? (C) and [Ni(H 2O) 6]² ? (D) is: 
A: A = B < D < C 
B: C = D < B < A 
C: C < B < D < A 
D: A = B < C < D 
Answer: C 
Explanation: 
Mn² ? : 3d 5 
In [MnBr 4]² ?, Mn is in +2 oxidation state, so configuration is 3d 5. 
Br ? is a weak field ligand and the complex is tetrahedral, so it is high spin. 
Hence number of unpaired electrons (n) = 5. 
Cu² ? : 3d ? 
In [Cu(H 2O) 6]² ?, Cu is in +2 oxidation state, so configuration is 3d ?. 
This is an octahedral complex (six water ligands). In d ?, there is always one unpaired electron. 
 
Ni² ? : 3d 8 
In [Ni(CN) 4]² ?, Ni is in +2 oxidation state, so configuration is 3d 8. 
CN ? is a strong field ligand, so electrons pair up and the complex becomes square planar. 
In square planar d 8, all electrons are paired, so (n) = 0. 
Ni² ? : 3d 8 (tetrahedral) 
In [Ni(H 2O) 6]² ?, Ni is in +2 oxidation state, so configuration is 3d 8. 
With H 2O (weak field), the complex is high spin; the distribution gives two unpaired electrons. 
 
 
Q2: The correct statement among the following is: 
A: [Ni(CN) 4]² ? and [NiCl 4]² ? are diamagnetic and Ni(CO) 4 is paramagnetic. 
B: Ni(CO) 4 is diamagnetic and [NiCl 4]² ? and [Ni(CN) 4]² ? are paramagnetic. 
C: Ni(CO) 4 and [NiCl 4]² ? are diamagnetic and [Ni(CN) 4]² ? is paramagnetic. 
D: Ni(CO) 4 and [Ni(CN) 4]² ? are diamagnetic and [NiCl 4]² ? is paramagnetic. 
Answer: D 
Explanation: 
Option D is correct. 
1. Ni(CO) 4 
Oxidation state of Ni: CO is a neutral ligand, so Ni is in 0 state. 
Electronic configuration: Ni(0) ? 3d¹ °4s ° 
Since it is d¹ °, all electrons are paired. 
Hence, Ni(CO) 4 is diamagnetic. 
2. [Ni(CN) 4]² ? 
Oxidation state of Ni: 
x + 4(-1) = -2 ? x = +2 
So, Ni² ? is 3d 8. 
CN ? is a strong field ligand, causes pairing of electrons. 
For Ni² ? (d 8) with strong field ligands, complex becomes square planar (inner orbital), and all 
electrons get paired. 
Hence, [Ni(CN) 4]² ? is diamagnetic. 
3. [NiCl 4]² ? 
Oxidation state of Ni: 
x + 4(-1) = -2 ? x = +2 
So, again Ni² ? is 3d 8. 
Cl ? is a weak field ligand, so it does not cause pairing. 
Such complexes are generally tetrahedral (outer orbital) and remain high spin, leaving 2 
unpaired electrons. 
Hence, [NiCl 4]² ? is paramagnetic. 
Therefore: 
Ni(CO) 4 ? diamagnetic 
[Ni(CN) 4]² ? ? diamagnetic 
[NiCl 4]² ? ? paramagnetic 
So, the correct statement is Option D. 
 
Q3: The wavelength of light absorbed for the following complexes are in the order 
 
A: III < IV < I < II < V 
B: III < I < IV < V < II 
C: III < I < IV < II < V 
D: III < I < II < IV < V 
Answer: C 
Explanation: 
Wavelength of light absorbed increases as C.F.S.E of complex decreases. 
[Co(CN) 6]³ ? has maximum CFSE 
[CoF 6]³ ? has least CFSE 
Ligand field strength ? : C.F.S.E ? 
Correct wavelength order. 
V > II > IV > I > III 
 
