Coordination Compounds
Multiple Choice Questions (MCQs)
1. Which of the following complexes formed by
(a)
(b)
(c)
(d)
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Answer
(b) Greater the value of
For this reaction,
-
(a)
The value of
for this complex is 11.6, which is lower than the value of 27.3 for the complex. A lower value indicates lower stability of the complex. -
(c)
The value of
for this complex is 15.4, which is lower than the value of 27.3 for the complex. A lower value indicates lower stability of the complex. -
(d)
The value of
for this complex is 8.9, which is the lowest among the given options. A lower value indicates lower stability of the complex.
2. The colour of the coordination compounds depends on the crystal field splitting. What will be the correct order of absorption of wavelength of light in the visible region, for the complexes,
(a)
(b)
(c)
(d)
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Thinking Process
This problem is based on the concept of crystal field splitting and colour of coordination compounds. Follow the steps to answer this question
(i) Arrange the given complexes in increasing order of their crystal field splitting energy.
(ii) Now arrange them in decreasing order of the wavelength of light.
(iii) As energy and wavelength are related as
Answer
(c) As we know that, strong field ligand split the five degenerate energy levels with more energy separation than weak field ligand, i.e., as strength of ligand increases crystal field splitting energy increases.
Hence,
As energy separation increases, the wavelength decreases.
Thus, the correct order is
Here, strength of ligand increases,
Hence, correct choice is (c).
-
Option (a)
: This option is incorrect because it suggests that absorbs the longest wavelength of light, which implies it has the smallest crystal field splitting energy ( ). However, CN⁻ is a strong field ligand and should have the largest and thus absorb the shortest wavelength of light. -
Option (b)
: This option is incorrect because it suggests that absorbs a longer wavelength than . However, NH₃ is a stronger field ligand than H₂O, meaning should have a larger and thus absorb a shorter wavelength of light compared to . -
Option (d)
: This option is incorrect for the same reason as option (a). It suggests that absorbs the longest wavelength of light, which is not true because CN⁻ is a strong field ligand and should have the largest and thus absorb the shortest wavelength of light.
3. When
(a)
(b)
(c)
(d)
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Answer
(b) One mole of
So, molecular formula of complex will be
Hence, option (b) is the correct.
-
Option (a)
electrolyte: This option is incorrect because a electrolyte would dissociate into four ions in total, with one cation and three anions. In the given problem, the complex dissociates into one cation and two anions , making it a electrolyte, not . -
Option (c)
electrolyte: This option is incorrect because a electrolyte would dissociate into two ions in total, with one cation and one anion. In the given problem, the complex dissociates into one cation and two anions , making it a electrolyte, not . -
Option (d)
electrolyte: This option is incorrect because a electrolyte would dissociate into four ions in total, with three cations and one anion. In the given problem, the complex dissociates into one cation and two anions , making it a electrolyte, not .
4. When 1 mole of
(a)
(b)
(c)
(d)
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Answer
(d) 1 mole of
Here, 3 moles of
So, the formula of the complex can be
-
(a)
: This formula suggests that all three chloride ions are coordinated to the chromium ion, leaving no free chloride ions to react with . Therefore, it would not result in the precipitation of 3 moles of . -
(b)
: This formula indicates that only one chloride ion is free (outside the coordination sphere) and can react with . This would result in the precipitation of only 1 mole of , not 3 moles. -
(c)
: This formula implies that two chloride ions are free and can react with . This would result in the precipitation of only 2 moles of , not 3 moles.
5. The correct IUPAC name of
(a) Diamminedichloridoplatinum (II)
(b) Diamminedichloridoplatinum (IV)
(c) Diamminedichloridoplatinum (0)
(d) Dichloridodiammineplatinum (IV)
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Thinking Process
This problem is based on IUPAC nomenclature of coordination compound. IUPAC nomenclature of any coordination compound can be done as follows
(i) Positively charged ions are named first.
(ii) Negatively charged ions are named in alphabetical order of ligands including their numbers followed by metal ending with -ium and oxidation state in the bracket.
Answer
(a) The complex compound is
The ligands present in the compound are
(i)
(ii)
The oxidation number of platinum in the compound is 2 . Hence, correct IUPAC name of
Diammine dichloridoplatinum (II)
So, (a) option is correct.
-
Option (b): Diamminedichloridoplatinum (IV) is incorrect because the oxidation state of platinum in the given complex
is +2, not +4. The correct oxidation state is determined by the charges of the ligands and the overall neutrality of the complex. -
Option (c): Diamminedichloridoplatinum (0) is incorrect because the oxidation state of platinum in the given complex
is +2, not 0. The correct oxidation state is determined by the charges of the ligands and the overall neutrality of the complex. -
Option (d): Dichloridodiammineplatinum (IV) is incorrect because the oxidation state of platinum in the given complex
is +2, not +4. Additionally, the correct naming convention places the ammine ligands before the chlorido ligands, making “diammine” come before “dichlorido” in the name.
