Test: Coordination Chemistry- 2 - Chemistry MCQ

Mn3O4 is a mixed oxide with a spinel structure. In this structure, Mn2+ ions typically occupy the tetrahedral (A) sites, while Mn3+ ions occupy the octahedral (B) sites. Therefore, in Mn3O4, there is one Mn2+ ion located at the tetrahedral site and two Mn3+ ions located at the octahedral sites. This distribution corresponds to option a) One Mn2+ and two Mn3+.

• Spinels have a general formula AB₂O₄ and are classified as either normal or inverse.


• In a normal spinel, A²⁺ ions occupy the tetrahedral sites, and B³⁺ ions are in octahedral sites.


• In an inverse spinel, half of the B³⁺ ions and all A²⁺ ions occupy octahedral sites, while the remaining B³⁺ ions occupy tetrahedral sites.


• CoFe₂O₄ is inverse because Co²⁺ prefers octahedral sites.


• FeFe₂O₄ (magnetite) is also inverse with Fe³⁺ ions distributed as described.


 

• The trigonal bipyramidal crystal field splits the d orbitals into different energy levels.


• The d_z² orbital has the highest energy because it points along the z-axis, which aligns with the axial positions of the trigonal bipyramidal geometry.


• These axial positions experience more repulsion, increasing the energy of the d_z² orbital.


• Other orbitals, like d_xy and d_yx, experience less direct interaction with ligands, leading to lower energy levels.

Magnetic moment of K₃[CoF₆]
- The magnetic moment of the complex K₃[CoF₆] is 5.0 B.

Stabilization energy
- The stabilization energy is the energy released during the formation of a complex from its constituent ions.
- The greater the stabilization energy, the more stable the complex.
- The stabilization energy can be calculated using the crystal field theory.

Crystal field theory
- According to the crystal field theory, the metal ion in a complex is surrounded by a set of negatively charged ligands.
- The ligands repel the electrons in the metal ion, causing the energy levels to split.
- The magnitude of the splitting depends on the nature of the ligands and the geometry of the complex.
- The energy difference between the highest and lowest energy levels determines the magnetic properties of the complex.
- The magnetic moment of the complex is related to the number of unpaired electrons in the highest energy level.

Calculation of stabilization energy
- The stabilization energy can be calculated using the following equation:
ΔE = -0.4Bn(unpaired)
where ΔE is the stabilization energy, B is the crystal field splitting energy, n is the number of ligands, and unpaired is the number of unpaired electrons in the highest energy level.
- In this case, the complex has a magnetic moment of 5.0 B, which indicates the presence of one unpaired electron in the highest energy level.
- Therefore, unpaired = 1 and n = 6 (since there are six ligands in the complex).
- The crystal field splitting energy for octahedral complexes with strong field ligands is 0.4 Δo, where Δo is the crystal field splitting parameter.
- For Co(III), Δo is approximately 10,000 cm^-1.
- Therefore, B = 0.4 x 10,000 = 4000 cm^-1.
- Substituting these values into the equation, we get:
ΔE = -0.4 x 4000 x 1 x 6 = -9600 cm^-1 = -0.96 eV.

Conclusion
- The total stabilization energy of the complex K₃[CoF₆] is -0.96 eV or -9600 cm^-1.
- The correct answer is option 'A'.

Having 6 ligands it must have octahedral structure but due to d9 electronic configuration ( unsymmetrically filled eg orbitals ( 3 electrons) ) it exhibits John teller distortion and thus have a tetragonally distorted octahedral structure.

- In [Cu(NH₃)₆]²⁺, there are six NH₃ ligands arranged in an octahedral geometry around the central Cu²⁺ ion.
- The bond distances are different due to the difference in the orientation of the ligands.
- There are two types of bond distances in this complex: axial and equatorial.
- Axial bonds: These are the bonds that are perpendicular to the plane of the equatorial bonds. There are two axial bonds in this complex.
- Equatorial bonds: These are the bonds that lie in the plane of the complex. There are four equatorial bonds in this complex.
- The repulsion between the lone pairs of electrons on the NH₃ ligands causes the axial bonds to be longer than the equatorial bonds.
- Therefore, the correct statement is that the axial bonds are longer than the equatorial ones (option B is correct).