Q4: Given below are two statements: 
Statement I: Hybridisation, shape and spin only magnetic moment of K 3[Co(CO 3) 3] is 
sp³d², octahedral and 4.9 BM respectively. 
Statement II: Geometry, hybridisation and spin only magnetic moment values (BM) of the 
ions [Ni(CN) 4]² ?, [MnBr 4]² ? and [CoF 6]³ ? respectively are square planar, tetrahedral, 
octahedral; dsp², sp³, sp³d² and 0, 5.9, 4.9. 
In the light of the above statements, choose the correct answer from the options given 
below: 
A: Both Statement I and Statement II are false 
B: Statement I is true but Statement II is false 
C: Both Statement I and Statement II are true 
D: Statement I is false but Statement II is true 
Answer: C 
Explanation: 
To decide hybridisation, geometry and magnetic moment, we mainly look at (i) the type of 
ligands (strong field or weak field) and (ii) the number of unpaired electrons. The spin-only 
magnetic moment depends on the number of unpaired electrons n. 
The given results for each complex are summarised below (hybridisation ? geometry ? 
number of unpaired electrons ? magnetic moment in B.M.). 
In K 3[Co(CO 3) 3]  
? sp³d² hybridized, octahedral 
? 4 unpaired electron 
? 4.9 B.M. 
[Ni(CN) 4]² ?  
? dsp² hybridized, square planar 
? 0 unpaired electron 
? 0 B.M. 
[MnBr 4]² ?  
? sp³ hybridized, tetrahedral 
? 5 unpaired electron 
? 5.9 B.M. 
[CoF 6]³ ?  
? sp³d² hybridized, octahedral 
? 4 unpaired electron 
? 4.9 B.M. 
So, the hybridisation/geometry and magnetic moments mentioned in the statements match 
these standard results (based on unpaired electrons in each case). 
 
Q5: Given below are two statements: 
Statement I: The number of paramagnetic species among [CoF 6]³ ?, [TiF 6]² ?, V 2O 5 and 
[Fe(CN) 6]³ ? is 3. 
Statement II: 
K 4[Fe(CN) 6] < K 3[Fe(CN) 6] < [Fe(H 2O) 6]SO 4 · H 2O < [Fe(H 2O) 6]Cl 3 is the correct order in terms 
of number of unpaired electron(s) present in the complexes. 
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 false 
C: Statement I is false but Statement II is true 
D: Both Statement I and Statement II are true 
Answer: D 
Explanation: 
Paramagnetic species are those which have one or more unpaired electrons. 
Let us check each species one by one. 
1. In [CoF 6]³ ?, oxidation state of Co is +3 (because 6 × (-1) = -6 and overall charge is -3). So 
Co is Co³ ? i.e. d 6. Since F ? is a weak field ligand, it forms a high spin octahedral complex, so 
unpaired electrons are present.  
Hence, [CoF 6]³ ? is paramagnetic. 
2. In [TiF 6]³ ?, oxidation state of Ti is +3. So Ti is Ti³ ? i.e. d¹. One electron will remain unpaired, so 
it is paramagnetic. 
3. In [Fe(CN) 6]³ ?, oxidation state of Fe is +3. So Fe is Fe³ ? i.e. d 5. Since CN ? is a strong field 
ligand, it gives a low spin octahedral complex, so only one electron remains unpaired.  
Hence, it is paramagnetic. 
4. In V 2O 5, V is in +5 oxidation state, so it is V 5 ? i.e. d °. There are no d-electrons, so there cannot 
be any unpaired electron.  
Hence, V 2O 5 is diamagnetic. 
So, the paramagnetic species are: 
[CoF 6]³ ?, [TiF 6]³ ?, [Fe(CN) 6]³ ? 
And the diamagnetic species is: V 2O 5 
Now, compare the complexes in Statement II by counting unpaired electrons. 
In K 4[Fe(CN) 6] ? Fe is +2 (because complex anion is 4-), so Fe² ? is d 6. With CN ? (strong field), it 
is low spin, so all electrons pair up. Therefore, number of unpaired electron = 0. 
 
