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Coordination Compounds Chemistry - Coordination Compounds

Gujarat Board (GSEB) - STD 12 Science - Chemistry (GSEB - English Medium). 349 questions in this chapter.

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Coordination Compounds question types

349 questions across 6 question groups — pick any mix to generate a Chemistry paper with step-by-step answer keys.

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Sample Questions

Coordination Compounds questions

One sample from each question group in this chapter. Select any group above to see the full set with answer keys.

Q 13 Marks Question3 Marks
  1. Using valence bond theory, predict the shape and magnetic character of

[Ni(CO)4]. [Ni = 28].

  1. Give one example of application of coordination compounds-in medicine.
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Q 23 Marks Question3 Marks
  1. For the complex [Fe(H2O)6]3+, write the hybridization, magnetic character and spin of the complex. (At. number: Fe = 26)
  2. Draw one of the geometrical isomers of the complex [Pt(en)2Cl2]2+ which is optically inactive.
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Q 33 Marks Question3 Marks
  1.  What type of isomerism is shown by [Co(NH3)5ONO]Cl2?
  2. On the basis of crystal field theory, write the electronic configuration for d4 ion if $\Delta$ o< P.
  3. Write the hybridization and shape of [Fe(CN)6]3-.

(Atomic number of Fe = 26)

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Q 43 Marks Question3 Marks
  1. What type of isomerism is shown by the complex [Co(NH3)5 (SCN)]2+?
  2. Why is [NiCl4]2- paramagnetic while [Ni(CN)4]2- is diamagnetic? (Atomic number of Ni = 28)
  3. Why are low spin tetrahedral complexes rarely observed?
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Q 53 Marks Question3 Marks
  1. Write the IUPAC name of the complex [Cr(NH3)4 Cl2]Cl.
  2. What type of isomerism is exhibited by the complex [Co(en)3]3+?

(en= ethane-1,2-diamine)

  1. Why is [NiCl4]2–paramagnetic but [Ni(CO)4] is diamagnetic?

(At. nos.: Cr= 24, Co=27, Ni= 28)

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Q 11Case study (4 Marks)4 Marks

Read the passage given below and answer the following questions:

The molecular compounds which are formed from the combination of two or more simple stable compounds and retain their identity in the solid as well as in the dissolved state are called coordination compounds. Their properties are completely different from the constituents. ln coordination compounds, the central metal atom or ion is linked to a number ofions or neutral molecules, called ligands, by coordinate bonds. For example, Dimethylglyoxime (dmg) is a bid en date ligand chelating large amounts of metals. When dimethyl glyoxime is added to alcoholic solution of NiCl2 and ammonium hydroxide is slowly added to it, a rosy red precipitate of a complex is formed.

The following questions are multiple choice questions. Choose the most appropriate answer:

  1. The structure of the complex is:

  1. Oxidation number of Ni in the given complex is:
  1. +3
  2. +1
  3. +2
  4. Zero
  1. Hybridisation and structure of the complex is:
  1. Sp3, tetrahedral.
  2. dsp2, square planar.
  3. Sp3, square planar.
  4. Sp3d, trigonal bipyramidal.
  1. Which of the following is true about this complex?
  1. It is paramagnetic, containing 2 unpaired electrons.
  2. It is paramagnetic, containing 1 unpaired electron.
  3. It is paramagnetic, containing 4 unpaired electrons.
  4. It is diamagnetic with no unpaired electron.
  1. Which one will give test for Fe3+ ions in the solution?
  1. [Fe(CN)6]3-
  2. [Fe(CN)6]2-
  3. (NH4)2SO4·FeSO4·6H2O
  4. Fe2(S04)3
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Q 12Case study (4 Marks)4 Marks
Read the passage given below and answer the following questions:

Arrangement of ligands in order of their ability to cause splitting $(\Delta)$ is called spectrochemical series. Ligands which cause large splitting (large $\Delta$) are called strong field ligands while those which cause small splitting (small $\Delta$) are called weak field ligands. When strong field ligands approach metal atom/ ion, the value of $\Delta_0$ is large, so that electrons are forced to get paired up in lower energy t2g orbitals. Hence, a low-spin complex is resulted from strong field ligand. When weak field ligands approach metal atom/ ion, the value of $\Delta_0$ is small, so that electrons enter high energy eg orbitals rather than pairing in low energy t2g orbitals. Hence, a high-spin complex is resulted from weak field ligands. Strong field ligands have tendency to form inner orbital complexes by forcing the electrons to pair up. Whereas weak field ligands have tendency to form outer orbital complex because inner electrons generally do not pair up.

