The Transition Elements (d-Block) MCQ Quiz in தமிழ் - Objective Question with Answer for The Transition Elements (d-Block) - இலவச PDF ஐப் பதிவிறக்கவும்

Concept:-

• The molar absorption coefficient, molar extinction coefficient, or molar absorptivity (ε), is a measurement of how strongly a chemical species absorbs light at a given wavelength.
• The electronic transition depends on two rules-

​(A) spin selection rule and (B) Laporte selection rule.

(A) The Spin Selection Rule :

• This rule forbids transitions between states with different total spin, and thus different spin multiplicity. This rule allows transitions only between states with the same total intrinsic spin (ΔS=0), and thus the same spin multiplicity value in the term symbol.

(B) La Porte Or orbital selection Rule: 

• In a centrosymmetric molecule, the transitions are allowed between orbitals having different symmetries with respect to inversion. Gerade to Ungerade are allowed and vice versa but 'g' to 'g' and 'u' to 'u' is not allowed.
• In other words, we can say that the transitions are allowed for 

Δ L = +/- 1.

• The La Porte selection rule is allowed in many cases where the center of inversion is relaxed. For example, when mixing of p and d orbitals occurs due to distortion in the molecule, transitions become allowed.
• The intensity of tetrahedral complexes is nearly 100 times as compared to similar octahedral complexes.

Explanation:

• In the case of [Ti(H2O)6]3+, the oxidation state of Ti is +3.
• The electronic configuration of Ti⁺³ is  d¹ (t_2g¹e_g⁰)
• On excitation, the single electron in the t2g orbital goes to the eg orbital and the spin number remains the same (follows ΔS=0).
• Thus, d-d transition observed in [Ti(H2O)6]3+ is spin allowded.
• During the d-d transition, it does not follows Δ L = +/- 1 rule.
• Thus, d-d transition is Laporte forbidden.

​Conclusion:-

• Hence, the d-d transition observed in [Ti(H2O)6]3+ is Laporte forbidden and spin allowed

Explanation:

Scandium (Sc) is a transition metal element that belongs to the 3rd period of the periodic table, and it has only one oxidation state, +3.

The colour of a transition metal ion is due to the presence of partially filled d orbitals, which can absorb certain wavelengths of visible light and reflect others, giving the ion its characteristic colour.

However, Scandium ion (Sc³⁺) does not have any partially filled d orbitals, as it has lost all three of its valence electrons to form the 3+ oxidation state. As a result, Sc³⁺ does not absorb any visible light and therefore appears colourless.

On the other hand, the other transition metal ions listed in the options all have partially filled d orbitals and exhibit characteristic colours in aqueous solutions or as solid compounds.

For example, V²⁺ (Vanadium ion) is blue-green, Mn²⁺ (Manganese ion) is pale pink, and Co³⁺ (Cobalt ion) is yellow.

Therefore, the correct answer is Option 1, Sc³⁺ (Scandium ion), which is the only colourless transition metal ion among the options given.

In general species having no unpaired electron is colourless. So Sc3+ has electronic configuration [Ar] 3d⁰4s⁰

So, it is colourless ion. 

CONCEPT:

dⁿ Electron Configuration in Coordination Complexes

• The electron configuration of the central metal ion in a coordination complex depends on its oxidation state.
• To determine the dⁿ configuration:
  • Identify the oxidation state of the metal ion in the complex.
  • Subtract the oxidation state from the total number of valence electrons in the neutral metal atom.
  • The remaining electrons represent the d-electron count (dⁿ configuration).

EXPLANATION:

• Analyzing each complex:
  • (A) [FeO₄]²⁻:
    • Fe oxidation state: +6
    • Fe atomic configuration: [Ar] 3d⁶ 4s²
    • After losing 6 electrons: [Ar] 3d² (d² configuration)
  • (B) [Mn(CN)₆]³⁻:
    • Mn oxidation state: +3
    • Mn atomic configuration: [Ar] 3d⁵ 4s²
    • After losing 3 electrons: [Ar] 3d⁴ (d⁴ configuration)
  • (C) [Fe(CN)₆]:
    • Fe oxidation state: +3
    • After losing 3 electrons: [Ar] 3d⁵ (d⁵ configuration)
  • (D) [Cr₂(O–C–Me)₄(H₂O)₂]:
    • Cr oxidation state: +2
    • Cr atomic configuration: [Ar] 3d⁵ 4s¹
    • After losing 2 electrons: [Ar] 3d⁴ (d⁴ configuration)
  • (E) [NiF₆]²⁻:
    • Ni oxidation state: +4
    • Ni atomic configuration: [Ar] 3d⁸ 4s²
    • After losing 4 electrons: [Ar] 3d⁶ (d⁶ configuration)
• d⁴ configuration is observed in:

  (B) [Mn(CN)₆]³⁻ and (D) [Cr₂(O–C–Me)₄(H₂O)₂]

Hence, the correct answer is: Option 3 (B and D only).

\( \text{Scandium (Z = 21)} \) forms colourless compounds.

\( \text{Scandium (Z = 21)} \) has valence shell electron configuration of \( 3d^1 \: 4s^2 \).

In its \( +3 \) oxidation state, it loses 3 electrons and has valence shell electron configuration of \( 3d^0 \: 4s^0 \). Since no unpaired electron is present, it forms colourless compounds.