 
Q6: Consider a mixture ‘X’ which is made by dissolving 0.4 mol of [Co(NH 3) 5SO 4]Br and 
0.4 mol of [Co(NH 3) 5Br]SO 4 in water to make 4 L of solution. When 2 L of mixture ‘X’ is 
allowed to react with excess of AgNO 3, it forms precipitate ‘Y’. The rest 2 L of mixture ‘X’ 
reacts with excess BaCl 2 to form precipitate ‘Z’. Which of the following statements is 
CORRECT? 
Page 5


JEE Main Previous Year Questions (2021-2026): 
Coordination Compounds  
 
(January 2026) 
 
 
Q1: The correct increasing order of spin-only magnetic moment values of the complex 
ions 
[MnBr 4]² ? (A), [Cu(H 2O) 6]² ? (B), [Ni(CN) 4]² ? (C) and [Ni(H 2O) 6]² ? (D) is: 
A: A = B < D < C 
B: C = D < B < A 
C: C < B < D < A 
D: A = B < C < D 
Answer: C 
Explanation: 
Mn² ? : 3d 5 
In [MnBr 4]² ?, Mn is in +2 oxidation state, so configuration is 3d 5. 
Br ? is a weak field ligand and the complex is tetrahedral, so it is high spin. 
Hence number of unpaired electrons (n) = 5. 
Cu² ? : 3d ? 
In [Cu(H 2O) 6]² ?, Cu is in +2 oxidation state, so configuration is 3d ?. 
This is an octahedral complex (six water ligands). In d ?, there is always one unpaired electron. 
 
Ni² ? : 3d 8 
In [Ni(CN) 4]² ?, Ni is in +2 oxidation state, so configuration is 3d 8. 
CN ? is a strong field ligand, so electrons pair up and the complex becomes square planar. 
In square planar d 8, all electrons are paired, so (n) = 0. 
Ni² ? : 3d 8 (tetrahedral) 
In [Ni(H 2O) 6]² ?, Ni is in +2 oxidation state, so configuration is 3d 8. 
With H 2O (weak field), the complex is high spin; the distribution gives two unpaired electrons. 
 
 
Q2: The correct statement among the following is: 
A: [Ni(CN) 4]² ? and [NiCl 4]² ? are diamagnetic and Ni(CO) 4 is paramagnetic. 
B: Ni(CO) 4 is diamagnetic and [NiCl 4]² ? and [Ni(CN) 4]² ? are paramagnetic. 
C: Ni(CO) 4 and [NiCl 4]² ? are diamagnetic and [Ni(CN) 4]² ? is paramagnetic. 
D: Ni(CO) 4 and [Ni(CN) 4]² ? are diamagnetic and [NiCl 4]² ? is paramagnetic. 
Answer: D 
Explanation: 
Option D is correct. 
1. Ni(CO) 4 
Oxidation state of Ni: CO is a neutral ligand, so Ni is in 0 state. 
Electronic configuration: Ni(0) ? 3d¹ °4s ° 
Since it is d¹ °, all electrons are paired. 
Hence, Ni(CO) 4 is diamagnetic. 
2. [Ni(CN) 4]² ? 
Oxidation state of Ni: 
x + 4(-1) = -2 ? x = +2 
So, Ni² ? is 3d 8. 
CN ? is a strong field ligand, causes pairing of electrons. 
For Ni² ? (d 8) with strong field ligands, complex becomes square planar (inner orbital), and all 
electrons get paired. 
Hence, [Ni(CN) 4]² ? is diamagnetic. 
3. [NiCl 4]² ? 
Oxidation state of Ni: 
x + 4(-1) = -2 ? x = +2 
So, again Ni² ? is 3d 8. 
Cl ? is a weak field ligand, so it does not cause pairing. 
Such complexes are generally tetrahedral (outer orbital) and remain high spin, leaving 2 
unpaired electrons. 
Hence, [NiCl 4]² ? is paramagnetic. 
Therefore: 
Ni(CO) 4 ? diamagnetic 
[Ni(CN) 4]² ? ? diamagnetic 
[NiCl 4]² ? ? paramagnetic 
So, the correct statement is Option D. 
 
