Bonding in Coordination Compounds MCQ Quiz - Objective Question with Answer for Bonding in Coordination Compounds - Download Free PDF

The correct answe is sp3, sp3, sd3

Concept:-

• Concept of Hybridization: Hybridization is the concept of mixing atomic orbitals to form a new hybrid orbitals suitable for the pairing of electrons to form chemical bonds in valence bond theory. Hybridization is used to model molecular geometry and to explain atomic bonding properties.
• Different Types of Hybridization: There are four simple types of hybridisation - sp, sp2, sp3 and sp3d, sp3d2. 's', 'p', 'd' indicate the orbital types. The superscript indicates the total number of those kinds of orbitals used.
• Achieving Stability: Atoms undergo hybridization to attain stability by minimizing electron repulsion in the process of molecule formation.
• Transition Metal Complexes: Transition metals often form complexes, in which the metal atom or ion is at the centre of several atoms or groups of atoms. These metal ions in complex ions are often involved in hybridization.

Explanation:-

• [NiCl₄]^2-: The nickel ion Ni²⁺ has an electron configuration of [Ar]3d8. In this complex, it can be considered to utilize 3d, 4s, and two 4p orbitals to form four sp3 hybrid orbitals which are half-filled and will form bonds with four chloride ions. So the hybridization state is sp³.
• Cl^- is weak ligand 
• 
• [CoCl₄]^2-: The cobalt ion Co²⁺ has an electron configuration of [Ar]3d7. Similar to the case of nickel, cobalt also utilizes the 3d, 4s, and two 4p orbitals creating four sp3 hybrid orbitals which will form bonds with four chloride ions. Hence, the hybridization state here is sp³.
• Cl- is weak ligand 
• 
• MnO_4^-: The manganese ion Mn⁷⁺ has an electron configuration of [Ne]3p6, which means it has no unpaired electron in the 3d subshell. However, for the bonding purposes with four oxygen atoms, it needs to engage the vacant 3d orbital, along with 4s and two 4p orbitals for hybridization. This would give the reference of the 3d subshell's involvement in hybridization, hence we use sd³ notation to specify it, instead of the general sp³ for tetrahedral geometry.
• 

Conclusion:-

So, The Hybridisiation of [NiCl4]2- , [CoCl4]2- , MnO4-  is sp3, sp3, sd3

The correct answer is 335 pm.

Key Points

• Graphite is composed of layers of carbon atoms arranged in a hexagonal lattice.
• The distance between the layers, also known as the interplanar distance, is approximately 335 pm (picometers).
• This layered structure allows graphite to act as a good lubricant and conductor of electricity.
• The weak van der Waals forces between the layers make it easy for them to slide over one another.

Important Points

• The interplanar distance in graphite is a crucial factor that contributes to its physical properties, such as electrical conductivity and lubricity.
• The hexagonal arrangement of carbon atoms within each layer is responsible for the material's strength and stability.

CONCEPT:

Thermal Decomposition and Molecular Orbital Theory (MOT)

• Thermal decomposition of AgNO₃ produces gases that exhibit paramagnetism.
• Paramagnetic gases have unpaired electrons.
• The molecular orbital theory (MOT) provides insight into the electronic structure of diatomic molecules, particularly the distribution of electrons in bonding and antibonding molecular orbitals.

EXPLANATION::

2 AgNO₃ (s) → 2 Ag (s) + 2 NO₂ (g) + O₂ (g)

Both NO₂ and O₂ gases are paramagnetic:

• NO₂ has 1 unpaired electron.
• O₂ has 2 unpaired electrons.

Since O₂ has the higher number of unpaired electrons, we focus on O₂:

Electronic configuration of O₂ based on Molecular Orbital Theory:

σ_1s²σ_1s^∗2σ_2s²σ_2s^∗2σ_2p²π_2p^x2π_2p^y2π_2p^x*1π_2p^y*1

The antibonding molecular orbitals (MOs) are indicated by the asterisk (*).