6. The stabilisation of coordination compounds due to chelation is called the chelate effect. Which of the following is the most stable complex species?
(a)
(b)
(c)
(d)
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Answer
(c) Chelation (formation of cycle by linkage between metal ion and ligand) stabilises the coordination compound. The ligand which chelates the metal ion are known as chelating ligand.
Here, only
-
(a)
: This complex does not involve chelation. The carbonyl (CO) ligands are monodentate, meaning they bind to the metal ion through a single donor atom, which does not provide the stabilizing chelate effect. -
(b)
: This complex also does not involve chelation. The cyanide (CN) ligands are monodentate, binding to the metal ion through a single donor atom, and thus do not provide the stabilizing chelate effect. -
(d)
: This complex does not involve chelation either. The water (H₂O) molecules are monodentate ligands, binding to the metal ion through a single donor atom, and do not provide the stabilizing chelate effect.
7. Indicate the complex ion which shows geometrical isomerism.
(a)
(b)
(c)
(d)
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Thinking Process
This problem includes concept of isomerism in coordination compound. Complex of
Answer
(a)
Hence, the possible geometrical isomers are
Hence, correct choice is (a).
-
(b)
: This complex does not show geometrical isomerism because it is a type coordination compound, which does not have the necessary arrangement of ligands to form different geometrical isomers. -
(c)
: This complex does not show geometrical isomerism because it is a type coordination compound, where all six ligands are identical, making it impossible to form different geometrical isomers. -
(d)
: This complex does not show geometrical isomerism because it is a type coordination compound, where the arrangement of ligands does not allow for the formation of different geometrical isomers.
8. The CFSE for octahedral
(a)
(b)
(c)
(d)
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Answer
(c) CFSE for octahedral and tetrahedral complexes are closely related to each other by formula
where,
Hence, correct choice is (c).
-
Option (a)
: This option is incorrect because it assumes that the CFSE for the tetrahedral complex is the same as that for the octahedral complex. However, the CFSE for a tetrahedral complex is generally less than that for an octahedral complex and is related by the formula . -
Option (b)
: This option is incorrect because it does not follow the correct relationship between the CFSE of octahedral and tetrahedral complexes. According to the formula , the CFSE for the tetrahedral complex should be significantly less than , and is too high. -
Option (d)
: This option is incorrect because it exceeds the CFSE of the octahedral complex, which is . The CFSE for a tetrahedral complex should be less than that of an octahedral complex, as given by the formula .
9. Due to the presence of ambidentate ligands coordination compounds show isomerism. Palladium complexes of the type
(a) linkage isomers
(b) coordination isomers
(c) ionisation isomers
(d) geometrical isomers
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Answer
(a) The ligand(s) which has two different bonding sites are known as ambident ligands e.g.,
Here, NCS has two binding sites at N and S.
Hence, NCS (thiocyanate) can bind to the metal ion in two ways
Thus, coordination compounds containing NCS as a ligand can show linkage isomerism i.e.,
Hence, correct choice is (a).
-
Coordination isomers: Coordination isomers occur when there is an interchange of ligands between the cationic and anionic parts of a coordination compound. In this case, both compounds have the same ligands and the same metal center, so they cannot be coordination isomers.
-
Ionisation isomers: Ionisation isomers occur when a compound can produce different ions in solution. Since both compounds have the same ligands and metal center, they will produce the same ions in solution, so they cannot be ionisation isomers.
-
Geometrical isomers: Geometrical isomers occur due to different spatial arrangements of ligands around the central metal atom. In this case, the isomerism is due to the different binding sites of the ambidentate ligand (NCS), not due to different spatial arrangements of the same ligands. Therefore, they cannot be geometrical isomers.
10. The compounds
(a) linkage isomerism
(b) ionisation isomerism
(c) coordination isomerism
(d) no isomerism
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Answer
(d) Compounds having same molecular formula but different structural formula are known as isomers.
-
(a) Linkage isomerism: This type of isomerism occurs when a ligand can coordinate to the metal through two different atoms. In the given compounds, the sulfate ligand (
) cannot coordinate through different atoms, so linkage isomerism is not possible. -
(b) Ionisation isomerism: This type of isomerism occurs when two compounds have the same composition but yield different ions in solution. The given compounds do not produce different ions in solution; they simply have different counterions (Br and Cl), so ionisation isomerism is not applicable.
-
(c) Coordination isomerism: This type of isomerism occurs when there is an interchange of ligands between the cationic and anionic entities of different coordination compounds. The given compounds do not involve such an interchange of ligands, so coordination isomerism is not relevant.
11. A chelating agent has two or more than two donor atoms to bind to a single metal ion. Which of the following is not a chelating agent?
(a) Thiosulphato
(b) Oxalato
(c) Glycinato
(d) Ethane-1, 2-diamine
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Answer
(a) A chelating ligand has two or more binding donor atoms to a single metal ion e.g.,