In these questions (Q. No. i-iv), a statement of assertion followed by a statement ofreason is given. Choose the correct answer out of the following choices.

  1. Assertion and reason both are correct statements and reason is correct explanation for assertion.
  2. Assertion and reason both are correct statements but reason is not correct explanation for assertion.
  3. Assertion is correct statement but reason is wrong statement.
  4. Assertion is wrong statement but reason is correct statement.
The following questions are multiple choice questions. Choose the most appropriate answer:

  1. Assertion: In tetrahedral coordination entity formation, the d orbital splitting is inverted and is smaller as compared to the octahedral field splitting.

Reason: Spectrochemical series is based on the absorption of light by complexes with different ligands.

  1. Assertion: In high spin situation, configuration of d5 ions will be $\text{t}^3_{2\text{g}}\text{e}^2_\text{g}.$

Reason: In high spin situation, pairing energy is less than crystal field energy.

  1. Assertion: F- ion is a weak field ligand and fonns outer orbital complex.

Reason: F- ion cannot force the electrons of dz2 and dx2-y2 orbitals of the inner shell to occupy dxy, dyz and dzx orbitals of the same shell.

  1. Assertion: The crystal field model is successful in explaining the formation, structures, colour and magnetic properties of coordination compounds.

Reason: In spectrochemical series, ligands are arranged in a series of increasing field strength.

  1. Assertion: NF3 is a weaker ligand than N(CH3)3.

Reason: NF3 ionizes to give F- ions in aqueous solution.