In O₂, the antibonding MOs are populated as follows: \(\sigma_{1 s}^{2} \sigma *_{1 s}^{2} \sigma_{2 s}^{2} \sigma{ }^{2}{ }_{2 s}^{2} \sigma_{2 p_{2}}^{2} \pi_{2 p_{x}}^{2}=\pi_{2 p_{y}}^{2} \pi *_{2 p_{x}}^{1}=\pi{ }_{2 p_{y}}^{1}\)
Adding the electrons in these antibonding orbitals:

2 (from σ_1s*) + 2 (from σ_2s*) + 1 (from π_2p*x) + 1 (from π_2p*y) = 6 electrons

Therefore, the total number of electrons present in the antibonding molecular orbitals of O₂ is 6.

Ans. (1)

Sol.

\(\mathrm{TiCl}_4 \Rightarrow \mathrm{Ti}^{+4} \quad \mathrm{e}^0 \mathrm{t}_2^0\)

\(\mathrm{MnO}_4^{-} \Rightarrow \mathrm{Mn}^{+7} \quad \mathrm{e}^0 \mathrm{t}_2^0\)

\(\mathrm{FeO}_4^{2-} \Rightarrow \mathrm{Fe}^{+6} \quad \mathrm{e}^2 \mathrm{t}_2^0\)

\(\mathrm{FeCl}_4^{2-} \Rightarrow \mathrm{Fe}^{+2} \quad \mathrm{e}^3 \mathrm{t}_2^3\)

\(\mathrm{CoCl}_4^{2-} \Rightarrow \mathrm{Co}^{+2} \quad \mathrm{e}^4 \mathrm{t}_2^3\)

Correct answer: 2)

Concept:

• A spectrochemical series is nothing but a list of ligands ordered on ligand strength (ability) and a list of metal ions based on oxidation number, group and its identity.
• In crystal field theory, ligands modify the difference in energy between the d orbital / triangle, called the ligand-field splitting parameter for ligands or the crystal-field splitting parameter, which is mainly observed in differences in colour of similar metal-ligand complexes.
• A strong field ligand has high crystal field stabilisation energy.
• The complexes formed in this are called low spin complexes.
• They are mostly diamagnetic or less paramagnetic than weak fields.
• A weak ligand has lower CFSE.
• The complexes formed with these ligands are also known as high spin complexes.
• They are mostly paramagnetic in nature.

Explanation:

• High spin and low spin complex are determined by magnitude of CFSE.
• The energy of t_2g orbital is lowered by 0.4\(\Delta _{o}\) and energy of e_g orbital is raised by 0.6\(\Delta _{o}\).
• If the ligand is strong field, then splitting will be greater. Thus, have high CFSE. 
• In that case, pairing energy cannot overcome the CFSE and the electrons get paired in lower orbitals. 
• So, as the electrons get paired, fewer electrons will be left (in case of d⁷ configuration) for e_g orbital.
• Thus, the d⁷ configuration for strong field ligands is as: 

• CFSE= energy difference between two sets of orbitals
• CFSE= Energy of lower set - energy of higher set
• CFSE= (0.4 x 6)-(0.6 x 1)
• CFSE= 1.8\(\Delta _{o}\)
• In case of weak field ligand, pairing will not be done.
• Thus, the d⁷ configuration for weak field ligands is as: 

• CFSE= (0.4 x 5)-(0.6 x 2)
• CFSE= 0.8\(\Delta _{o}\)

Conclusion:

Thus, in weak and strong fields the (CFSE), Δ0 for a metal ion with d7 electronic configuration are 0.8 Δ0 and 1.8 Δ0, respectively

Ans. (1)

Sol.

