Transition Elements and Inner Transition Elements MCQ Quiz in తెలుగు - Objective Question with Answer for Transition Elements and Inner Transition Elements - ముఫ్త్ [PDF] డౌన్‌లోడ్ కరెన్

Concept: 

• The red color of Ruby is due to the d-d transition of Cr³⁺ ion in the  Al₂O₃ lattice. The mineral corundum is a crystalline form of alumina Al2O3. A pure crystal of alumina is colorless. However, if just 1% of the Al3³⁺+ ions are replaced with Cr³+ ions, the mineral becomes deep red in color. 
• Rubies are scientifically known as corundum, a rock-forming mineral and crystalline form of aluminium oxide Al2O3 which is two aluminium atoms and three oxygen atoms  in a close packed hexagonal structure.

Explanation:

• In pure corundum, all electrons are paired and there is no absorption of light. Once one out of every hundred aluminum atoms is replaced by chromium atoms, negatively charged oxygen ions surround the aluminum ion (which has donated 3 electrons), so a chromium atom must donate three electrons to become Cr³⁺ , replacing Al³⁺, in order for the charge to remain the same. In Al³⁺ there are no partially filled energy levels or orbitals. However, in Cr³⁺ there are partially filled energy levels or orbitals. It is these electrons that can be excited and that cause absorption of certain wavelengths of light, resulting in color.
• Orbitals have specific shapes and geometric configurations in space. In Cr3+ , there are three unpaired electrons in the outer energy level (which we will call 3d). This energy level can accommodate a total of 10 electrons, so there are unoccupied energy levels. So in this case there are three out of five orbitals partially filled. In an isolated Cr3+ ion, the fact that the five 3d orbitals point in different directions does not matter, and it would be colorless. However, in a ruby, the five 3d orbitals of the Cr3+ ion are surrounded by six oxygen ions in a distorted octahedral configuration. Interactions with the orbitals of the six oxygen ligands produce shifts in the energy levels of the individual orbitals.

Conclusion:

Allowed transitions between the energy levels of atoms of Cr3+ in a distorted octahedral field, the energy levels and transitions in ruby, and the resulting absorption spectrum and fluorescence of ruby. Thus, the correct answer is option 3.

Concept:

• Mercury is a lustrous liquid metal at room temperature.
• Hg(I) is paramagnetic and unstable. So, it exists in dimeric form,  [Hg2]²⁺
• Mercurous ion, that is, [Hg₂]²⁺ is a diamagnetic species

 .

Explanation:

Electronic configuration of Hg = [Xe] 4f¹⁴ 5d¹⁰ 6s²

Electronic configuration of Hg(I) = [Xe] 4f¹⁴5d¹⁰6s¹

Hg(I) has only one unpaired electron so its spin pairs with another Hg(I) unit to form dimer, [Hg2]²⁺

clearly, orbitals involved in bonding are s of both Hg(I) and bond order of species is 1

Conclusion:

Hence, the bond order and the orbitals involved in bonding in [Hg2]²⁺ are one; s and s respectively.

Concepts:

• Elements whose f-orbital getting filled up by electrons are called f-block elements.
  • They are divided into lanthanoids and actinoids. 
• The first series of elements are called lanthanides and include elements with atomic numbers beginning from 57 and ending at 71. These elements are non-radioactive.
• The second series of elements are called actinides and include elements with atomic numbers beginning from 89 and ending at 103. They are radioactive.
• Actinides are the elements whose atomic number is from 90 to 103.
• The members of the series are Thorium, Protactinium, Uranium, Neptunium, Plutonium, Americium, Curium, Berkelium, Californium, Einsteinium, Fermium, Mendelevium, Nobelium, Lawrencium.

