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:

The Tanabe-Sugano (TS) diagram is used to predict the electronic transitions of transition metal complexes and helps in understanding the effects of crystal field splitting and Racah parameters. TS diagrams are especially useful for high-spin and low-spin octahedral complexes of dⁿ configurations.

• The diagram shows the energy levels of excited states relative to the ground state as a function of the ligand field strength (represented as Dq/B).
• Each curve corresponds to a particular electronic state of the metal ion in the complex.
• For octahedral complexes, the transitions between energy levels are represented, and their corresponding absorption spectra can be studied.
• The splitting of d-orbitals in an octahedral field leads to different possible transitions between the ground state and excited states, influencing the peaks observed in the absorption spectrum.

Explanation: 

• In the Tanabe-Sugano diagram for d⁸ metal ions, such as [Ni(NH₃)₆]²⁺, we can observe several transitions due to the splitting of energy levels in the octahedral field.
• For this complex, the primary transition is from the ground state ³A_2g to the excited state ³E_g, but the shoulder peak is attributed to the transition from ³A_2g to the singlet state ¹E_g.
• The shoulder peak is weaker compared to the main absorption band because the transition from a triplet state to a singlet state is spin-forbidden, leading to lower intensity in the spectrum.
• In the case of [Ni(NH₃)₆]²⁺, the transition from ³A_2g to ¹E_g is observed as a shoulder peak in the absorption spectrum, confirming the presence of a weak transition between these states.

• 

Conclusion:

According to the TS diagram, the [Ni(NH₃)₆]²⁺ complex shows a shoulder peak due to the transition from ³A_2g to ¹E_g. Hence, the correct answer is Option 3.

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:

• Orgel diagrams are correlation diagrams that show the relative energies of electronic terms in transition metal complexes.
• Orgel diagrams will, however, show the number of spin-allowed transitions, along with their respective symmetry designations.

Explanation:

• The observed electronic spectra of d², d³, d⁷, and d⁸ octahedral and tetrahedral complexes ion, is as follows:

• In case of [Cr(en)3]3+ ion, the oxidation state of Cr is +3 and the electronic configuration is d3. En (Ethylenediamine) is a bidentate ligand thus the complex is octahedral.
• Hence, the complex [Cr(en)3]3+ will show 3 electronic transitions. These are ⁴A_2g →⁴T_2g, ⁴A_2g→⁴T_1g(F), and ⁴A_2g→4T1g(P).
• In case of trans-[Cr(en)2F2]+ ion, the oxidation state of Cr is also +3 and the electronic configuration is d3. But it changes it symmetry to D4h due to axial F.
• Splitting of energy level takes place due to change into symmetry.

  • Oh	D4h
    A1g	A1g
    A2g	B1g
    Eg	A1g + B1g
    T1g	A2g + Eg
    T2g	B2g + Eg

     

• 
• Thus, the complex ion trans-[Cr(en)2F2]+ will show 6 electronic transitions.

​Conclusion:

• ​Hence, the number of expected electronic transitions in [Cr(en)3]3+ and trans-[Cr(en)2F2]+ at 4 K is, respectively 3 and 6.