Spectral MCQ Quiz in தமிழ் - Objective Question with Answer for Spectral - இலவச PDF ஐப் பதிவிறக்கவும்

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:

• [V(H2O)6]3+ is a weak field octahedral complex as H₂O is weak filed ligand
• Now, the ground state of V3+(d2) is t2g2eg0
• Total Orbital Momentum, L = 1+2 = 3 ,i.e., F
• Total Spin Momentum, S = \(\frac{2}{2}\) = 1 (number of unpaired electron is 2)
• Spin Multiplicity = 2S+1 = \(2\times1+1\)= 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 d² octahedral complex is T_1g.
• Therefore, any transition other than the transition from T1g will not be considered.
• So the two bands one at 17, 800 (v1) and the second at 25, 700 (v2) cm–1 will be

v1 = 3T1g (F) → 3T2g (F),

v2 = 3T1g (F) → 3T1g (P)

​Conclusion:

In its electronic spectrum, [V(H2O)6]3+ exhibits two absorption bands, one at 17, 800 (v1) and the second at 25, 700 (v2) cm–1. The correct assignment of these bands, respectively, is

v1 = 3T1g (F) → 3T2g (F),

v2 = 3T1g (F) → 3T1g (P)

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 the case of Cr3+ ion, the oxidation state of Cr is +3 and the electronic configuration is d3 (high spin complex).
• First, we will find the ground state of Cr3+(d3) 

• Total Orbital Momentum, L = 1+2+0 =3 ,i.e., F
• Total Spin Momentum, S = \(\frac{3}{2}\)

Spin Multiplicity = 2S+1 = \(2\times\frac{3}{2}+1\)= 4

ground state term is = 4F

• Thus, the ground state of high-spin octahedral Cr(III) complexes is 4A2g 
• Thus, the ground state of high-spin tetrahedral Cr(III) complexes is 4T1 

​Conclusion:

• ​Hence, the ground state of high-spin octahedral and tetrahedral Cr(III) complexes are respectively:  4A2g and 4T1 

Concept:

Russell-Saunders Coupling: -

• The atomic term symbol for a particular electronic configuration is 2S+1LJ. Where 2S+1 is the spin multiplicity, L is the total orbital angular momentum and J is the total angular momentum.
• The total spin angular momentum (S) measures the entire resultant z component of the electron’s spins. For two electrons having spin quantum number s1 and s2, the quantum number S can take the values

| s1+ s2|, | s1+ s2-1|……. | s1- s2|.

• For a metal complex, the spin-only magnetic moment value can be calculated by using the formula:

​\({\rm{\mu = }}\sqrt {{\rm{n(n + 2)}}} {\rm{ B}}{\rm{.M}}\) , where n is the number of unpaired electrons and B.M is Bohr magneton a unit of the magnetic moment.

Explanation:

• The observed electronic spectra of d1, d4, d6, and d9 octahedral and tetrahedral complexes ion, is as follows:

• The electron configuration for the high-spin d6 (octahedral complex) ion is (t2g4eg2).
• The number of unpaired electrons is 4.
• So, S = \({4 \over 2}\)

= 2

• Spin, multiplicity

= (2S+1)

\( = 2 \times 2 + 1\) 

= 5

• From the above orgel diagram for the d6 octahedral complex, we can conclude that the ground state of the high-spin d6 octahedral complex is 5T2g.
• The electronic transition for this complex will be 5Eg ←  5T2g
• Now, the magnetic moment of the d6 octahedral complex is

\({\rm{\mu = }}\sqrt {{\rm{n(n + 2)}}} {\rm{ B}}{\rm{.M}}\)

\({\rm{\mu = }}\sqrt {{\rm{4(4 + 2)}}} {\rm{ B}}{\rm{.M}}\) (As, n = 4)

\({\rm{\mu = }}\sqrt {{\rm{24}}} {\rm{ B}}{\rm{.M}}\)

= 4.9 B.M

Conclusion:

• Hence, the spin‐only magnetic moment (B.M.) and the electronic transition for this complex, respectively, are

4.9 and 5Eg ←  5T2g

Concept:

Russell-Saunders Coupling: -

• The atomic term symbol for a particular electronic configuration is 2S+1LJ. Where 2S+1 is the spin multiplicity, L is the total orbital angular momentum and J is the total angular momentum.
• The total spin angular momentum (S) measures the entire resultant z component of the electron’s spins. For two electrons having spin quantum number s1 and s2, the quantum number S can take the values

| s1+ s2|, | s1+ s2-1|……. | s1- s2|.

