Coordination Compounds MCQ Quiz in मल्याळम - Objective Question with Answer for Coordination Compounds - സൗജന്യ PDF ഡൗൺലോഡ് ചെയ്യുക

Concept:

• The reaction of NiBr2 with two equivalents of PPh3 in CS2 is giving a red-coloured diamagnetic complex, [NiBr2 (PPh3)2 ].
• Nickel (Ni) in the complex [NiBr2 (PPh3)2 ]. Ni is in a +2 oxidation state. Hence the electronic configuration is [Ar]3d8.
• Out of eight 3d electrons, 6 are paired and two are unpaired.
• Octahedral and tetrahedral complexes of Ni(II) is thus paramagnetic, whereas the square planar ones are diamagnetic with no unpaired electrons.
• NiBr₂ is a yellow coloured complex. During its preparation, it is dissolved in ethanol and heated, the colour of the solution changes to green.
• Phosphine is dissolved in alcohol and then NiBr₂ is added to the solution, an immediate dark green precipitate is formed.

Explanation:

• Nickel (Ni) in the complex [NiBr2 (PPh3)2 ] is in a +2 oxidation state and is red coloured. It is also said that it is diamagnetic which indicates that the complex is square planar in nature.

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• The complex  [NiBr2 (PPh3)2 ], immediately transforms to a green-coloured complex and it is said that the complex now is paramagnetic.
• So, we know that Ni (II) complexes will be paramagnetic only when they are tetrahedral and octahedral, So as this was a four coordinated complex, the complex now formed is also four coordinated and tetrahedral.
• Electron pairs from Br and PPh3 are donated into one 4s and three 4p empty orbitals of Nickel giving us sp3 hybridization.

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• The structure is tetrahedral, and the number of unpaired electrons is two.

Hence, the geometry and the number of unpaired electrons in the green-coloured complex, respectively, are tetrahedral and 2.

Concept:

Water exchange reactions involve the replacement of coordinated water molecules in the inner coordination sphere of a metal complex with other water molecules from the surrounding environment. The rate of these reactions depends on various factors, which influence how quickly water molecules are exchanged. Key factors affecting the rate of water exchange include:

• Charge on the Metal Ion: Higher positive charges on the metal ion generally result in slower exchange rates due to stronger electrostatic attraction between the metal ion and the water molecules. For example, [Cr(OH₂)₆]³⁺ has a slower exchange rate than [Fe(OH₂)₆]²⁺ due to the higher charge.

• Size of the Metal Ion: Smaller metal ions with high charge density tend to form stronger bonds with water molecules, resulting in slower exchange rates. Larger ions or ions with lower charge densities allow for faster exchange.

• Electronic Configuration: The electronic structure of the metal ion can also affect the exchange rate. For example, transition metals with high field stabilization energy (HFSE) tend to have slower exchange rates due to more stable metal-water bonds.

• Ligand Field Strength: Complexes with strong ligand fields often exhibit slower exchange rates, as the bonding with water molecules becomes more stable, making it harder for water to be replaced.

Explanation: 

• [V(OH₂)₆]³⁺: Vanadium(III) has a relatively fast water exchange rate among the given options due to lower ligand field stabilization energy and larger ionic radius, which reduces the strength of the metal-water bond.

• [Fe(OH₂)₆]³⁺: Iron(III) has a moderate exchange rate as it has a smaller radius and higher charge than vanadium, leading to a stronger metal-water bond.

• [Cr(OH₂)₆]³⁺: Chromium(III) has the slowest exchange rate among the three because it has a higher ligand field stabilization energy, resulting in a very stable metal-water bond and thus a slower water exchange rate.

Conclusion:

The correct option is: 4) [V(OH₂)₆]³⁺ > [Fe(OH₂)₆]³⁺ > [Cr(OH₂)₆]³⁺. This order reflects the rate of water exchange, with [V(OH₂)₆]³⁺ having the fastest exchange rate and [Cr(OH₂)₆]³⁺ the slowest.

Concept:

Crystal Field Splitting (CFS) is the process where degenerate orbitals of metal cations split into sets of different energy levels when placed in a ligand field. In trigonal bipyramidal geometry, the metal ion is surrounded by five ligands, resulting in distinct energy levels for the d-orbitals due to electrostatic interactions with the ligands.

• In trigonal bipyramidal geometry: The d-orbitals split into three sets: one non-bonding set and two sets of bonding orbitals with different energy levels.