- In K₂[NiBr₄], the nickel ion is surrounded by four bromide ions.
- The electronic configuration of Ni(II) is 3d⁸ 4s².
- To form the complex, the nickel ion utilizes one 4s and three 4p orbitals to hybridize into four sp³ hybrid orbitals.
- The hybrid orbitals are then used to form sigma bonds with the four bromide ions.
- Therefore, the hybridization of Ni(II) in K₂[NiBr₄] is sp³.

The spinels have the general chemical formula AB₂X₄. 

    Where:

    A^II = a divalent cation like Mg, Cr, Mn, Fe, Co, Ni, Cu, Zn, Cd, Sn 

    B^III = a trivalent cation like Al, Ga, In, Ti, V, Cr, Mn, Fe, Fe, Co, Ni 

    X = O, S, Se etc.

Examples of Normal Spinels: MgAl₂O₄ (known as spinel), Mn₃O₄, ZnFe₂O₄, FeCr₂O₄ (chromite) etc.

UV-Visible spectroscopy is a powerful technique that can be used to determine the CFSE of transition metal complexes. This method provides valuable information about the electronic structure of the complex, which can be used to design new materials with specific properties.

Jahn-Teller Distortion in Highspin Transition Metal Ions
The Jahn-Teller effect is a phenomenon observed in high-spin transition metal ions with partially filled d-orbitals in an octahedral field. It occurs due to the interaction between electrons and the degenerate molecular orbitals in the octahedral crystal field.

Explanation:
- Highspin transition metal ions have partially filled d-orbitals and exhibit strong electron-electron repulsion.
- When placed in an octahedral field, the d-orbitals split into two sets of degenerate orbitals, known as t2g and eg.
- In highspin ions, the electrons occupy the t2g orbitals before the eg orbitals due to the Hund's rule.
- However, this arrangement is not energetically favorable as the t2g orbitals are more stable than the eg orbitals.
- The Jahn-Teller effect occurs when the degeneracy of the eg orbitals is lifted by a distortion of the octahedral field, leading to a lowering of the energy of the system.
- The distortion can be either elongation or compression of the octahedral field along one of the axes.
- The distortion causes the electrons to occupy the lower energy eg orbital, resulting in a non-degenerate ground state.
- Only those pairs of electronic configurations of highspin transition metal ions that exhibit the Jahn-Teller effect undergo substantial distortion.

- Option 'B' is correct as it represents the pair of electronic configurations d4 and d9, which exhibit the Jahn-Teller effect. The d4 ion has two electrons in the eg orbitals and two in the t2g orbitals, while the d9 ion has one electron in the eg orbitals and four in the t2g orbitals. Both these ions have partially filled eg orbitals, which are energetically unstable and undergo Jahn-Teller distortion to lower the energy.

Jahn-Teller Distortions. The Jahn-Teller effect is a geometric distortion of a non-linear molecular system that reduces its symmetry and energy. This distortion is typically observed among octahedral complexes where the two axial bonds can be shorter or longer than those of the equatorial bonds.

Crystal Field Theory and Unpaired Electrons

• The crystal field theory is a model used to explain the behavior of transition metal ions in coordination compounds. It is based on the idea that the ligands surrounding the central metal ion create a crystal field that affects the energy levels of the metal ion's d orbitals. In particular, the crystal field splits the d orbitals into two sets of energy levels, with some orbitals raised in energy and others lowered.
• One consequence of this splitting is that some of the d orbitals may become partially filled, leading to unpaired electrons. These unpaired electrons can have important consequences for the properties of the compound, such as its magnetic behavior.

Ni²⁺ and Unpaired Electrons

• Ni²⁺ is a transition metal ion with a d⁸ electron configuration. This means it has eight electrons in its d orbitals, and in the absence of ligands, these electrons would pair up in the lower energy levels.
• However, when Ni²⁺ is coordinated to ligands, the crystal field splits the d orbitals into two sets of energy levels. In particular, the t2g set of orbitals is lowered in energy, while the eg set is raised. This means that if Ni²⁺ is in a geometry where the ligands are close enough to cause significant splitting, some of the d orbitals may be partially filled, leading to unpaired electrons.