Explanation:

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

• The electron configuration for high-spin Co(II) (octahedral complex) ion is d⁷ (t_2g⁵e_g²).
• The number of unpaired electrons is 3
• So, S = \({3 \over 2}\).
• Spin, multiplicity

= (2S+1)

\( = 2 \times {3 \over 2} + 1\) 

= 4 

• From the above orgel diagram for d7 octahedral complex, we can conclude that the ground state of high-spin d7 octahedral complex is 4T1g.
• The electron configuration for high-spin Co(II) (tetrahedral complex) ion is also d7 (e4t3).
• The number of unpaired electrons is 3
• So, S = \({3 \over 2}\).
• Spin, multiplicity

= (2S+1)

\( = 2 \times {3 \over 2} + 1\) 

= 4 

• From the above orgel diagram for d7 tetrahedral complex, we can conclude that the ground state of high-spin d7 tetrahedral complex is 4A2.

Conclusion:

• Hence, the ground state of high-spin octahedral and tetrahedral Co(II) complexes are respectively: 4T1g and 4A2

Charge Transfer complex:

• A charge transfer complex or an electron donor-acceptor complex describes a type of assembly of two or more molecules or ions.
• Charge transfer complex is of 4 types, these are 
  • Ligand to metal charge transfer (LMCT),
  • Metal to ligand charge transfer (MLCT),
  • Ligand-to-ligand charge transfer (LLCT), and
  • Metal-to-metal charge transfer (MMCT).​
• It is a correlation between the energies of charge transfer absorptions and the electrochemical properties of metals and ligands.
• Charge transfer transitions are not restricted by the selection rules that involve ‘d–d ’ transitions, The probability of these electronic transitions is very high, and the absorption bands are therefore intense.

​LMCT:

• The transfer of an electron from an orbital primarily ligand character to one with primarily metal character is known as ligand-to-metal charge transfer or LMCT.
• LMCT occurs if a ligand that can be easily oxidized is bound to a metal center in a high oxidation state, which is readily reduced.

Explanation:-

• In [IrBr6]2−, the oxidation state of Ir is +4 and Ir⁴⁺ is in low spin state.
• [IrBr6]2− is an octahedral complex and the electronic configuration, of Ir4+(d5) is t_2g⁵e_g⁰.
• Given below is a MO diagram for ML₆ octahedral complex

• The 5 t2g electrons of Ir4+(d5 or  t2g5eg0) will occupy the t_2g non-bonding orbitals (HOMO).
• This will result in two LMCT transitions from 

σ → t2g and σ → eg.

Conclusion:-

• Hence, Two, ligand → metal (σ → t2g and σ → eg)  is the correct answer.

Concept:

Phosphorescence is a type of photoluminescence where light emission occurs after a delay, due to forbidden transitions in the excited states of a molecule or complex. In transition metal complexes, phosphorescence often arises from intersystem crossing or specific electronic transitions between d-orbitals. Key concepts for understanding phosphorescence in transition metal complexes include:

• Electronic Transitions: Certain transitions, such as those involving a change in spin multiplicity (e.g., from triplet to singlet), are forbidden but can occur with a low probability, resulting in delayed emission (phosphorescence).
• Intersystem Crossing: This is a non-radiative transition between states of different multiplicity (e.g., singlet to triplet), allowing for the transition to a lower-energy state, which then emits phosphorescent light.
• Internal Conversion: This is a non-radiative process in which an excited state of a molecule relaxes to a lower electronic state without emitting radiation.

Explanation: 

• Cr³⁺ shows red phosphorescence due to the transition from ²E_g to ⁴A_2g. This is a correct statement as such transitions in Cr³⁺ often emit red light.
  • ​
• [Fe(bpy)₃]²⁺ shows orange phosphorescence due to intersystem crossing. This is also correct; intersystem crossing allows Fe(II) complexes to exhibit phosphorescence due to delayed emission.
  • ​

Conclusion:

The correct option is: 1) A and B. Statements A and B are correct, as they accurately describe the phosphorescent behaviors of Cr³⁺ and [Fe(bpy)₃]²⁺ due to specific transitions and intersystem crossing.