Explanation:

• All the actinide elements are radioactive, hence the statement All the actinides are radioactive is correct.
• In the actinoids, the last electron goes into the 5f orbital.
• Now, the 5f orbital due to its large size and diffused nature has a very poor shielding effect.
• As a result of the poor shielding effect, much of the nuclear charge leaks from the electron cloud, as a result, the outer electrons face more nuclear charge of the nucleus.
• As the atomic number increases, the nuclear charge increases and as a result, the ionic radii or the atomic radii decrease from Thorium to Lawrencium.
• This is known as Actinoid contraction. Thus, the statement Actinides do not exhibit actinide contraction is incorrect.
• The poor shielding effect of 4f electrons causes Lanthanide contraction is Lanthanides.
• Comparing it to 5f orbital, as 5f orbitals are larger in size, its shielding effect is much poor, as a result of which the electrons in 5forbitals are less tightly bound. Also, it is much farther than the nucleus compared to 4f orbitals.
• Hence, the statement: The 5f electrons of actinides are bound less tightly than the 4f electrons is also correct.
• The transuranium compounds are the elements with an atomic number greater than 92.
• The transuranium compounds are all radioactive and are man-made elements.
• Hence, the statement: The trans uranium elements are prepared artificially is also correct.
• Hence, the correct options are: 1,2 and 3.

Concept:

Oxidation States in Compounds

• Oxidation state represents the charge of an atom in a compound, considering electrons are completely transferred in bonds.
• In compounds containing multiple oxidation states for the same element, the sum of oxidation states equals the net charge of the compound.
• Calcium hypochlorite, Ca(OCl)₂, consists of calcium ions (Ca²⁺) and two hypochlorite ions (OCl^-).

Explanation:

• Structure of Ca(OCl)₂:
  • Calcium ion: Oxidation state = +2.
  • Hypochlorite ion (OCl^-): Consists of oxygen and chlorine.
• Oxidation states of chlorine in OCl^-:
  • Oxygen typically has an oxidation state of -2.
  • For OCl^-, the sum of oxidation states must equal -1.
  • Let the oxidation state of chlorine be x. Solving: x + (-2) = -1 → x = +1.
• Overall Oxidation States in Ca(OCl)₂:
  • One chlorine in OCl^- has an oxidation state of +1.
  • Another chlorine has an oxidation state of -1 (in ionic form Cl^-).

Correct Answer: 4. +1, -1

Concept:

Lanthanides are a series of 15 metallic elements from lanthanum (La) to lutetium (Lu) with atomic numbers ranging from 57 to 71. These elements, also known as rare earth elements, have unique chemical properties that make them useful in various applications. Here are some key points:

• Oxidation States: The most common oxidation state for lanthanides is +3. However, some elements can also exhibit +2 and +4 oxidation states, depending on their specific electronic configurations and the chemical environment.
• Chemical Reactivity: Lanthanides are highly reactive metals, especially when finely divided. They react with water, acids, and most nonmetals. Their reactivity decreases gradually from lanthanum to lutetium.
• Coordination Chemistry: Lanthanides form complexes with a variety of ligands. Lanthanide(III) ions have a strong preference for oxygen-donor ligands due to their hard acid nature, according to the hard and soft acids and bases (HSAB) principle.
• Magnetic and Optical Properties: Many lanthanides exhibit unique magnetic and optical properties due to their unfilled 4f orbitals, making them valuable for applications in electronic devices, lasers, and phosphors.

Explanation:

1. Lanthanide organometallic compounds do not undergo oxidative addition reactions: This statement is correct. Lanthanide organometallic compounds typically do not participate in oxidative addition reactions due to the stable +3 oxidation state of lanthanides and the limited availability of low-energy acceptor orbitals.

   • ​​Also, lanthanides does not have two stable oxidation state differing by 2, so the lanthanides organometallic compounds does not show oxidative addition or reductive elimination but shows σ- bond metathesis type reactivity.

2. The oxidation state of Ln in LnI₃ is +3: This statement is incorrect. In lanthanide triiodides (LnI₃), the oxidation state of the lanthanide ion is +1 as the compound exist as \(Ln^{+1}I_3^-\).

3. Ln³⁺ prefers soft Lewis bases to form complexes: This statement is incorrect. According to the HSAB principle, Ln³⁺ ions are hard acids and thus prefer hard Lewis bases (such as oxygen donors) to form complexes rather than soft Lewis bases (such as sulfur or phosphorous donors).

4. Ce³⁺ is a stronger oxidizing agent than Ce⁴⁺: This statement is incorrect. Ce⁴⁺ is a stronger oxidizing agent than Ce³⁺ because Ce⁴⁺ has a higher positive charge, making it more likely to accept electrons and be reduced to Ce³⁺.