Here
thiosulphato

-
Oxalato (b): Oxalato (C₂O₄²⁻) is a chelating agent because it has two oxygen atoms that can donate electron pairs to bind to a single metal ion, forming a five-membered ring.
-
Glycinato (c): Glycinato (NH₂CH₂COO⁻) is a chelating agent because it has two donor atoms: the nitrogen atom in the amine group and the oxygen atom in the carboxylate group, which can bind to a single metal ion.
-
Ethane-1,2-diamine (d): Ethane-1,2-diamine (NH₂CH₂CH₂NH₂) is a chelating agent because it has two nitrogen atoms that can donate electron pairs to bind to a single metal ion, forming a five-membered ring.
12. Which of the following species is not expected to be a ligand?
(a)
(b)
(c)
(d)
Show Answer
Answer
(b) Ligand must donate a pair of electron or loosely held electron pair to metal and form a
e.g.
Among
Hence
-
(a) NO: Nitric oxide (
) can act as a ligand because it has an unpaired electron and can donate this electron to form a bond with a metal center. -
(c) NH₂CH₂CH₂NH₂: Ethylenediamine (
) is a bidentate ligand, meaning it has two nitrogen atoms each with a lone pair of electrons that can be donated to a metal center to form coordinate bonds. -
(d) CO: Carbon monoxide (
) is a strong ligand because the carbon atom has a lone pair of electrons that can be donated to a metal center, forming a coordinate bond.
13. What kind of isomerism exists between
(a) Linkage isomerism
(b) Solvate isomerism
(c) lonisation isomerism
(d) Coordination isomerism
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Answer
(c) Soluate isomerism to shown when two compounds having same molecular formula differ by whether or solvent molecule is directly bonded to metal ion or is present as free solvent molecules in the crystal lattice.
When water is present as solvent and show this type of isomerism then it is known as hydrate isomerism.
Coordination compound
-
Linkage isomerism: This type of isomerism occurs when a ligand can coordinate to the metal ion through two different atoms. For example, the ligand NO₂⁻ can bind through the nitrogen or the oxygen atom. In the given compounds, there is no such ligand that can bind through different atoms, so linkage isomerism is not applicable.
-
Ionisation isomerism: This type of isomerism occurs when two compounds have the same composition but yield different ions in solution. For example, [Co(NH₃)₅Br]SO₄ and [Co(NH₃)₅SO₄]Br are ionisation isomers. In the given compounds, the ions produced in solution are not different, so ionisation isomerism is not applicable.
-
Coordination isomerism: This type of isomerism occurs when there is an interchange of ligands between the cationic and anionic entities of different metal ions in a complex. For example, [Co(NH₃)₆][Cr(CN)₆] and [Cr(NH₃)₆][Co(CN)₆] are coordination isomers. In the given compounds, there is no such interchange of ligands between different metal ions, so coordination isomerism is not applicable.
14. IUPAC name of
(a) Platinum diaminechloronitrite
(b) Chloronitrito-N-ammineplatinum (II)
(c) Diamminechloridonitrito-N-platinum (II)
(d) Diamminechloronitrito-N-platinate (II)
Show Answer
Answer
(c) Correct IUPAC name can be written as The ligands present in the given coordination compound are
(i)
(ii)
(iii)
According to IUPAC rule, ligands are named in an alphabetical order before central atom. Prefex di-will be used to indicate the number of
Oxidation state of metal is indicated by Roman numeral in parenthesis.
So, IUPAC name will be
diamminechloronitrito-N-platinum (II)