Tetrahedral and Octahedral Geometries

• The crystal field splitting is affected by the geometry of the ligands around the central metal ion. In particular, the splitting is greater for octahedral geometries than for tetrahedral geometries. This means that for Ni²⁺, the eg set of orbitals is more likely to be raised above the t2g set in an octahedral geometry, leading to unpaired electrons.
• However, even in a tetrahedral geometry, there can still be some splitting of the d orbitals, leading to the possibility of unpaired electrons. This means that Ni²⁺ can have two unpaired electrons in both tetrahedral and octahedral geometries.

Conclusion

• In summary, the crystal field theory can be used to predict whether a transition metal ion will have unpaired electrons in a coordination compound. For Ni²⁺, the geometry of the ligands around the central metal ion affects the magnitude of the crystal field splitting, but even in a tetrahedral geometry, there can still be some splitting, leading to unpaired electrons. Therefore, the correct answer is option D, both tetrahedral and octahedral geometries.

Correct Answer :- d

Explanation : [Cr(H₂O)₆]³⁺ shows the strongest distortion from the regular octahedral structure.

The central Cr atom in +2 oxidation state has d⁷ outer electronic configuration. It has degenerate electronic states occupying the eg orbital set and tend to show stronger Jahn-Teller effects.

Enthalpy of Hydration

• Enthalpy of hydration is the amount of energy released when one mole of an ion in the gaseous state dissolves in water to form an aqueous solution.

Order of Enthalpies of Hydration

• The order of enthalpies of hydration of Ca²⁺, Mn²⁺, Zn²⁺ can be determined by considering the size and charge of the ions.
• Ca²⁺ has a larger size than Mn²⁺ and Zn²⁺, so it has a weaker hydration energy than the other two.
• Mn²⁺ has a smaller size than Zn²⁺, but it has a greater charge density due to its higher charge. This results in a stronger hydration energy for Mn²⁺ than Zn²⁺.
• Zn²⁺ has the smallest size and charge, resulting in the strongest hydration energy.
• Therefore, the order of the enthalpies of hydration is Zn²⁺ > Mn²⁺ > Ca²⁺.

The correct answer is option D, which lists the order of enthalpies of hydration as Zn²⁺, Mn²⁺, Ca²⁺.

Acidity of Aquo Complexes

• Aquo complexes are formed when metal ions are surrounded by water molecules. They are called aquo complexes because water molecules act as ligands in these complexes. The acidity of these aquo complexes depends on the charge on the metal ion and its size.

Order of Acidity

• The order of acidity among the given aquo complexes is determined by the charge on the metal ion. The higher the charge on the metal ion, the more acidic the complex.
• The correct order of acidity among the given species is: [Sc(H₂O)₆]_{3 ,}[Ni(H₂O)₆]₂ ,[Mn(H₂O)₆]₂ [Na(H₂O)₆]

Explanation

The correct answer is option B, which places [Sc(H₂O)₆]₃ as the most acidic and [Na(H₂O)₆] as the least acidic. This is because:

1. [Sc(H₂O)₆]₃ has the highest charge (+3) among the given species, making it the most acidic.

2. [Ni(H₂O)₆]₂ and [Mn(H₂O)₆]₂ have a charge of +2 each, making them less acidic than [Sc(H₂O)₆]₃.

3. [Na(H₂O)₆] has the lowest charge (+1) among the given species, making it the least acidic.

Hence, the correct order of acidity among the given species is option B.

Octahedral means "on the axis " orbitals i.e. dz² and dx²- y² have higher energy than "between the axis" orbitals i.e. dxy ,dxz, dyz .

Correct Answer :- D

Explanation : The general trend of increasing electronegativity of the element when moving from left to right ia any horizontal row. The electronegativity increases due to which charge separation and ionization of hydrides increases.This increases the acidic strength of hydrides. 

The correct order is :

Ca²⁺, Mn²⁺, Fe³⁺, Cu²⁺

• Spinels have a general formula AB₂O₄, where A and B are metal ions.


• In a simple spinel, A²⁺ ions occupy tetrahedral sites, and B³⁺ ions occupy octahedral sites.


• In an inverse spinel, B³⁺ ions occupy half of the tetrahedral sites and all octahedral sites, while A²⁺ ions occupy the remaining tetrahedral sites.


• Mn₃O₄ is a simple spinel. NiFe₂O₄ is an inverse spinel.


• Thus, option C correctly identifies the pair Mn₃O₄ (simple) and NiFe₂O₄ (inverse).