Conclusion:

Among the provided statements, the correct answer is option 1:

• Lanthanide organometallic compounds do not undergo oxidative addition reactions.

Concept:

Actinides are a series of 15 metallic elements from actinium to lawrencium in the periodic table. These elements have unique properties due to their f-orbital electron configuration.

• Actinides exhibit a wide range of oxidation states, typically ranging from +3 to +6, due to the availability of 5f, 6d, and 7s electrons for bonding.
• They have strong interactions with ligands due to the radial extension of their 5f orbitals, leading to significant covalent character in their compounds.
• Actinides are generally radioactive, with half-lives that vary widely depending on the element and isotope.
• Their chemistry is complex, often involving coordination complexes with high coordination numbers.

Explanation: 

• Statement A: Actinide metals interact more strongly with ligand orbitals compared to lanthanides due to the greater spatial extension and covalent nature of their 5f orbitals. This is a correct statement.

• Statement B: The coordination number of uranium in [UO₂(NO₃)₂]^- is indeed 8, as few of the nitrates act as a bidentate ligand. This statement is correct.

• Statement C: Isoelectronic lanthanum (La) and actinium (Ac) differ in their magnetic moment values, with Ac having a smaller value due to the lower number of unpaired 5f electrons. This statement is correct.

• Statement D: ThO₂ (thorium dioxide) has a coordination number of 8 due to its fluorite crystal structure. This is a correct statement.

Conclusion:

The correct answer is Option 4, as all the statements (A, B, C, and D) are correct and accurately describe the properties of actinides and their compounds.

CONCEPT:

Chromium Compounds and Color Changes in Oxidation States

• Certain chromium compounds undergo color changes based on their oxidation states and reactions with different reagents.
• Chromium trioxide (\( \text{CrO}_3 \)) is a red compound formed from potassium dichromate in acidic conditions. When it reacts with sodium carbonate, it forms chromate ions (\( \(\text{CrO}_4^{2-} \)), which appear green. Adding hydrogen peroxide in acidic conditions further oxidizes the chromium to form a violet-colored complex, chromium peroxide (\( \text{CrO}_5 \)).

REACTIONS:

• Step 1: Formation of Compound X
  • When potassium dichromate is treated with concentrated sulfuric acid, chromium trioxide (\( \text{CrO}_3 \)) is formed.
  • Reaction: \( \text{K}_2\text{Cr}_2\text{O}_7 + \text{H}_2\text{SO}_4 \rightarrow 2 \text{CrO}_3 + \text{K}_2\text{SO}_4 + \text{H}_2\text{O} \)
  • Compound X : Chromium trioxide (\( \text{CrO}_3\)), which is red.
  • Structure: 
• Step 2: Formation of Compound Y 
  • Passing \( \text{CrO}_3 \) through a sodium carbonate solution produces chromate ions (\( \text{CrO}_4^{2-} \)), which turn the solution green.
  • Reaction: \( \text{CrO}_3 + \text{Na}_2\text{CO}_3 \rightarrow \text{Na}_2\text{CrO}_4 \)
  • Compound Y : Chromate ion (\(\text{CrO}_4^{2-} \)), which is green.
  • Structure: 
• Step 3: Formation of Compound Z
  • Adding hydrogen peroxide in acidic conditions to chromate solution forms chromium peroxide (\( \text{CrO}_5 \)), a violet-colored complex.
  • Reaction: \( \text{CrO}_4^{2-} + \text{H}_2\text{O}_2 + 2\text{H}^+ \rightarrow \text{CrO}_5 + 2\text{H}_2\text{O} \)
  • Compound Z : Chromium peroxide (\(\text{CrO}_5 \)), which is violet.
  • Structure:

CONCLUSION:

The correct option is: Option 3.

Explanation:

• In the first reaction, chromyl chloride (P) is formed by the reaction of CrO₃ with HCl gas:

  • CrO₃ + 2HCl → CrO₂Cl₂ (P) + H₂O

  •  

  • 

• When P is treated with NaOH, sodium chromate (Q) is formed:

  • CrO₂Cl₂ + 4NaOH → Na₂CrO₄ (Q) + 2NaCl + 2H₂O

  • 

• When Q is treated with hydrogen peroxide in an acidic medium, a blue peroxo complex of chromium (R) is formed:

  • Na₂CrO₄ + H₂O₂ + H⁺ → [CrO(O₂)₂]^2- (R) + H₂O

  • 

Conclusion:

The correct option containing the structures of P (CrO₂Cl₂), Q ([CrO₄] ^2-), and R ([CrO(O₂)₂]^2-) is option 4.