Hence, option (c) is correct.
-
Option (a): “Platinum diaminechloronitrite” is incorrect because:
- The term “diamine” should be “diammine” to correctly represent the two ammonia (
) ligands. - The term “nitrite” should be “nitrito-N” to specify the binding through the nitrogen atom.
- The oxidation state of the metal (II) is not indicated.
- The term “diamine” should be “diammine” to correctly represent the two ammonia (
-
Option (b): “Chloronitrito-N-ammineplatinum (II)” is incorrect because:
- The ligands are not listed in alphabetical order. “Ammine” should come before “chloronitrito-N”.
- The prefix “di-” is missing to indicate the presence of two ammonia (
) ligands.
-
Option (d): “Diamminechloronitrito-N-platinate (II)” is incorrect because:
- The suffix “platinate” is used for anionic complexes, but the given complex is neutral. The correct term should be “platinum”.
Multiple Choice Questions (More Than One Options)
15. Atomic number of Mn. Fe and Co are 25,26 and 27 respectively. Which of the following inner orbital octahedral complex ions are diamagnetic?
(a)
(b)
(c)
(d)
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Thinking Process
This problem is based on magnetic property of coordination compound. Coordination compound containing at least one unpaired electron(s) are paramagnetic and coordination compounds all containing all paired electrons are diamagnetic in nature.
Answer
Molecular orbital electronic configuration of
Number of unpaired electron
Magnetic property = Diamagnetic
Molecular orbital electronic configuration of
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Number of unpaired electrons
Magnetic property
Molecular orbital electronic configuration of

Number of unpaired electron
Magnetic property
Molecular orbital electronic configuration of

Number of unpaired electron
Magnetic property
Thus,
Hence, correct choices are options (a) and (c).
-
Option (b)
:- The molecular orbital electronic configuration of
in shows that it has 2 unpaired electrons. - Magnetic property: Paramagnetic (due to the presence of unpaired electrons).
- The molecular orbital electronic configuration of
-
Option (d)
:- The molecular orbital electronic configuration of
in shows that it has 1 unpaired electron. - Magnetic property: Paramagnetic (due to the presence of unpaired electrons).
- The molecular orbital electronic configuration of
16. Atomic number of
(a)
(b)
(c)
(d)
Show Answer
Answer
(a, c)
Molecular orbital electronic configuration of

Number of unpaired electrons
Magnetic property
Molecular orbital electronic configuration of

Number of unpaired electrons
Magnetic property
Molecular orbital electronic configuration of

Number of unpaired electrons
Magnetic property
Molecular orbital electronic configuration of

Number of unpaired electrons
Magnetic property
Thus,
Hence, correct choices are (a) and (c).
-
Option (b)
:- The molecular orbital electronic configuration of
in shows that it has 5 unpaired electrons. This is different from the 4 unpaired electrons found in and .
- The molecular orbital electronic configuration of
-
Option (d)
:- The molecular orbital electronic configuration of
in shows that it has 2 unpaired electrons. This is different from the 4 unpaired electrons found in and .
- The molecular orbital electronic configuration of
17. Which of the following options are correct for
(a)
(b)
(c) Paramagnetic
(d) Diamagnetic
Show Answer
Answer
According to VBT, the molecular orbital electronic configuration of