• Non-bonding orbitals remain unaffected by the ligand field, as they do not interact with the ligands directly. These orbitals generally have lower energy and are not involved in bonding or interactions with ligands.

• In trigonal bipyramidal geometry, the d_yz and d_zx orbitals are the non-bonding orbitals, as they are oriented in such a way that they do not face the ligands directly, thus experiencing minimal repulsion.

Explanation: 

• In a trigonal bipyramidal geometry, the five d-orbitals of a transition metal ion split into three sets: bonding orbitals, anti-bonding orbitals, and non-bonding orbitals. The non-bonding orbitals d_yz and d_zx do not face the ligands directly, resulting in minimal interaction with the ligands i.e. orbital with least energy after crystal field splitting act as non bonding orbitals.

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• As these orbitals remain non-bonding, their energy levels are lower compared to bonding orbitals, which face the ligands and experience higher repulsion.

Conclusion:

The non-bonding orbitals in trigonal bipyramidal geometry are d_yz and d_zx. Hence, the correct answer is Option 3.

The correct answer is 6 and Paramagnetic 

Concept:-

• Atomic Radii / Atomic Size: The size of an atom is a fundamental property that influences its behavior in chemical reactions. Generally, atoms increase in size down a group in the periodic table due to the addition of electron shells.
• Molecular Orbital Theory: It can be used to predict the bond order or the number of bonds in a molecule, as well as its magnetic properties. In general, a molecule is paramagnetic if its electron configuration includes unpaired electrons and diamagnetic if all electrons are paired.
• Valence Electrons: The electrons in the outermost shell of an atom are known as valence electrons. These electrons are involved in forming chemical bonds with other atoms. In transition metals, the valence electrons can be in both the outermost shell and in the inner d or f orbitals.
• Complex Ions and Coordination Chemistry: A complex ion has a metal ion at its center with several other molecules or ions surrounding it. These can use d-orbitals for binding with ligands which is a concept in coordination chemistry.

Explanation:-

The number of Metal-Metal (M-M) bonds (σ-bonds only), generally known as the "18-electron rule" for-transition metals.

M-M bond = (18n - TVE) / 2

where, n = total number of metal atoms TVE = Total Valence Electrons

• [Cr₂Cl₉]₃- : Cr is in group 6 so it has 6 valence electrons. For two Cr atoms, we therefore have 26 = 12 valence electrons. There are nine negatively charged Cl- ligands, which contributes 9 electrons. The overall complex has a -3 charge which contributes an additional 3 electrons. So, TVE for this complex ion is 12 + 9 + 3 = 24 electrons.
• Substituting into the formula gives M-M bond = (18x2 - 24) / 2 = 6.

Conclusion:-

So, the correct answer is 6 and Paramagnetic.

Concept:

• Orgel Diagrams are actually correlation diagram that represents relative energy levels of electronic terms for octahedral and tetrahedral complexes.
• Orgel Diagrams are used to represent a transition in metal complexes.
• It is only applicable for high-spin complexes
• Transitions represented by the orgel Diagram occur only from the ground state.

Explanation:

• [Ni(H2O)6]2+ is a weak field octahedral complex as H2O is weak filed ligand
• Now, the ground state of Ni2+(d8) is t2g6eg2
• Total Orbital Momentum, L = |-1-2| = 3 ,i.e., F
• Total Spin Momentum, S =  = 1 (number of unpaired electron is 2)
• Spin Multiplicity = 2S+1 = = 3
• The, ground state term is = 3F
• The observed electronic spectra of d2, d3, d7, and d8 octahedral and tetrahedral complexes ion, is as follows:

• According to the Orgel diagram, a transition occurs only from the ground state.
• The ground state for the d8 octahedral complex is A2g.
• Therefore, the possible transitions are,

³T2g(F)←³A2g(F),

³T1g(F)← ³A2g(F),

and ³T1g (P)← ³A2g(F)

• The energy gap is highest for the transition ³T1g (P)←3A2g(F), followed by 3T1g(F)← 3A2g(F), and is lowest for 3T2g(F)←3A2g(F).
• Now, the energy gap between the two energy levels is inversely proportional to wavelength.
• So the three bands A (~400 nm), B (~690 nm) and C (~1070 nm) will be

A (~400 nm) = 3T1g (P)←3A2g(F),

B (~690 nm) = 3T1g(F)← 3A2g(F)

C (~1070 nm) = 3T2g(F)←3A2g(F)

​Conclusion:

• Hence, the transitions assigned to A, B, and C, respectively, are

T1g(P) ← A2g, T1g ← A2g, and T2g ← A2g

Concept:

The magnetic moment of coordination compounds is a property that provides insight into the magnetic behavior of these complex molecules. Coordination compounds are composed of a central metal atom or ion bonded to surrounding ligands. The magnetic moment is a measure of the magnetic strength and orientation of the electron distribution within the compound.