The highest oxidation states of transition metals are stabilized by strong electronegative ligands due to:

• Strong Electron Withdrawing Ability: Electronegative ligands, like fluorine or oxygen, have a high affinity for electrons. They can stabilize metals in high oxidation states by balancing the electron deficiency.


• Effective Electron Pairing: These ligands can form multiple bonds (like pi bonds), helping stabilize high oxidation states through effective electron sharing and pairing

Therefore, option C is correct as strong electronegative ligands stabilize high oxidation states by these mechanisms.

Crystal Field Stabilization Energy (CFSE):
CFSE is the energy required to place electrons in the d-orbitals of a metal ion in a ligand field. It depends on the oxidation state and coordination number of the metal ion. CFSE is proportional to the number of electrons in the t2g and eg​ orbitals. Metal ions with zero CFSE distribute their electrons such that the energy difference between these orbitals is zero.

Metal ions with zero CFSE:

Ca²⁺, Mn²⁺, Zn²⁺

• Ca²⁺: With configuration [Ar] 4s² 3d⁰, it has no d-electrons to split, resulting in zero CFSE.
• Mn²⁺: With [Ar] 3d⁵, its electrons evenly occupy t2g​ and eg​, neutralizing the energy difference, resulting in zero CFSE.
• Zn²⁺: With [Ar] 3d¹⁰, all t2g​ and eg orbitals are fully occupied, leading to zero energy difference and zero CFSE.

Non-zero CFSE options:

Ca³⁺, Mn³⁺, Zn³⁺:

• Ca³⁺ and Mn³⁺: With configurations [Ar] 3d³ and [Ar] 3d⁴, their electrons do not neutralize the energy difference, resulting in non-zero CFSE.
• Zn³⁺: Does not exist.

In summary, Ca²⁺, Mn²⁺, Zn²⁺ have zero CFSE.

- To find the volume of 0.1 M AgNO₃ needed, start by determining the moles of chloride ions from [Cr(H₂O)₅Cl]Cl₂.
- The concentration of chloride ions is 0.01 M in 30 mL, giving 0.01 x 0.030 = 0.0003 moles of Cl^-.
- Each mole of Cl^- reacts with 1 mole of AgNO₃ to form AgCl.
- Therefore, you need 0.0003 moles of AgNO₃, which is equivalent to 0.0003 moles x 10 mL (0.1 M) = 3 mL.
- Thus, the correct answer is approximately 3 mL.

The pale color of [Mn(H₂O)₆]²⁺ arises from spin-forbidden d-d transitions. In Mn²⁺ (3d⁵), all d-electrons are unpaired in a high-spin state due to weak field ligands like water. Transitions between d-orbitals require a spin flip, which is forbidden by selection rules, resulting in weak absorption and a pale color.

In the normal spinel structure, MnCr₂O₄ is arranged as (Mn²⁺) tetrahedral (Cr³⁺)₂ Octahedral. The O²⁻ ions act as weak field ligands.

• For Mn²⁺ (d⁵ configuration), the ion is in a high-spin state. Therefore, its CFSE = 0.
• For Cr³⁺ (d³ configuration) in an octahedral field, the CFSE = -12 Dq per Cr³⁺ ion.
  Since there are two Cr³⁺ ions, the total CFSE contributed by Cr³⁺ ions is:
  2 × (-12 Dq) = -24 Dq.

Adding these contributions:
CFSE of Mn²⁺ + CFSE of Cr³⁺ = 0 + (-24 Dq) = -24 Dq.

Thus, MnCr₂O₄ is a normal spinel with a total CFSE of -24 Dq, making Option C correct.

1. Dimethyl glyoxime: Bidentate ligand with 2 donor sites (two nitrogen atoms).
2. Glycinato: Bidentate ligand with 2 donor sites (nitrogen and oxygen).
3. Diethylene triamine: Tridentate ligand with 3 donor sites (three nitrogen atoms).
4. EDTA: Hexadentate ligand with 6 donor sites (two nitrogen and four oxygen atoms).

The donor sites are 2, 2, 3, and 6, making Option B correct.

Jahn Teller Distortion in Complexes

• Jahn Teller distortion is a phenomenon that occurs in octahedral complexes with degenerate electronic states. It is caused by the interaction between the electrons in the degenerate orbitals and the molecular vibrations, leading to a distortion in the complex.