Hybridisation
Number of unpaired electron
Magnetic property
Hence, correct choices are options (a) and (c).
-
Option (b)
hybridisation is incorrect because the correct hybridisation for is , not . This is due to the involvement of inner d-orbitals in the hybridisation process, which leads to the formation of an octahedral complex. -
Option (d) Diamagnetic is incorrect because
has one unpaired electron, making it paramagnetic, not diamagnetic. A diamagnetic substance would have all electrons paired, resulting in no net magnetic moment.
18. An aqueous pink solution of cobalt(II) chloride changes to deep blue on addition of excess of
(a)
(b)
(c) tetrahedral complexes have smaller crystal field splitting than octahedral complexes
(d) tetrahedral complexes have larger crystal field splitting than octahedral complex
Show Answer
Answer
Aqueous pink solution of cobalt (II) chloride is due to electronic transition of electron from
(i)
(ii) Tetrahedral complexes have smaller crystal field splitting than octahedral complexes because
Hence, options (b) and (c) are correct choices.
- Option (a) is incorrect because the transformation that occurs is from
to , not to . - Option (d) is incorrect because tetrahedral complexes have smaller crystal field splitting than octahedral complexes, not larger.
19. Which of the following complexes are homoleptic?
(a)
(b)
(c)
(d)
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Answer
(a, c)
Homoleptic complex The complex containing only one species or group as ligand is known as homoleptic ligand.
e.g.,
Here,
While other two complexes
Hence, options (a) and (c) are correct choices.
-
Option (b)
: This complex is not homoleptic because it contains two different types of ligands, and . -
Option (d)
: This complex is not homoleptic because it also contains two different types of ligands, and .
20. Which of the following complexes are heteroleptic?
(a)
(b)
(c)
(d)
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Answer
(
Heteroleptic complexes Coordination complexes which contain more than one type of ligands are known as heteroleptic complexes.
e.g.,
Hence, optons (b) and (d) are correct choices.
-
(a)
: This complex is homoleptic because it contains only one type of ligand, . -
(c)
: This complex is homoleptic because it contains only one type of ligand, .
21. Identify the optically active compounds from the following
(a)
(b) trans
(c) cis
(d)
Show Answer
Answer
(a, c)

Hence, (a) and (c) are correct choices.
-
(b) trans
: This compound is not optically active because it has a plane of symmetry. The trans configuration allows for the mirror image to be superimposable on the original molecule, thus making it optically inactive. -
(d)
: This compound is not optically active because it does not have any chiral centers. The arrangement of the ligands around the central metal ion is such that the mirror image can be superimposed on the original molecule, making it optically inactive.
22. Identify the correct statements for the behaviour of ethane-1, 2diamine as a ligand.
(a) It is a neutral ligand
(b) It is a didentate ligand
(c) It is a chelating ligand
(d) It is a unidentate ligand
Show Answer
Answer
Molecular formula of ethane-1, 2-diamine is
(a) Ethane-1, 2-diamine is a neutral ligand due to absence of any charge.
(b) It is a didentate ligand due to presence of two donor sites one at each nitrogen atom of amino group.
(c) It is a chelating, ligand due to its ability to chelate with the metal.
Hence, options (a), (b) and (c) are correct choices.
- Option (d) is incorrect because ethane-1, 2-diamine is not a unidentate ligand. A unidentate ligand has only one donor site that can bind to a metal atom, whereas ethane-1, 2-diamine has two donor sites (one at each nitrogen atom of the amino groups), making it a didentate ligand.
23. Which of the following complexes show linkage isomerism?
(a)
(b)
(c)
(d)
Show Answer
Answer
Coordination compounds containing a ligand with more than one non-equivalent binding position (known as ambident ligand) show linkage isomerism.
e.g.,
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Hence,
Hence, options (a) and (c) are correct choices.
-
Option (b):
does not show linkage isomerism because it does not contain an ambident ligand. The ligands present (water and carbon monoxide) do not have multiple non-equivalent binding positions. -
Option (d):
does not show linkage isomerism because it does not contain an ambident ligand. The ligands present (ethylenediamine and chloride) do not have multiple non-equivalent binding positions.
Short Answer Type Questions
24. Arrange the following complexes in the increasing order of conductivity of their solution
Show Answer
Thinking Process
This problem is based on the concept of conductivity of coordination compound. Greater the number of ions, greater the conductivity of coordination compound.
Answer
Ions or molecules present outside the coordination sphere are ionisable. A complex which gives more ions on dissolution, is more conducting.
Here, number of ions increases and conductivity increases.
25. A coordination compound
Show Answer
Answer
Formation of white precipitate with
Thus, the formula of the complex is
26. A complex of the type
Show Answer
Answer
An optically active complex of the type
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Non-superimposable isomers of
27. Magnetic moment of
Show Answer
Answer
The magnetic moment
28. 0n the basis of crystal field theory explain why Co(III) forms paramagnetic octahedral complex with weak field ligands whereas it forms diamagnetic octahedral complex with strong field ligands.
Show Answer
Answer
With weak field ligands;