Explanation:

\(u = {g\sqrt{J(J+1)}}\)

The above equation is used to calculate the magnetic moment of heavier atoms. Eg. F block elements.

Where, \(g = {1+{S(S+1) - L(L+1) +J(J+1)\over 2J(J+1)}}\)

S = 1/2 +1/2 +1/2+ 1/2 +1/2

S = 5/2

L = (3 x 1) +(2 x 1) + (1 x 1) + (0 x 1) + (-1 x 1)

L = 5

J = |L - S|.  As f orbital is less than half filled

J = |5 - 5/2|

J = 5/2

Putting the values of S, L and J in the above equation of g

Hence, g = 2/7

Now, putting the value of g to calculate magnetic moment of f⁵ ion.

\(u ={ {2 \over7}}{\sqrt{{5\over2}({5 \over2}+1)}}\)

\(u = {\sqrt {35}\over 7}\)

Conclusion:

The magnetic moment of f⁵ ion is \({\sqrt {35}\over 7}\).

Concept:-

Electronic spectra: emission

The energy of the absorbed radiation corresponds to the energy of a transition from ground to an excited state. Selection rules for electronic spectroscopy only allow transitions between states of the same multiplicity. Thus, excitation may occur from a singlet ground state (S0) to the singlet first excited state (S1). Decay of the excited state back to the ground state may take place by:

• Radiative decay (i.e. the emission of electromagnetic radiation),
• Non-radiative decay in which thermal energy is lost, or
• Non-radiative intersystem crossing to a triplet state (T1 in the Figure below represents the lowest energy triplet state)

• These processes compete with each other.
• Emission without a change in multiplicity is called fluorescence, while phosphorescence refers to an emission in which there is a change in multiplicity.
• The excitation and decay processes are represented using a Jablonski diagram in which radiative transitions are represented by straight arrows and nonradiative transitions by wavy arrows.
• It follows from the Jablonski diagram that the wavelength of light emitted in a phosphorescence will be longer wavelength (red-shifted) than the absorbed radiation.
• In fluorescence, the emitted light is also red-shifted, but to a much smaller extent; while the absorption involves the ground vibrational state of S0 and an excited vibrational state of S1, the emission is from the lowest vibrational state of S1.
• Since phosphorescence involves a spin-forbidden transition, the lifetime of the excited state is often relatively long (nanoseconds to microseconds or longer). Fluorescence lifetimes (typically between singlet states) are shorter and usually lie in the picosecond to nanosecond range.

Explanation:-

In the Jablonski diagram given below, the initial excitation takes place from the singlet ground state to the second singlet excited state (S0 → S2).

• Event A involves non-radiative intersystem crossing from a singlet state (S₁) to a triplet state (T₂).
• Event B involves radiative emission from a  triplet state (T2) to another triplet state (T1) without a change in multiplicity. Thus, event B is Fluorescence.
• Event C involves radiative emission from a  triplet state (T1) to a singlet state (S0) with a change in multiplicity. Thus, event B is Phosphorescence.

Conclusion:-

• Hence, A: Inter system crossing, B: Fluorescence, C: Phosphorescence

Concept -

• In coordination compounds, valence bond theory (VBT) accounts for the hybridization type and inner or outer orbital complex type of a particular complex.

• Inner orbital complexes are composed of metal atoms or ions that use (n-1)d orbitals or inner shell d orbitals for the hybridization in the central metal atom or ion. Outer orbital complexes are composed of metal atoms or ions that use orbitals or outer shell d orbitals for the hybridization in the central metal atom or ion.

• Depending upon the type of hybridization, there are two possible ways in which the complexes with coordination number 4 can be formed- 
  • If the complex involves sp³ hybridization, it would have tetrahedral geometry
  • if it involves dsp² hybridization, it would have square planar geometry.