Concept:

The number of peaks in the electronic spectrum of a transition metal ion depends on the possible electronic transitions within the d-orbital split under different environments. For V³⁺, which has a d² configuration, three d-d transitions are theoretically possible. However, the observed number of peaks may vary based on the surrounding environment (aqueous or solid) and interference from other types of transitions, such as charge transfer (CT) bands.

In an aqueous solution, additional factors, such as the presence of water molecules and charge transfer bands, can obscure certain peaks, reducing the number of observed spectral bands.

Key points about the Orgel diagram for a d² electronic configuration:

• Electronic Transitions: Orgel diagrams illustrate the energy levels and possible transitions for high-spin d² systems, such as V³⁺. For a d² configuration, three primary transitions are possible, corresponding to excitations within the t_2g and e_g orbitals in an octahedral field.
• High-Spin Configurations: Orgel diagrams are particularly useful for high-spin configurations, where electron-electron repulsions within the d-orbitals create specific energy levels. In a d² system, the ground term is ³T_1g, with excited terms ³T_2g and ³A_2g.
• Interpretation of Spectral Bands: The Orgel diagram helps predict the number of spectral bands in d² systems, as it visually represents the possible electronic transitions that contribute to the observed spectrum.
  • ​

Explanation: 

• V³⁺(aq): In the aqueous state, V³⁺ is surrounded by water ligands, which can create a charge transfer (CT) band that overlaps with one of the d-d transitions. This obscuring effect leads to the observation of only 2 spectral bands in the visible region.
  • ​
• V³⁺(solid): In the solid state, there is no overlapping charge transfer band, and all three d-d transitions are visible, resulting in the observation of 3 spectral bands.
  • ​

Conclusion:

The correct option is: 4) 2 and 3. This indicates that V³⁺(aq) has 2 observable peaks, while V³⁺(solid) has 3 observable peaks due to the absence of CT interference in the solid state.

Concept:

In Ligand-to-Metal Charge Transfer (LMCT) transitions, electrons are transferred from the ligand (usually a donor atom such as halide or oxide) to the metal center. The energy of the LMCT band depends on various factors, including the nature of the ligand and the metal ion. Higher LMCT energy generally corresponds to ligands that are stronger electron donors or metals with higher oxidation states. Key factors affecting LMCT energy include:

• Electronegativity of Ligand: Ligands with lower electronegativity (e.g., I^- < Br^- < Cl^-) have a higher tendency to donate electrons, resulting in lower LMCT energy. Thus, LMCT energy decreases in the order of Cl > Br > I.
• Oxidation State of Metal: A higher oxidation state on the metal ion increases the energy of the LMCT band, as the metal ion attracts electrons more strongly, requiring higher energy for electron transfer from the ligand.
• Metal-Ligand Bond Strength: Stronger metal-ligand bonds result in higher LMCT energy because more energy is required to transfer an electron from the ligand to the metal. In octahedral complexes, bond strength often depends on the ligand type.
• Polarizability of the Ligand: More polarizable ligands (e.g., I^- is more polarizable than Cl^-) have lower LMCT energy as they can more easily donate electron density to the metal center.

Explanation: 

• For the given complexes, the LMCT energy follows the trend based on the electronegativity and polarizability of the halide ligands.
• [OsCl₆]^2-: Chloride (Cl^-) is less polarizable and more electronegative than Br^- and I^-, resulting in the highest LMCT energy for this complex.
• [OsBr₆]^2-: Bromide (Br^-) is less electronegative and more polarizable than chloride, but more electronegative than iodide, giving it intermediate LMCT energy.
• [OsI₆]^2-: Iodide (I^-) is the most polarizable and least electronegative, leading to the lowest LMCT energy among the three complexes.

Conclusion:

The correct option is: 3) [OsCl₆]^2- > [OsBr₆]^2- > [OsI₆]^2-. This order reflects the relative LMCT energy for the complexes, with chloride having the highest energy and iodide the lowest.

Concept:

The spectroscopic ground state term symbols of transition metal complexes can be determined using crystal field theory and Orgel diagrams. Orgel diagrams are useful in understanding the electronic transitions in transition metal complexes, particularly for d² to d⁸ electronic configurations.