Q3: The wavelength of light absorbed for the following complexes are in the order 
 
A: III < IV < I < II < V 
B: III < I < IV < V < II 
C: III < I < IV < II < V 
D: III < I < II < IV < V 
Answer: C 
Explanation: 
Wavelength of light absorbed increases as C.F.S.E of complex decreases. 
[Co(CN) 6]³ ? has maximum CFSE 
[CoF 6]³ ? has least CFSE 
Ligand field strength ? : C.F.S.E ? 
Correct wavelength order. 
V > II > IV > I > III 
 
Q4: Given below are two statements: 
Statement I: Hybridisation, shape and spin only magnetic moment of K 3[Co(CO 3) 3] is 
sp³d², octahedral and 4.9 BM respectively. 
Statement II: Geometry, hybridisation and spin only magnetic moment values (BM) of the 
ions [Ni(CN) 4]² ?, [MnBr 4]² ? and [CoF 6]³ ? respectively are square planar, tetrahedral, 
octahedral; dsp², sp³, sp³d² and 0, 5.9, 4.9. 
In the light of the above statements, choose the correct answer from the options given 
below: 
A: Both Statement I and Statement II are false 
B: Statement I is true but Statement II is false 
C: Both Statement I and Statement II are true 
D: Statement I is false but Statement II is true 
Answer: C 
Explanation: 
To decide hybridisation, geometry and magnetic moment, we mainly look at (i) the type of 
ligands (strong field or weak field) and (ii) the number of unpaired electrons. The spin-only 
magnetic moment depends on the number of unpaired electrons n. 
The given results for each complex are summarised below (hybridisation ? geometry ? 
number of unpaired electrons ? magnetic moment in B.M.). 
In K 3[Co(CO 3) 3]  
? sp³d² hybridized, octahedral 
? 4 unpaired electron 
? 4.9 B.M. 
[Ni(CN) 4]² ?  
? dsp² hybridized, square planar 
? 0 unpaired electron 
? 0 B.M. 
[MnBr 4]² ?  
? sp³ hybridized, tetrahedral 
? 5 unpaired electron 
? 5.9 B.M. 
[CoF 6]³ ?  
? sp³d² hybridized, octahedral 
? 4 unpaired electron 
? 4.9 B.M. 
So, the hybridisation/geometry and magnetic moments mentioned in the statements match 
these standard results (based on unpaired electrons in each case). 
 
Q5: Given below are two statements: 
Statement I: The number of paramagnetic species among [CoF 6]³ ?, [TiF 6]² ?, V 2O 5 and 
[Fe(CN) 6]³ ? is 3. 
Statement II: 
K 4[Fe(CN) 6] < K 3[Fe(CN) 6] < [Fe(H 2O) 6]SO 4 · H 2O < [Fe(H 2O) 6]Cl 3 is the correct order in terms 
of number of unpaired electron(s) present in the complexes. 
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 false 
C: Statement I is false but Statement II is true 
D: Both Statement I and Statement II are true 
Answer: D 
Explanation: 
Paramagnetic species are those which have one or more unpaired electrons. 
Let us check each species one by one. 
1. In [CoF 6]³ ?, oxidation state of Co is +3 (because 6 × (-1) = -6 and overall charge is -3). So 
Co is Co³ ? i.e. d 6. Since F ? is a weak field ligand, it forms a high spin octahedral complex, so 
unpaired electrons are present.  
Hence, [CoF 6]³ ? is paramagnetic. 
2. In [TiF 6]³ ?, oxidation state of Ti is +3. So Ti is Ti³ ? i.e. d¹. One electron will remain unpaired, so 
it is paramagnetic. 
3. In [Fe(CN) 6]³ ?, oxidation state of Fe is +3. So Fe is Fe³ ? i.e. d 5. Since CN ? is a strong field 
ligand, it gives a low spin octahedral complex, so only one electron remains unpaired.  
Hence, it is paramagnetic. 
4. In V 2O 5, V is in +5 oxidation state, so it is V 5 ? i.e. d °. There are no d-electrons, so there cannot 
be any unpaired electron.  
Hence, V 2O 5 is diamagnetic. 
So, the paramagnetic species are: 
[CoF 6]³ ?, [TiF 6]³ ?, [Fe(CN) 6]³ ? 
And the diamagnetic species is: V 2O 5 
Now, compare the complexes in Statement II by counting unpaired electrons. 
In K 4[Fe(CN) 6] ? Fe is +2 (because complex anion is 4-), so Fe² ? is d 6. With CN ? (strong field), it 
is low spin, so all electrons pair up. Therefore, number of unpaired electron = 0. 
 