With strong field ligands,
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29. Why are low spin tetrahedral complexes not formed?
Show Answer
Answer
In tetrahedral complex, the
Hence, the orbital splitting energies are not enough to force pairing. As a result, low spin configurations are rarely observed in tetrahedral complexes.
30. Give the electronic configuration of the following complexes on the basis of crystal field splitting theory.
Show Answer
Answer
According to spectrochemical series, ligands can be arranged in a series in the order of increasing field strength i.e.,
Hence,
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
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31. Explain why
Show Answer
Answer
As we know, where,
[Ar]
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One unpaired electron
For

Five unpaired electron
Hence,
32. Arrange following complex ions in increasing order of crystal field splitting energy
Show Answer
Answer
CFSE is higher when the complex contains strong field ligand. Thus, crystal field splitting energy increases in the order
Because according to spectrochemical series the order of field strength is
33. Why do compounds having similar geometry have different magnetic moment?
Show Answer
Answer
It is due to the presence of weak and strong field ligands in complexes. If CFSE is high, the complex will show low value of magnetic moment and vice-versa, e.g.
Q.34
Show Answer
Answer
In
35. Name the type of isomerism when ambidentate ligands are attached to central metal ion. Give two examples of ambidentate ligands.
Show Answer
Answer
Ligand having more than one different binding position are known as ambidentate ligand. e.g., SCN has two different binding positions S and N. Coordination compound containing ambidentate ligands are considered to show linkage isomerism due to presence of two different binding positions.
e.g., (i)
(ii)
Matching The Columns
36. Match the complex ions given in Column I with the colours given in Column II and assign the correct code.
Column I (Complex ion) |
Column II (Colour) |
||
---|---|---|---|
A. | 1. | Violet | |
B. | 2. | Green | |
C. | 3. | Pale blue | |
D. | 4. | Yellowish orange |
Codes
A | B | C | D | |
---|---|---|---|---|
(a) | 1 | 2 | 4 | 5 |
(b) | 4 | 3 | 2 | 1 |
(c) | 3 | 2 | 4 | 1 |
(d) | 4 | 1 | 2 | 3 |
Show Answer
Answer
A.
B.
C.
D.
Colour of coordination compound is closely related to CFSE of coordination compound. Depending upon the CFSE of given coordination compounds. Correct matching will be as follows
Column I (Complex ion) |
Column II (Colour) |
||
---|---|---|---|
A. | 4. | Yellowish orange | |
B. | 3. | Pale blue | |
C. | 2. | Green | |
D. | 1. | Violet |
Hence, correct choice is (b).
37. Match the coordination compounds given in Column I with the central metal atoms given in Column II and assign the correct code.
Column I (Coordination compound) |
Column II (Central metal atom) |
||
---|---|---|---|
A. | Chlorophyll | 1. | Rhodium |
B. | Blood pigment | 2. | Cobalt |
C. | Wilkinson catalyst | 3. | Magnesium |
D. | Vitamin B 12 | 4. | Iron |
Codes
A | B | C | D | |
---|---|---|---|---|
(a) | 3 | 4 | 1 | 2 |
(b) | 3 | 4 | 5 | 1 |
(c) | 4 | 3 | 2 | 1 |
(d) | 3 | 4 | 1 | 2 |
Show Answer
Answer
A.
B.
C.
D.
Central metal ions present on coordination compounds determine the properties of coordination compound and their biological role.
Column I (Coordination compound) |
Column II (Central metal atom) |
|||
---|---|---|---|---|
A. | Chlorophyll | 3. | Magnesium | |
B. | Blood pigment | 4. | Iron | |
C. | Wilkinson catalyst | 1. | Rhodium | |
D. | Vitamin B 12 | 2. | Cobalt |
Hence, correct choice is (a).
38. Match the complex ions given in Column I with the hybridisation and number of unpaired electrons given in Column II and assign the correct code.
Column I (Complex ion) |
