250+ TOP MCQs on Bonding in Coordination Compounds and Answers

Chemistry Multiple Choice Questions on “Bonding in Coordination Compounds – 1”.

1. Which of the following does not explain the nature of bonding in coordination compounds?
a) Crystal Field Theory
b) Molecular Orbital Theory
c) Valence Bond Theory
d) VSEPR Theory
Answer: d
Clarification: The VSEPR Theory explains the structure of individual molecules based on the electron pairs in their atoms. VBT, CFT, LFT and MOT are theories that explain the nature of bonding in coordination compounds.

2. How many types of hybridisation are possible for complexes with a coordination number of 4?
a) 1
b) 2
c) 3
d) 4
Answer: b
Clarification: According to VBT, a complex with CN=4 can have two possible types of hybridisation and hence geometries. The sp³ hybridisation results in tetrahedral geometry and dsp² hybridisation results in square planar geometry.

3. How many empty orbitals are available in the central metal ion of a complex that has a trigonal bipyramidal geometry?
a) 2
b) 3
c) 4
d) 5
Answer: d
Clarification: One of the assumptions of VBT is that the number of empty orbitals in the metal ion for making bonds with the ligands is equal to its CN. A trigonal bipyramidal geometry has a hybridisation of sp³d and hence a CN=5.

4. A complex having _________ geometry can have more than one type of hybridisation.
a) tetrahedral
b) square planar
c) trigonal bipyramidal
d) octahedral
Answer: d
Clarification: Complexes having octahedral geometry can have either sp³d² or d²sp³ hybridisation depending on whether the outer or inner d orbitals are involved in hybridisation respectively.

5. Which statement regarding [Cr(NH₃)₆]³⁺ is incorrect?
a) It has octahedral geometry
b) It has d²sp³ hybridisation
c) It is diamagnetic
d) It is a low spin complex
Answer: c
Clarification: Cr³⁺ has six empty orbitals- two 3d, one 4s and three 4p orbitals, which are occupied by six electron pairs from six NH₃ molecules to form six d²sp³ hybridised orbitals. This leaves, three unpaired 3d electrons, making it paramagnetic. Also, since the inner d orbitals are involved, it is a low spin complex.

6. The ferricyanide complex ion is _________
a) paramagnetic
b) outer orbital
c) spin free
d) tetrahedral
Answer: a
Clarification: The oxidation state of Fe in [Fe(CN)₆]^3- ion is +3, which leaves it with five unpaired electrons in 3d orbitals. Two of these pair up leaving one unpaired electron in 3d and six vacant orbitals to form d²sp³ hybridised octahedral inner orbital complex.

7. What is the number of unpaired electrons in [Fe(H₂O)₆]²⁺ if it is known to be a high spin complex?
a) 0
b) 2
c) 3
d) 4
Answer: d
Clarification: The configuration of Fe²⁺ is 3d⁶. The complex has a CN=6 has should have sp³d² hybridisation. This leaves the 3d orbitals as it is with 4 unpaired electrons.

8. Identify the magnetic nature of the complex from its electronic configuration as shown.

a) Strongly paramagnetic
b) Weakly paramagnetic
c) Diamagnetic
d) Ferromagnetic
Answer: b
Clarification: There is one unpaired electron in the 3d orbitals. The presence of unpaired electrons makes a complex paramagnetic. In this case, since only 1 unpaired electron is present, it is a weak paramagnet.

9. Coordination entities with only three ligand groups attached to the metal ion can have d²sp³ hybridisation.
a) True
b) False
Answer: a
Clarification: This happens in the case of didentate ligands which have six pairs of electrons from three molecules. For example, [Co(ox)₃]^3- has d²sp³ hybridisation and is diamagnetic.

10. Identify the correct geometry and magnetic behaviour of the complex from the configuration shown.

a) Square planar, diamagnetic
b) Square planar, paramagnetic
c) Tetrahedral, diamagnetic
d) Tetrahedral, paramagnetic
Answer: a
Clarification: The hybridisation type of the complex is dsp² which corresponds to square planar geometry. Because there are no unpaired electrons, it is diamagnetic.

11. Predict the geometry of the complex from the electronic configuration shown.

a) Linear
b) Square planar
c) Tetrahedral
d) Octahedral
Answer: a
Clarification: The hybridisation type is sp which involves only two orbitals, so the geometry has to be linear and the coordination number is 2.

12. [Co(en)₂(NH₃)Cl]²⁺ is known to be a diamagnetic complex. What is the type of hybridisation it shows?
a) dsp²
b) sp³
c) d²sp³
d) sp³d²
Answer: c
Clarification: Co³⁺ has 3d⁶ configuration. For a complex to be diamagnetic, it should have no unpaired electrons. So, the single electrons in the 3d orbital will have to pair up and form 3 pairs leaving two 3d orbitals vacant. This results in d²sp³ hybridisation.

13. The compound K₄[Mn(CN)₆] is diamagnetic.
a) True
b) False
Answer: b
Clarification: The atomic number of Mn is 25 and the outer shell configuration of Mn²⁺ is 3d⁵. Four out of these five electrons pair up leaving 1 unpaired electron in 3d orbital and forming d²sp³ hybridisation. Thus, it is paramagnetic.

14. Identify the incorrect statement regarding VBT.
a) It does not explain the colour of coordination compounds
b) It is unreliable in the prediction of geometries of 4-coordinate complexes
c) It does not explain the kinetic stabilities of coordination compounds
d) It can distinguish between strong and weak ligands
Answer: d
Clarification: One of the limitations of VBT is that it cannot differentiate between weak and strong ligands. The other options are also its limitations.