Explanation-

• In [Ni(CO)4], Ni is in the zero oxidation state i.e. it has the configuration of 3d⁸ 4s². But since CO is a strong field ligand, all the 10 electrons are pushed into a 3d orbital and get paired up. The empty 4s and three 4p orbitals undergo sp³ hybridization and thus [Ni(CO)4] has tetrahedral geometry. The complex [Ni(CO)4] is diamagnetic.
• In [Ni(CN)4]2-, Ni is in a +2 oxidation state i.e it has the configuration of 3d⁸ 4s⁰. So when a strong ligand such as CN^- approaches Ni⁺², it has one vacant d orbital available for hybridization and thus it undergoes dsp² hybridization and will have square planar geometry.
• The hybridization for [Ni(CN)4]2- complex is shown below:

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• In [NiCl4]2-, Ni is in the +2 oxidation state i.e. it has the configuration of 3d⁸ 4s⁰. But since Cl^- is a weak field ligand, the 3d electrons do not pair up and therefore d orbitals are not available for hybridization. The empty 4s and three 4p orbitals undergo sp³ hybridization and, thus [NiCl4] ^2- has a tetrahedral geometry.
• In [MnCl4]-, Mn is in the +3 oxidation state i.e. it has the configuration of 3d4 4s0. But since Cl^- is a weak field ligand, the 3d electrons do not pair up and therefore d orbitals are not available for hybridization. The empty 4s and three 4p orbitals undergo sp3 hybridization and, thus [MnCl4] 2- has a tetrahedral geometry and it is paramagnetic.

Conclusion -

• [Ni(CN)4]2- (diamagnetic) undergoes dsp² hybridization and thus does not have tetrahedral geometry instead has square planar geometry.
• Hence, the correct option is option B.

Concept:

• The cobalt ore [Co3(AsO4)2∙H2O] given will dissociate to give us metallic cobalt as:

\([Co_3(AsO_4)_2 \cdot H_2O] → 3Co^{2+} + 2AsO_4^{3-}\)

1 mol ore → 3 mol of Co

3 mol Co → 1 mol ore

• So from the above equation, we see that we get three moles of cobalt from one mole of the ore, so the quantity of ore required to form 1 kg of pure cobalt can easily be calculated.

Calculation:

• Let us calculate the molar mass of the ore first.
• The molar mass of the ore [Co3(AsO4)₂∙H2O] = 3 × 59 + (2 × 75 + 4 × 16) × 2 + 18 = 623 g/mol.
• Now mass of 3 moles of Cobalt = 3 × 59 = 177g
• We know that one mole of ore gives three moles of Cobalt, which means that 177 gms of cobalt is obtained from 623g/mol of ore.

Hence, 1 kg of cobalt will be obtained from:

\(Mass= {473\over 177 }{\times 1000\over1000} =2.67kg\)

Hence, the mass of ore required = 2.67kg.

Concept:

• The reaction at an octahedral metal site in an aqueous medium involves the:
  • Either replacement of the coordinated ligand with water-hydrolysis, or,
  • Replacement of coordinated water by other ligands, including an exchange in water molecules with the same ligand.
• The substitution reaction may involve two pathways:
  • Dissociative:  When the leaving group leaves the reaction before any incoming ligand occurs.

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• Associative: The entering group binds to the reaction centre before any bond weakening takes place of the Leaving group.

• Interchange mechanism: The leaving group and the entering group exchanges in a single step forming an activated complex but not a transition step.
• The 1st transition series metals involve the given pathways:

Explanation:

• We see that the Cobalt 2+ complex will follow the dissociative mechanism. Metal complexes with 18 electrons or more generally undergo a dissociative mechanism.
• In the Dissociative mechanism, an anion first dissociates from the metal complex resulting in an intermediate with a lower coordination number.
• The lost ligand is then replaced by the incoming nucleophile. The dissociative step is the rate-determining step.
• The reaction mechanism is similar to the SN¹ reaction and does not depend upon the concentration of the incoming nucleophile.
• The general reaction mechanism can be represented as follows:

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• The dissociative step is the rate-determining step and the rate varies with the strength of the Metal to leaving group bond or M-X bond strength and steric factors.
• The rate thus only depends on the starting complex and not the incoming nucleophile Y. The rate equation is:

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• Hence, our complex [Co(CN)5Cl]3– will follow a Dissociative mechanism as: 

[Co(CN)5Cl]3– → [Co(CN)5]²– + OH^- →  [Co(CN)5 (OH)]3– 

• The rate thus depends on the complex [Co(CN)5Cl]3–  and not on the concentration of OH^- ion.

Hence, the rate of the substitution reaction of [Co(CN)5Cl]3– with OH– to give [Co(CN)5 (OH)]3– depends on the concentration of [Co(CN)5Cl]3– only.