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Q 13Case study (4 Marks)4 Marks
Read the passage given below and answer the following questions:
Coordination compounds are formulated and named according to the IUPAC system.
Few rules for naming coordination compounds are:
  1. In ionic complex, the cation is named first and then the anion.
  2. In the coordination entity, the ligands are named first and then the central metal ion.
  3. When more than one type of ligands are present, they are named in alphabetical order of preference without any consideration of charge.
The following questions are multiple choice questions. Choose the most appropriate answer:
  1. The IUPAC name of the complex [Pt(NH3)3Br(NO2)Cl]Cl is:
  1. Triamminechlorobromonitroplatinum (IV) chloride.
  2. Triamminebromonitrochloroplatinum (IV) chloride.
  3. Triamminebromidochloridonitroplatinum (IV) chloride.
  4. Triamminenitrochlorobromoplatinum (IV) chloride.
  1. The IUPAC name of [Ni(CO)4] is:
  1. Tetracarbonylnickel (II).
  2. Tetracarbonylnickel (0).
  3. Tetracarbonylnickelate (II).
  4. Tetracarbonylnickelate (0).
  1. As per IUPAC nomenclature, the name of the complex [Co(H2O)4(NH3)2]Cl3 is:
  1. Tetraaquadiamminecobalt (II) chloride.
  2. Tetraaquadiamminecobalt (III) chloride.
  3. Diamminetetraaquacobalt (II) chloride.
  4. Diamminetetraaquacobalt (III) chloride.
  1. Which of the following represents correct formula of dichloridobis(ethane -1, 2-diamine)cobalt (III) ion?
  1. [CoCl2(en)]2+
  2. [CoCl2(en)2]2+
  3. [CoCl2(en)]+
  4. [CoCl2(en)2]+
  1. Correct formula of pentaamminenitro-O-cobalt (III) sulphate is:
  1. [Co(NO2)(NH3)5]SO4
  2. [Co(ONO)(NH3)5]SO4
  3. [Co(NO2)(NH3)4](SO4)2
  4. [Co(ONO)(NH3)4](SO4)2
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Q 14Case study (4 Marks)4 Marks
Read the passage given below and answer the following questions:
Valence bond theory considers the bonding between the metal ion and the ligands as purely covalent. On the other hand, crystal field theory considers the metal-ligand bond to be ionic arising from electrostatic interaction between the metal ion and the ligands. In coordination compounds, the interaction between the ligand and the metal ion causes the five d-orbitals to split-up. This is called crystal field splitting and the energy difference between the two sets of energy level is called crystal field splitting energy. The crystal field splitting energy $(\Delta_0)$ depends upon the nature of the ligand. The actual configuration of complexes is divided by the relative values of $\Delta_0$ and P (pairing energy).
If $\Delta_0<\text{P},$ then complex will be high spin.
If $\Delta_0>\text{P},$ then complex will be low spin
The following questions are multiple choice questions. Choose the most appropriate answer:
  1. Which of the following ligand has lowest $\Delta_0$ value?
  1. CN-
  2. CO
  3. F-
  4. NH3
  1. The crystal field splitting energy for octahedral $(\Delta_0)$ and tetrahedral $(\Delta_t)$ complex is related as:
  1. $\Delta_\text{t}=\frac{1}{2}\Delta_0$
  2. $\Delta_\text{t}=\frac{4}{9}\Delta_0$
  3. $\Delta_\text{t}=\frac{3}{5}\Delta_0$
  4. $\Delta_\text{t}=\frac{2}{5}\Delta_0$
  1. On the basis of crystal field theory, the electronic configuration of d4 in two situations : (i) t.0 > P and (ii) t.0
  (i) (ii)
(a) $\text{t}^4_{2\text{g}}\text{e}^0_\text{g}$ $\text{t}^3_{2\text{g}}\text{e}^1_\text{g}$
(b) $\text{t}^3_{2\text{g}}\text{e}^1_\text{g}$ $\text{t}^4_{2\text{g}}\text{e}^0_\text{g}$
(c) $\text{t}^3_{2\text{g}}\text{e}^1_\text{g}$ $\text{t}^3_{2\text{g}}\text{e}^1_\text{g}$
(d) $\text{t}^4_{2\text{g}}\text{e}^0_\text{g}$ $\text{t}^4_{2\text{g}}\text{e}^0_\text{g}$
  1. Using crystal field theory, calculate magnetic moment of central metal ion of [FeF6]4-.
  1. 1.79B.M.
  2. 2.83B.M.
  3. 3.85B.M.
  4. 4.9B.M.
  1. Electronic configuration of d-orbitals in [Ti(H2O)6]3+ ion in an octahedral crystal field is:
  1. $\text{t}^1_{2\text{g}}\text{e}^0_\text{g}$
  2. $\text{t}^2_{2\text{g}}\text{e}^0_\text{g}$
  3. $\text{t}^0_{2\text{g}}\text{e}^1_\text{g}$
  4. $\text{t}^1_{2\text{g}}\text{e}^1_\text{g}$
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Q 15Case study (4 Marks)4 Marks
Read the passage given below and answer the following questions:
To explain bonding in coordination compounds various theories were proposed. One of the important theory was valence bond theory. According to that, the central metal ion in the complex makes available a number of empty orbitals for the formation of coordination bonds with suitable ligands. The appropriate atomic orbitals of the metal hybridise to give a set of equivalent orbitals of definite geometry.
The d-orbitals involved in the hybridisation may be either inner d-orbitals i.e., (n - 1)d or outer d-orbitals i.e., nd. For example, Co3+ forms both inner orbital and outer orbital complexes, with ammonia it forms [Co(NH3)6]3+ and with fluorine it forms [CoF6]3- complex ion.
The following questions are multiple choice questions. Choose the most appropriate answer:
  1. Which of the following is not true for [CoF6]3-?
  1. It is paramagnetic.
  2. It has coordination number of 6.
  3. It is outer orbital complex.
  4. It involves d2sp3 hybridisation.
  1. [Cr(H2O)6]Cl3 (at. no. of Cr = 24) has a magnetic moment of 3.83B.M. The correct distribution of 3d-electrons in the central metal of the complex is:
  1. $3\text{d}^1_\text{xy},3\text{d}^1_{\text{x}^2-\text{y}^2},3\text{d}^1_\text{yz}$
  2. $3\text{d}^1_\text{xy},3\text{d}^1_{\text{yz}},3\text{d}^1_\text{zx}$