Column II (Hybridisation, number of unpaired electrons) |
||
---|---|---|---|
A. | 1. | ||
B. | 2. | ||
C. | 3. | ||
D. | 4. |
Codes
A | B | C | D | A | B | C | D | ||
---|---|---|---|---|---|---|---|---|---|
(a) | 3 | 1 | 4 | 2 | (b) | 4 | 3 | 2 | 1 |
(c) | 3 | 2 | 4 | 1 | (d) | 4 | 1 | 2 | 3 |
Show Answer
Answer
A.
B.
C.
D.
Formation of inner orbital complex and outer orbital complex determines hybridisation of molecule which inturn depends upon field strength of ligand and number of vacant
(i) Strong field ligand forms inner orbital complex with hybridisation
(ii) Weak field ligand forms outer orbital complex with hybridisation
According to VBT, hybridisation and number of unpaired electrons of coordination compounds can be calculated as
A.
MOEC (Molecular orbital electronic configuration) of
Hybridisation
B.
MOEC of
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Hybridisation
C.
MOEC of
Hybridisation
D.
MOEC of
Hybridisation
Hence, correct choice can be represented by (a).
39. Match the complex species given in Column I with the possible isomerism given in Column II and assign the correct code.
Column I (Complex species) |
Column II (Isomerism) |
||
---|---|---|---|
A. | 1. | Optical | |
B. | cis |
2. | Ionisation |
C. | 3. | Coordination | |
D. | 4. | Geometrical |
Codes
A | B | C | D | |
---|---|---|---|---|
(a) | 1 | 2 | 4 | 3 |
(b) | 4 | 3 | 2 | 1 |
(c) | 4 | 2 | 1 | 3 |
(d) | 4 | 1 | 2 | 3 |
Show Answer
Answer
A.
B.
C.
D.
Isomerism in coordination compound is decided by type of ligands and geometry of coordination and arrangement of ligands.
A.
B. cis
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C.
D.
Hence, correct choice is (d).
40. Match the compounds given in Column I with the oxidation state of cobalt present in it (given in column II) and assign the correct code.
Column I |
Column II (Oxidation state of Co) |
||
---|---|---|---|
A. | 1. | +4 | |
B. | 2. | 0 | |
C. | 3. | +2 | |
D. | 4. | +3 |
Code
A | B | C | D | |
---|---|---|---|---|
(a) | 1 | 2 | 4 | 3 |
(b) | 4 | 3 | 2 | 1 |
(c) | 3 | 1 | 4 | 2 |
(d) | 4 | 1 | 3 | 2 |
Show Answer
Answer
A.
B.
C.
D.
Oxidation state of
A.
Let oxidation state of
B.
Let oxidation state of
C.
Let oxidation state of
D.
Let oxidation state of
Hence, correct choice is (d).
Assertion and Reason
In the following questions a statement of assertion (A) followed by a statement of reason
(a) Assertion and reason both are true, reason is correct explanation of assertion.
(b) Assertion and reason both are true but reason is not the correct explanation of assertion.
(c) Assertion is true, reason is false.
(d) Assertion is false, reason is true.
41. Assertion (A) Toxic metal ions are removed by the chelating ligands.
Reason (
Show Answer
Answer
(a) Assertion and reason both are correct and reason is the correct explanation of assertion.
Toxic metal ions are removed by chelating ligands. When a solution of chelating ligand is added to solution containing toxic metals ligands chelates the metal ions by formation of stable complex.
42. Assertion (A)
Show Answer
Answer
(b) Assertion and reason both are true but reason is not correct explanation of assertion. Correct reason is
43. Assertion (A) Linkage isomerism arises in coordination compounds containing ambidentate ligand.
Reason (R) Ambidentate ligand has two different donor atoms.
Show Answer
Answer
(a) Assertion and reason both are correct and reason is correct explanation of assertion. Linkage isomerism arises in coordination compounds containing ambidentate ligands because ambidentate ligand has two different donor atoms.
e.g.,
44. Assertion (A) Complexes of
Reason (R) Geometrical isomerism is not shown by complexes of coordination number 6.
Show Answer
Answer
(b) Assertion and reason both are correct and reason is not correct explanation of assertion.
Complexes of