Crystal Field Theory:

• Transition metal ions in a crystal field experience splitting of their d-orbitals due to the interaction with surrounding ligands.
• The nature of the splitting depends on the geometry of the complex (octahedral, tetrahedral, etc.).

Orgel Diagrams:

• Orgel diagrams provide a way to visualize the electronic transitions between different term symbols (energy states) of metal ions in an octahedral or tetrahedral ligand field.
• For d⁷ electronic configuration, cobalt(II) in different geometries will have different term symbols representing its ground state and excited states.
• In an octahedral field (such as in Co(H₂O)₆²⁺), the ground state term symbol is typically identified as the lowest energy state after d-orbital splitting.
• In a tetrahedral field (such as in CoCl₄²⁻), the ground state term symbol is similarly identified according to the splitting pattern in tetrahedral geometry.
• Orgel Diagram:
  • ​

Explanation:

For Co(H₂O)₆²⁺, which is an octahedral complex:

• Cobalt(II) has a t_2g⁵e_g² configuration with 3 unpaired electrons.
• s = 3/2, S = 2s +1 = 4
• In an octahedral crystal field, the d-orbitals split into t_2g and e_g orbitals.
• According to the orgel diagram, the ground state term symbol for a d⁷ configuration in an octahedral field is typically ⁴T_1g.

For CoCl₄²⁻, which is a tetrahedral complex:

• Cobalt(II) also has a e⁴t₂³ configuration with 3 unpaired electrons.
• Spin Multiplicity = 2s+1 = 4(s = 3/2).
• In a tetrahedral crystal field, the d-orbitals split into e and t₂ orbitals.
• The ground state term symbol for a d⁷ configuration in a tetrahedral field is typically ⁴A₂.

Conclusion:

The spectroscopic ground state term symbols of cobalt ions in Co(H₂O)₆²⁺ and CoCl₄²⁻, respectively, are ⁴T_1g and ⁴A₂.

The correct answer is option 2.

Concept:

Term symbols provide information about the quantum state of an electron in an atom. They are represented in the form ^2S+1L_J, where:

• S: Total spin quantum number, which is the sum of the spins of all electrons in the atom or ion.

• L: Total orbital angular momentum quantum number, represented by letters (S, P, D, F, etc.) corresponding to L values 0, 1, 2, 3, and so on.

• J: Total angular momentum quantum number, which is the vector sum of L and S (J = L + S, L + S - 1, ..., |L - S|).

The term symbol provides information about the energy, spin, and angular momentum of the electron configuration in an atom.

Ground State Term Symbol:

The ground state term symbol represents the lowest energy state of an atom's electron configuration, determined by applying Hund's rules:

• Maximize S: The term with the highest spin multiplicity (2S + 1) has the lowest energy.
• Maximize L: For terms with the same spin multiplicity, the term with the highest L value has the lowest energy.
• J Value: For subshells less than half-filled, the lowest J value has the lowest energy; for more than half-filled, the highest J value is the lowest in energy.

Selection Rules for Spectral Transitions:

In atomic transitions, certain selection rules must be satisfied for a transition to be allowed:

• Change in L: ΔL = ±1 (the orbital angular momentum quantum number must change by 1).
• Change in J: ΔJ = 0, ±1, but J = 0 ↔ J = 0 transitions are forbidden.
• Change in S: ΔS = 0 (the spin quantum number must remain the same).

Explanation: 

• In the given question, the term symbol ⁴D_5/2 indicates:

  • L = 2 (D state), S = 3/2 (multiplicity = 4), and J = 5/2.

• Option 1 states it corresponds to 𝐿 = 2, 𝑆 = 1/2, and 𝐽 = 5/2 which is incorrect as S = 3/2 for the given state.

• Option 2 states it can originate from s1 p2 electronic configuration. From s1p2, L= 1. Thus this is also incorrect as L = 2 for the given state.

• Option 3 states it splits into five levels in the presence of magnetic field. This is incorrect as according to Zeeman effect it will splits into six levels.

• Option 4 states that the ⁴D_5/2 state can show a spectral transition to the ⁴P_3/2 state, which is correct based on the selection rules (ΔL = ±1, ΔJ = 0 or ±1, and ΔS = 0).

Conclusion:

The correct answer is: Option 4