\(\mathrm{TiCl}_4 \Rightarrow \mathrm{Ti}^{+4} \quad \mathrm{e}^0 \mathrm{t}_2^0\)

\(\mathrm{MnO}_4^{-} \Rightarrow \mathrm{Mn}^{+7} \quad \mathrm{e}^0 \mathrm{t}_2^0\)

\(\mathrm{FeO}_4^{2-} \Rightarrow \mathrm{Fe}^{+6} \quad \mathrm{e}^2 \mathrm{t}_2^0\)

\(\mathrm{FeCl}_4^{2-} \Rightarrow \mathrm{Fe}^{+2} \quad \mathrm{e}^3 \mathrm{t}_2^3\)

\(\mathrm{CoCl}_4^{2-} \Rightarrow \mathrm{Co}^{+2} \quad \mathrm{e}^4 \mathrm{t}_2^3\)

The correct answer is option 1 that is [Ni(CN)4]2-

Concept:

• Hybridisation is the process of mixing of orbitals of different shapes and energies.
• When two atomic orbital combines, it forms a hybrid orbital.
• When one ‘s’ orbital and ‘3’ 3p orbitals are mixed, then the formed hybridization is sp3 and geometry is tetrahedral.
• When one ‘d’ orbital, one ‘s’ orbital, and two ‘p’ orbitals are mixed, then the formed hybridization is square planar.
• The hybridization of square planar is  dsp2 and for tetrahedral is sp3.
• A strong ligand helps in pairing up the valence shell electron to create the space for ligand whereas a weak field ligand won’t help in the pairing of electrons.

Explanation:

• In [Ni(CN)4​]2−, Ni is in +2 state
• Ni2+⇒[Ar]4s03d8⇒   
• The ligand (CN-) is strong it will pair up single electrons in the d-orbital.
• Hence the hybridization would be dsp2 and the structure is  square planar structure.

Conclusion:

Thus, [Ni(CN)4]2- has square planar structure.

Explanation-

Let us observe some definitions and facts of different carbon allotropes

Buckminsterfullerene-

• These are carbon allotropes that are arranged in shape particle .
• They contain 60 carbon atoms arranged in pentagons and hexagon geometry similar to soccer ball.
• In this structure all carbons are equivalent and each atom has sp² bonding.

Haeckelite- 

• These are 3 fold coordinated networks of carbon atoms which are generated by periodic arrangements of different polygons.
• These structures are only proposed theory and yet to be synthesised in laboratory  

Carbon Nanocone- 

• These are conical shaped structures that are predominantly made from carbon.they have the height and the base diameter of the equal magnitude.
• The cone apex angle of the nanocones are not arbitrary but preferred angles are approximately in the range 20°,40°and 60°.

Carbon Nanotube -

• These are cylindrical tubes like structure made up of carbon.
• The structure of an ideal single walled carbon nanotube is that like a regular hexagonal lattice structure drawn on a cylindrical surface,where the vertices indicates the position of carbon atoms.

let us observe the shape of buckminsterfullerene

containing 60 carbon atoms with sp² bonding.

• also known as Bucky Balls
• it is most common naturally occurring fullerene
• it is insoluble in water

Therefore from given options only buckminsterfullerene satisfies the required statement.

Hence the correct option is 1

Correct answer: 2)

Concept:

• A complex that has the valence electrons of the central metal atom in the outer d orbitals is called the outer orbital complex, while in the metal atom when the valence electrons are present in the inner d orbital it is called the inner orbital complex.
•  If the field is strong, it will have few unpaired electrons and thus low spin. 
• If the field is weak, it will have more unpaired electrons and thus a high spin.

Explanation:

• The magnetic moment of [Fe (F)6]3- is 5.9 B.M.
• \(\mu =\sqrt{n(n+2)} B.M.\)
• where, \(\mu = \) magnetic moment, n = number of unpaired electron 
• B. M. = Bohr Magneton (unit of magnetic moment)
• 5.9= \(\sqrt{n(n+2)} \)
• n= 5
• [Fe (F)6]3- involve sp³d² hybridisation with 5 unpaired electrons.
• Cl- is a weak field ligand so, no pairing with takes place in outer d-orbital.
• Therefore, [Fe (F)6]3- is outer orbital complex : high spin complex.

Conclusion:

Thus, [Fe (F)6]3- is outer orbital complex : high spin complex.