45. Assertion (A)
Reason (
Show Answer
Answer
(d) Assertion is false but reason is true.
According to VBT, MOEC of

Hybridisation
Hence, correct assertion is
i.e.,
Long Answer Type Questions
46. Using crystal field theory, draw energy level diagram, write electronic configuration of the central metal atom/ion and determine the magnetic moment value in the following
(a)
(b)
Show Answer
Thinking Process
This problem includes conceptual mixing of crystal field theory and magnetic moment
Answer
(a)

Configuration of
Number of unpaired electrons
Magnetic moment

Configuration of
Number of unpaired electrons

There is no unpaired electron, so it is diamagnetic.
(b)

Number of unpaired electrons,

Number of unpaired electrons,

Since,
Because there is no unpaired electron, so it is diamagnetic in nature.
47. Using valence bond theory, explain the following in relation to the complexes given below
(a) Type of hybridisation
(b) Inner or outer orbital complex
(c) Magnetic behaviour
(d) Spin only magnetic moment value.
Show Answer
Answer
(a)

(i)
(ii) Inner orbital complex because
(iii) Paramagnetic, as two unpaired electrons are present.
(iv) Spin only magnetic moment
(b)
(i)
(ii) Inner orbital complex because of the involvement of
(iii) Diamagnetic, as no unpaired electron is present.
(iv)
(c)

(i)
(ii) Inner orbital complex (as
(iii) Paramagnetic (as three unpaired electrons are present.)
(iv)
(d)

(i)
(ii) Outer orbital complex because
(iii) Paramagnetic (because of the presence of four unpaired electrons).
(iv)
Q.48
(a) Identify ’
(b) Name the type of isomerism involved.
(c) Give the IUPAC name of ’
Show Answer
Thinking Process
This problem is based on chemical properties of coordination compounds, ionisation isomerism, and nomenclature of coordination compounds.
Answer
’
’
(a)
(b) Ionisation isomerism (as give different ions when subjected to ionisation.)
(c)
[B], Pentaamminechloridocobalt (III) sulphate.
49. What is the relationship between observed colour of the complex and the wavelength of light absorbed by the complex?
Show Answer
Answer
When white light falls on the complex, some part of it is absorbed. Higher the crystal field splitting energy, lower will be the wavelength absorbed by the complex. The observed colour of complex is the colour generated from the wavelength left over.
e.g., if green light is absorbed, the complex appears red.
In terms of crystal field theory, suppose there is an octahedral complex with empty
In absence of ligand, crystal field splitting does not occur and the substance is colourless.
e.g., anhydrous
50. Why are different colours observed in octahedral and tetrahedral complexes for the same metal and same ligands?
Show Answer
Answer
Extent of splitting of
where,
Wavelength of light and CFSE are related to each other by formula
So, higher wavelength of light is absorbed in octahedral complexes than tetrahedral complexes for same metal and ligands. Thus, different colours are observed.