 
Q6: Consider a mixture ‘X’ which is made by dissolving 0.4 mol of [Co(NH 3) 5SO 4]Br and 
0.4 mol of [Co(NH 3) 5Br]SO 4 in water to make 4 L of solution. When 2 L of mixture ‘X’ is 
allowed to react with excess of AgNO 3, it forms precipitate ‘Y’. The rest 2 L of mixture ‘X’ 
reacts with excess BaCl 2 to form precipitate ‘Z’. Which of the following statements is 
CORRECT? 
A: 0.1 mol of ‘Y’ is formed. 
B: 0.4 mol of ‘Z’ is formed. 
C: 0.2 mol of ‘Z’ is formed. 
D: ‘Y’ is BaSO 4 and ‘Z’ is AgBr. 
Answer: C 
Explanation: 
For coordination compounds, only the ions outside the coordination sphere dissociate in water 
and give precipitates. 
1. Identify free ions in solution 
(i) [Co(NH 3) 5SO 4]Br 
Here Br ? is outside the bracket, so in water: 
[Co(NH 3) 5SO 4]Br ? [Co(NH 3) 5SO 4] ? + Br ? 
So it gives 1 mol Br ? per mol of salt. 
(ii) [Co(NH 3) 5Br]SO 4 
Here SO 4² ? is outside the bracket, so: 
[Co(NH 3) 5Br]SO 4 ? [Co(NH 3) 5Br]² ? + SO 4² ? 
So it gives 1 mol SO 4² ? per mol of salt. 
2. Moles present in 2 L of mixture X 
Total solution volume = 4 L, and we take 2 L, i.e. half. 
So in 2 L: 
From 0.4 mol of [Co(NH 3) 5SO 4]Br, we have 0.2 mol of this salt ? 0.2 mol Br ?. 
From 0.4 mol of [Co(NH 3) 5Br]SO 4, we have 0.2 mol of this salt ? 0.2 mol SO 4² ?. 
3. Precipitates formed 
With excess AgNO 3 (in 2 L): 
Ag ? precipitates Br ? as AgBr: 
Ag ? + Br ? ? AgBr(s) 
So moles of precipitate Y = 0.2 mol (as AgBr). 
With excess BaCl 2 (in remaining 2 L): 
Ba² ? precipitates SO 4² ? as BaSO 4: 
Ba² ? + SO 4² ? ? BaSO 4(s) 
So moles of precipitate Z = 0.2 mol (as BaSO 4). 
Correct option 
Option C: 0.2 mol of Z is formed. 
(Also, Y = AgBr and Z = BaSO 4.) 
Q7: Identify the CORRECT set of details from the following: 
A. [Co(NH 3) 6]³ ? : Inner orbital complex; d²sp³ hybridized 
B. [MnCl 6]³ ? : Outer orbital complex; sp³d² hybridized 
C. [CoF 6]³ ? : Outer orbital complex; d²sp³ hybridized 
D. [FeF 6]³ ? : Outer orbital complex; sp³d² hybridized 
E. [Ni(CN) 4]² ? : Inner orbital complex; sp³ hybridized 
Choose the correct answer from the options given below: 
A: C & D Only 
B: A, C & E Only