  3. $3\text{d}^1_\text{xy},3\text{d}^1_{\text{zy}},3\text{d}^1_{\text{z}^2}$
  4. $3\text{d}^1_{\text{x}^2-\text{y}^2},3\text{d}^1_{\text{z}^2},3\text{d}^1_\text{xz}$
  1. Which of the following is true for [Co(NH3)6]3+?
  1. It is an octahedral, di magnetic and outer orbital complex.
  2. It is an octahedral, paramagnetic and outer orbital complex.
  3. It is an octahedral, paramagnetic and inner orbital complex.
  4. It is an octahedral, di magnetic and inner orbital complex.
  1. The paramagnetism of [CoF6]3- is due to.
  1. 3 electrons.
  2. 4 electrons.
  3. 2 electrons.
  4. 1 electron.
  1. Which of the following is an inner orbital or low spin complex?
  1. [Ni(H2O)6]3+
  2. [FeF6]3-
  3. [Co(CN)6]3-
  4. [NiCl4]2-
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Q 18M.C.Q (1 Marks)1 Mark
The correct order of ligands for writing the formula of complex compounds is _________.
  • A
    Neutral, anionic, cationic
  • B
    Anionic, neutral, cationic
  • C
    Anionic, cationic, neutral
  • D
    Cationic, neutral, anionic
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Q 19M.C.Q (1 Marks)1 Mark
The co-ordination number of a metal in co-ordination compound is:
  • A
    Same as primary valency.
  • B
    Sum of primary and secondary valences.
  • C
    Same as secondary valency.
  • D
    None of the above.
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Q 20M.C.Q (1 Marks)1 Mark
IUPAC name of [Pt(NH3)2 Cl(NO2)] is:
  • A
    Platinum diaminechloronitrite.
  • B
    Chloronitrito-N-ammineplatinum (II).
  • C
    Diamminechloridonitrito-N-platinum (II).
  • D
    Diamminechloronitrito-N-platinate (II).
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Q 21Assertion (A) & Reason (B) MCQ1 Mark
Two statements are given- one labelled Assertion (A) and the other labelled Reason (R). Select the correct answer to these questions from the codes (a), (b), (c) and (d) as given below:
  1. Both Assertion (A) and Reason (R) are correct statements, and Reason (R) is the correct explanation of the Assertion (A).
  2. Both Assertion (A) and Reason (R) are correct statements, but Reason (R) is not the correct explanation of the Assertion (A).
  3. Assertion (A) is correct, but Reason (R) is incorrect statement.
  4. Assertion (A) is incorrect, but Reason (R) is correct statement.
Assertion (A): Linkage isomerism arises in coordination compounds because of ambidentate ligand.
Reason (R): Ambidentate ligand like NO2 has two different donor atoms i.e., N and O.
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Q 22Assertion (A) & Reason (B) MCQ1 Mark
Two statements are given- one labelled Assertion (A) and the other labelled Reason (R). Select the correct answer to these questions from the codes (a), (b), (c) and (d) as given below:
  1. Both Assertion (A) and Reason (R) are correct statements, and Reason (R) is the correct explanation of the Assertion (A).
  2. Both Assertion (A) and Reason (R) are correct statements, but Reason (R) is not the correct explanation of the Assertion (A).
  3. Assertion (A) is correct, but Reason (R) is incorrect statement.
  4. Assertion (A) is incorrect, but Reason (R) is correct statement.
Assertion (A): Linkage isomerism arises in coordination compounds because of ambidentate ligand.
Reason (R): Ambidentate ligand like NO2 has two different donor atoms i.e., N and O.
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Q 23Assertion (A) & Reason (B) MCQ1 Mark
In these questions, a statement of assertion followed by a statement of reason is given. Choose the correct answer out of the following choices.
  1. Assertion and reason both are correct statements and reason is correct explanation for assertion.
  2. Assertion and reason both are correct statements but reason is not correct explanation for assertion.
  3. Assertion is correct statement but reason is wrong statement.
  4. Assertion is wrong statement but reason is correct statement.
Assertion: [Cu(NH3)4]2+ is coloured while [Cu(CN)4]3- ion is colourless.
Reason: [Cu(NH3)4]2+ has dsp2 hybridisation.
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Q 24Assertion (A) & Reason (B) MCQ1 Mark
In these questions, a statement of assertion followed by a statement of reason is given. Choose the correct answer out of the following choices.
  1. Assertion and reason both are correct statements and reason is correct explanation for assertion.
  2. Assertion and reason both are correct statements but reason is not correct explanation for assertion.
  3. Assertion is correct statement but reason is wrong statement.
  4. Assertion is wrong statement but reason is correct statement.
Assertion: The ligands nitro and nitrito are called ambidentate ligands.
Reason: An ambidentate ligand can attach to metal through two different atoms.
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Q 25Assertion (A) & Reason (B) MCQ1 Mark
In these questions, a statement of assertion followed by a statement of reason is given. Choose the correct answer out of the following choices.
  1. Assertion and reason both are correct statements and reason is correct explanation for assertion.
  2. Assertion and reason both are correct statements but reason is not correct explanation for assertion.
  3. Assertion is correct statement but reason is wrong statement.
  4. Assertion is wrong statement but reason is correct statement.
Assertion: [Pt(NH3)2Cl2] is square planar.
Reason: The oxidation state of platinum is +2.
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Q 275 Marks Questions5 Marks
 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 (Compound)
 