The correct answer is Population of electrons at higher J level(s) via thermal excitation.

Concept:- 

1. Effective Magnetic Moment (\( \mu_{{eff}}\) ):
   - Calculated using both spin and orbital contributions.
   - For lanthanides, spin-orbit coupling must be considered.

2. Thermal Excitation:
   - At higher temperatures, electrons can populate higher energy J levels.
   - This affects the observed magnetic properties, leading to deviations from theoretical predictions.

3. Lanthanide Chemistry:
   - f-orbitals are shielded from the ligand field effects.
   - Magnetic moments of lanthanides are complex due to the contributions from unpaired electrons in f-orbitals.

Explanation:-

The magnetic moment ( \mu ) of a lanthanide ion like Eu³⁺ is influenced by both spin ( S ) and orbital ( L ) contributions, often expressed using the formula for the effective magnetic moment \(( \mu_{{eff}} )\):

\(\mu_{{eff}} = \sqrt{4S(S+1) + L(L+1)}\)

For f-block elements, particularly lanthanides, the magnetic moment is also affected by spin-orbit coupling, which combines the effects of spin and orbital angular momentum into a total angular momentum ( J ).

Europium (Eu³⁺)

1. Electronic Configuration:
   - The atomic number of Europium (Eu) is 63.
   - In the +3 oxidation state, Eu3+ has the electronic configuration [Xe] 4f⁶.
   - This results in 6 f-electrons.

2. Magnetic Moment Calculation:
   - For Eu³⁺, which is a 4f6 configuration:
     - Spin quantum number ( S ) = 3 (since there are 6 unpaired electrons each with \(S = \frac{1}{2} )\).
     - Orbital quantum number ( L ) = 3 (sum of the orbital contributions of the 6 electrons in the f-orbitals).

   Using the formula:
\( \mu_{{eff}} = \sqrt{4S(S+1) + L(L+1)} \\ \mu_{{eff}} = \sqrt{4 x 3 x 4 + 3 x 4} \\ \mu_{{eff}} = \sqrt{48 + 12} \\ \mu_{{eff}} = \sqrt{60} \approx 7.75 \, {BM}\)

This calculated value is significantly higher than the experimental value of 3.4 BM. 

Conclusion:-

So, The difference is due to Population of electrons at higher J level(s) via thermal excitation.

The correct answer is A, B, and D only.

Explanation:-

• Fe³⁺ ions are indeed highly colored in solutions due to d-d transitions. These transitions involve the promotion of an electron from one d-orbital to another, which requires energy in the visible spectrum, thus imparting color to the solution. Therefore, Statement A is correct.
• MnO₄- (permanganate ion) is known to be a very strong oxidizing agent in acidic medium. It can take part in various redox reactions, often resulting in the reduction of MnO4- to Mn2+ (in strongly acidic solutions) and the oxidation of other species. Thus, Statement B is correct.
• NH₃ acts as a strong ligand (Lewis base) in the formation of the tetraamminecopper(II) complex, [Cu(NH₃)₄]²⁺. It donates electron pairs to the central metal ion (Cu²⁺), forming coordinate bonds, not a weak base in this context. Therefore, Statement C is incorrect in implying NH₃ is a weak base in this scenario; it is actually acting strongly as a Lewis base to stabilize the complex ion.
• Cl- ions are not purely spectator ions in the reactions involving silver ions (Ag+); for example, Ag+ reacts with Cl- to form AgCl(s), a white precipitate, in a well-known reaction. However, in the specific formation of [Ag(NH₃)₂]+ from Ag+ ions, the Cl- ions do not participate directly in the formation of this complex. But in the statement's narrow context regarding the direct formation of the [Ag(NH₃)₂]+ complex, Cl- doesn't interact with the silver ammine complex directly. 

Conclusion:-

So, the correct statements for the given chemical species are A, B, and D only.