Correct answer: 4)

Concept:

• The Crystal Field Theory is more widely accepted than the valence bond theory.
• It assumes that the attraction between the central metal and the ligands in a complex is purely electrostatic.
• In the crystal field the following assumptions are made : (i) Ligands are treated as point charges. (ii) There is no interaction between metal orbitals and ligand orbitals. (iii) The d orbitals on the metal all have the same energy (that is degenerate) in the free atom.
• However, when a complex is formed the ligands destroy the degeneracy of these orbitals, i.e. the orbitals now have different energies. 

Explanation:

• The power for splitting the metal d orbitals is higher for strong ligands than for weaker ligands. 
• Spectrochemical series arranges the ligands based on their splitting powers.
• Ligands are classified as strong or weak based on the spectrochemical series:
• I^- < Br^- < Cl^- < SCN^- < F^- < OH^- < ox^2-< ONO^- < H₂O < SCN^- < EDTA^4- < NH₃ < en < NO₂^- < CN^-
• The crystal field-splitting for Cr³⁺ ion in octahedral field increases for ligands I⁻,H₂​O,NH₃​,CN⁻ and the order is I⁻2​O3​−.
• This is in accordance with spectrochemical series.

Conclusion:

Thus, In the octahedral crystal field, the correct order of splitting in Cr3+ complexes for I- is H2O, NH3 and CN- is I−2​O3​−

The correct answer is due to charge transfer from Fe2+ to orbitals of bpy ligand

Concept:-

• Electronic transitions: The origin of color in complex ions comes from electronic transitions from one energy level (or orbital) to another.
• Ligand to Metal Charge Transfer (LMCT) and Metal to Ligand Charge Transfer (MLCT): In these transitions, an electron moves from a metal-based orbital to a ligand-based orbital or vice versa.
• Properties of particular ligands: Ligands can have pi donor, pi acceptor, or sigma donor properties that can influence the electronic transitions and hence the color of a complex.
• d-d transitions: These are transitions where an electron is excited from one d orbital to another.

Explanation:-

Now, let's go onto the detailed explanation:

In the complex [Fe(bpy)]2+ (where bpy stands for 2,2'-bipyridyl, a bidentate ligand), the Fe is in the +2 oxidation state and hence in the d⁶ configuration. Now, the bipyridyl ligand (bpy) is a pi-accepting ligand. This means it can accept electrons in its empty pi* (pi star) orbitals.

One might think at first that the color comes from d-d transitions within the Fe²⁺ ion. However, given the presence of the bpy ligand, which can accept electron density into its unoccupied π* orbitals, the color is instead due to electrons being excited from the Fe2+(d) orbitals to the π* orbitals of the bpy ligand. This type of transition is known as a Metal to Ligand Charge Transfer (MLCT) transition, which can often result in intense colors.

Hence, the correct answer is 3) due to charge transfer from Fe²⁺ to π∗ orbitals of bpy ligand.

Conclusion:-

So, In complex [Fe(bpy)]2+, colour arises due to charge transfer from Fe2+ to π∗ orbitals of bpy ligand.

Correct answer: 2)

Concept:

• A complex that has the valence electrons of the central metal atom in the outer d orbitals is called the outer orbital complex, while in the metal atom when the valence electrons are present in the inner d orbital it is called the inner orbital complex.
•  If the field is strong, it will have few unpaired electrons and thus low spin. 
• If the field is weak, it will have more unpaired electrons and thus a high spin.

Explanation:

• The magnetic moment of [Fe (F)6]3- is 5.9 B.M.
• \(\mu =\sqrt{n(n+2)} B.M.\)
• where, \(\mu = \) magnetic moment, n = number of unpaired electron 
• B. M. = Bohr Magneton (unit of magnetic moment)
• 5.9= \(\sqrt{n(n+2)} \)
• n= 5
• [Fe (F)6]3- involve sp³d² hybridisation with 5 unpaired electrons.
• Cl- is a weak field ligand so, no pairing with takes place in outer d-orbital.
• Therefore, [Fe (F)6]3- is outer orbital complex : high spin complex.