Column II (Oxidation state of Co)
a.
[Co(NCS)(NH3)5](SO3)
1.
+4
b.
[Co(NH3)4Cl2]SO4
2.
0
c.
Na4[Co(S2O3)3]
3.
+1
d.
[Co2(CO)8]
4.
+2
 
 
5.
+3

Code:

  1. A (1), B (2), C (4), D (5).
  2. A (4), B (3), C (2), D (1).
  3. A (5), B (1), C (4), D (2).
  4. A (4), B (1), C (2), D (3).
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Q 285 Marks Questions5 Marks
 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.
[Co(NH3)4Cl2]+
1.
Optical
b.
cis-[Co(en)2Cl2]+
2.
Ionisation
c.
[Co(NH3)5 (NO2)]Cl2
3.
Coordination
d.
[Co(NH3)6][Cr(CN)6]
4.
Geometrical
 
 
5.
Linkage

Code:

  1. A (1), B (2), C (4), D (5).
  2. A (4), B (3), C (2), D (1).
  3. A (4), B (1), C (5), D (3).
  4. A (4), B (1), C (2), D (3).
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Q 295 Marks Questions5 Marks
CoSO4Cl.5NH3 exists in two isomeric forms ‘A’ and ‘B’. Isomer ‘A’ reacts with AgNO3 to give white precipitate, but does not react with BaCl2. Isomer ‘B’ gives white precipitate with BaCl2 but does not react with AgNO3. Answer the following questions.
  1. Identify ‘A’ and ‘B’ and write their structural formulas.
  2. Name the type of isomerism involved.
  3. Give the IUPAC name of ‘A’ and ‘B’.
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Q 305 Marks Questions5 Marks
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.
[Cr(H2O)6]3+
1.
dsp2, 1
b.
[Co(CN)4]2-
2.
sp3d2, 5
c.
[Ni(NH3)6]2+
3.
d2sp3, 3
d.
[MnF6]4-
4.
sp3, 4
 
 
5.
sp3d2, 2
Code:
  1. A (3), B (1), C (5), D (2).
  2. A (4), B (3), C (2), D (1).
  3. A (3), B (2), C (4), D (1).
  4. A (4), B (1), C (2), D (3).
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