Complexes with Jahn Teller Distortion

• Among the given complexes, K₄[Fe(CN)₆], K₃[Co(CN)₆], and K₄[Mn(CN)₆], Jahn Teller distortion is expected only in K₄[Mn(CN)₆].
• K₄[Fe(CN)₆] and K₃[Co(CN)₆] have d⁶ electronic configuration which is not degenerate and hence do not show Jahn Teller distortion. On the other hand, K₄[Mn(CN)₆] has d⁵ electronic configuration which is degenerate. Hence, it shows Jahn Teller distortion.

In conclusion, Jahn Teller distortion is a phenomenon that occurs only in octahedral complexes with degenerate electronic states. Among the given complexes, K₄[Mn(CN)₆] is the only one with degenerate electronic states and hence shows Jahn Teller distortion.

Which matches Option a) 2.4 ∆_o, 0.00 BM and 0.4 ∆_o, 4.90 BM

Electronic Configurations with Orbital Angular Momentum Contribution in Octahedral Environment

In an octahedral environment, the d-orbitals split into two sets of different energies due to the crystal field effect. The lower energy set, called the t²g set, consists of three orbitals (dxy, dyz, and dxz), while the higher energy set, called the eg set, consists of two orbitals (dx²-y² and dz²).

The electronic configurations that have orbital angular momentum contribution in an octahedral environment are d¹ and d². Let's understand why:

A. d¹ and high spin d⁴:
- In an octahedral environment, a d¹ configuration means that only one electron occupies the t²g set.
- The electron occupies one of the three t²g orbitals (dxy, dyz, or dxz).
- Since there is only one electron, there is no orbital angular momentum contribution.

B. d¹ and d²:
- In an octahedral environment, a d² configuration means that two electrons occupy the t²g set.
- The two electrons can occupy any two of the three t²g orbitals (dxy, dyz, or dxz).
- The presence of two electrons in the t²g set introduces orbital angular momentum contribution.
- The two electrons have opposite spins, resulting in a net orbital angular momentum.

C. d² and high spin d⁵:
- In an octahedral environment, a d² configuration means that two electrons occupy the t²g set.
- The two electrons can occupy any two of the three t²g orbitals (dxy, dyz, or dxz).
- The presence of two electrons in the t²g set introduces orbital angular momentum contribution.
- The two electrons have opposite spins, resulting in a net orbital angular momentum.
- A d⁵ configuration means that five electrons occupy the t²g and eg sets.
- The five electrons can occupy any combination of the t²g and eg orbitals.
- However, the presence of five electrons in the d orbitals does not introduce additional orbital angular momentum contribution.

D. high spin d⁴ and high spin d⁶:
- A high spin d⁴ configuration means that four electrons occupy the t²g and eg sets.
- The four electrons can occupy any combination of the t²g and eg orbitals.
- However, the presence of four electrons in the d orbitals does not introduce additional orbital angular momentum contribution.
- Similarly, a high spin d⁶ configuration means that six electrons occupy the t²g and eg sets, but it also does not introduce additional orbital angular momentum contribution.

Therefore, the only electronic configurations that have orbital angular momentum contribution in an octahedral environment are d¹ and d², making option B the correct answer.

Explanation:

The colour of potassium dichromate is due to Ligand to metal charge transfer.

- Potassium dichromate (K2Cr2O7) is an inorganic compound that contains the dichromate ion (Cr2O7 2-) as a ligand and potassium ions (K+) as the metal.
- When light interacts with a compound, it can be absorbed by the electrons present in the compound's atoms or ions.
- In the case of potassium dichromate, the dichromate ion (Cr2O7 2-) acts as a ligand and donates a pair of electrons to the central chromium ion (Cr).
- The dichromate ion has a deep orange color due to the presence of multiple double bonds between the oxygen and chromium atoms.
- When light passes through potassium dichromate, the electrons in the oxygen atoms of the dichromate ion absorb certain wavelengths of light.
- These absorbed wavelengths correspond to the complementary color of orange, which is blue.
- As a result, the transmitted light appears blue, giving potassium dichromate its characteristic color.

To summarize:

- The colour of potassium dichromate is due to ligand to metal charge transfer.
- The dichromate ion acts as a ligand and donates electrons to the central chromium ion.
- The absorption of specific wavelengths of light by the ligand-metal complex leads to the observed color of potassium dichromate.