Conclusion:

Thus, [Fe (F)6]3- is outer orbital complex : high spin complex.

Explanation:

• In the case of the complex ion [Co(NH3)6]3+, ammonia (NH₃) is a neutral ligand and the oxidation state of cobalt (Co) in the complex is +3.  The electronic configuration of Co³⁺ is [Ar]3d⁶
• NH₃ is a moderately strong field ligand and will form a low-spin complex with Co3+ ion. Therefore, crystal field splitting energy will be higher than the pairing energy (P), i.e Δ>P, and the pairing of electrons takes place in the t_2g orbital.
• The crystal field splitting diagram is shown below.

Conclusion:

Correct answer: 2)

Concept:

• A spectrochemical series is nothing but a list of ligands ordered on ligand strength (ability) and a list of metal ions based on oxidation number, group and its identity.
• In crystal field theory, ligands modify the difference in energy between the d orbital / triangle, called the ligand-field splitting parameter for ligands or the crystal-field splitting parameter, which is mainly observed in differences in colour of similar metal-ligand complexes.
• A strong field ligand has high crystal field stabilisation energy.
• The complexes formed in this are called low spin complexes.
• They are mostly diamagnetic or less paramagnetic than weak fields.
• A weak ligand has lower CFSE.
• The complexes formed with these ligands are also known as high spin complexes.
• They are mostly paramagnetic in nature.

Explanation:

• High spin and low spin complex are determined by magnitude of CFSE.
• The energy of t_2g orbital is lowered by 0.4\(\Delta _{o}\) and energy of e_g orbital is raised by 0.6\(\Delta _{o}\).
• If the ligand is strong field, then splitting will be greater. Thus, have high CFSE. 
• In that case, pairing energy cannot overcome the CFSE and the electrons get paired in lower orbitals. 
• So, as the electrons get paired, fewer electrons will be left (in case of d⁷ configuration) for e_g orbital.
• Thus, the d⁷ configuration for strong field ligands is as: 

• CFSE= energy difference between two sets of orbitals
• CFSE= Energy of lower set - energy of higher set
• CFSE= (0.4 x 6)-(0.6 x 1)
• CFSE= 1.8\(\Delta _{o}\)
• In case of weak field ligand, pairing will not be done.
• Thus, the d⁷ configuration for weak field ligands is as: 

• CFSE= (0.4 x 5)-(0.6 x 2)
• CFSE= 0.8\(\Delta _{o}\)

Conclusion:

Thus, in weak and strong fields the (CFSE), Δ0 for a metal ion with d7 electronic configuration are 0.8 Δ0 and 1.8 Δ0, respectively

Correct answer: 2)

Concept:

• Hybridisation is the process of mixing of orbitals of different shapes and energies.
• When two atomic orbital combines, it forms a hybrid orbital.
• When one ‘s’ orbital and ‘3’ 3p orbitals are mixed, then the formed hybridization is sp³ and geometry is tetrahedral.
• When one ‘d’ orbital, one ‘s’ orbital, and two ‘p’ orbitals are mixed, then the formed hybridization is square planar.
• The hybridization of square planar is  dsp² and for tetrahedral is sp³.
• A strong ligand helps in pairing up the valence shell electron to create the space for ligand whereas a weak field ligand won’t help in the pairing of electrons.

Explanation:

• In [Ni(CN)₄​]²⁻, Ni is in +2 state
• Ni²⁺⇒[Ar]4s⁰3d⁸⇒\(4s^{0}3d_{xy}^{2},3d_{yz}^{2},3d_{xz}^{2},3d_{z^{2}}^{2},3d_{x^{2}-y^{2}}^{0} \)   
• The ligand (CN^-) is strong it will pair up single electrons in the d-orbital.
• Hence the hybridization would be dsp² and the structure is  square planar structure.

Conclusion:

Thus, [Ni(CN)4]2- has square planar structure.