CONCEPT:

Chemical Equilibrium and pH Influence

• This equilibrium between dichromate ion (Cr₂O₇²⁻) and chromate ion (CrO₄²⁻) is pH-dependent.
• The equilibrium can be represented by the following chemical equation: [\( \text{Cr}{2}\text{O}{7}^{2-} + \text{H}{2}\text{O} \leftrightarrow 2\text{CrO}{4}^{2-} + 2\text{H}^+\) ]
• According to Le Chatelier's Principle:
  • In acidic medium: High concentration of (H+) ions will shift the equilibrium to the left to produce more (\( \text{Cr}{2}\text{O}{7}^{2-}\) ) and reduce (H+) ions.
  • In basic medium: Presence of (OH-) ions will react with ( H+ ) ions to form water, thereby reducing (H+ ) concentration and shifting the equilibrium to the right, favoring the formation of ( \(\text{CrO}_4^{2-} \)).

EXPLANATION:

• In an acidic medium:
  • The equilibrium shifts to the left, favoring the formation of dichromate ions ((\( \text{Cr}{2}\text{O}{7}^{2-}\)).
• In a basic medium:
  • The equilibrium shifts to the right, favoring the formation of chromate ions \( \text{CrO}{4}^{2-}\).
• In a weakly acidic or neutral medium:
  • The equilibrium will be somewhere between the shifts observed in strongly acidic or basic media, but the shift is more pronounced in a basic medium for the formation of ( \(\text{CrO}{4}^{2-}\) ).

Conclusion:
The equilibrium ( \( \text{Cr}{2}\text{O}{7}^{2-} \leftrightarrow 2\text{CrO}_{4}^{2-}\) ) is shifted to the right in a basic medium.

Explanation:

Determining the Number of Unpaired Electrons

• Scandium (Sc): [Ar] 4s² 3d¹
  • Unpaired electrons: 1
• Titanium (Ti): [Ar] 4s² 3d²
  • Unpaired electrons: 2
• Vanadium (V): [Ar] 4s² 3d³
  • Unpaired electrons: 3
• Chromium (Cr): [Ar] 4s¹ 3d⁵
  • Unpaired electrons: 6
• Manganese (Mn): [Ar] 4s² 3d⁵
  • Unpaired electrons: 5

So, Sc (1) < Ti (2) < V (3) < Mn (5) < Cr (6)

The correct answer is: 4) (A) < (D) < (C) < (E) < (B)

The correct answer is Both A and R are true and R is the correct explanation of A

Concept:

Mixtures are combinations of two or more substances that are physically combined but not chemically bonded. In a mixture, the substances retain their individual properties and can be separated by physical means. Mixtures can vary widely in composition and properties, depending on the types and proportions of substances involved. There are two main types of mixtures:

• Homogeneous Mixtures: Also known as solutions, homogeneous mixtures have a uniform composition throughout. The individual components are evenly distributed and not easily distinguishable by the naked eye.
  • Examples include salt dissolved in water (creating a saline solution
• Heterogeneous Mixtures: Heterogeneous mixtures have a non-uniform composition, meaning that the individual components are not evenly distributed. The different substances may exist as distinct phases or layers within the mixture.
  • Examples include a mixture of sand and water.

Explanation:

• Chromatography is a separation technique based on differential affinities of components in a mixture for a stationary phase and a mobile phase. In chromatography, the mixture to be separated is dissolved in a solvent (the mobile phase) and passed through a stationary phase. The stationary phase can be a solid or a liquid, depending on the type of chromatography being used.
• The unique interactions between the components and the stationary and mobile phases enable their separation. Each component interacts differently with the stationary phase (which may be a solid support or a liquid) and the mobile phase (usually a solvent), leading to varying rates of movement and ultimately separation of the components.

Explanation:

As the atomic number of lanthanides increases, size of the lanthanide ion (La³⁺) decreases and a smaller lanthanide ion forms a stable complex with an o-hydroxy iso-butyrate ion. 

Hence, a Lu³⁺  o-hydroxy iso-butyrate ion complex comes out first and Eu³⁺comes out last, i.e., the order of seperation is Lu3+, Yb3+, Dy3+, Eu3+

Conclusion: the correct answer is option 1.

Explanation:

¹D ⇒ ¹D₂ ⇒ 5 electron arrangements

{|2, 0〉 |1, 0〉 |0, 0〉 |-1, 0〉 |-2, 0〉}

³P ⇒ ³P_2,1,0 ⇒ 9 electron arrangements

\(\rm \left\{\begin{matrix}|1,1〉&|1,0〉&|1,-1〉\\\ |0,1〉&|0,0〉&|0,-1〉\\\ |-1,1〉&|-1,0〉&|-1,-1〉\end{matrix}\right\}\)

¹S ⇒ ¹S₀ ⇒ 1 electron arrangements

{|0, 0〉}

According to Hund's rules, the p² configuration has a ground state term component ³P₀, and p⁴ configuration has a ground state from term component ³P₂.

Conclusion: Thus, the correct answer is option 4.

Concept:

The number of microstates (N) of a system corresponds to the total number of distinct arrangements for “e” number of electrons to be placed in “n” number of possible orbital positions.

The number of arrangements is possible in many orbitals s, p, d, etc. If we have electrons in the same orbitals as if we have two electrons in p- orbital, then we can write the formula as:

 ⁿC_r where n is the total number of electrons that orbital can possess and r is the number of electrons it is having for that case. These numbers of arrangements are called microstates.

\(^{n}C_{r}=\frac{n!}{r!\times\left ( n-r \right )!}\)

Explanation

For  6C2: \(^{n}C_{r}=\frac{6!}{2!\times\left ( 6-2 \right )!}\)

\(^{n}C_{r}=\frac{6\times5\times4\times3\times2\times1}{2\times1\times\left ( 4\times3\times2\times1 \right )}\) ⇒ 15