d And f - Block Elements MCQ Quiz in मल्याळम - Objective Question with Answer for d And f - Block Elements - സൗജന്യ PDF ഡൗൺലോഡ് ചെയ്യുക

Concept:

• All the ions in the option are d block elements.
• Most of the ions of d block elements in the periodic table are colored.
• This is due to the absorption of radiation in the visible light region to excite electrons from lower energy level d-orbital to higher energy level d-orbitals.
• This is also known as the d-d transition.
• The color of the ion is complimentary of the color absorbed by it.

Explanation:

• The d-d transition occurs only when there is vacant d orbital in the ions.
• Vacant d orbitals can be found from the electronic configuration of ions.
• Among the given ions, only Ti3+, Cr3+, and V3+ have vacant d orbitals.

Hence the set of colored ions are Ti³⁺, Cr³⁺, and V³⁺.

CONCEPT:

Oxidizing Agents

• An oxidizing agent (or oxidant) is a substance that has the ability to accept electrons and thereby get reduced in a chemical reaction.
• The strength of an oxidizing agent is determined by its ability to accept electrons readily, making it more likely to be reduced.
• The higher the oxidation state of an element, the stronger its oxidizing power, as it has a greater tendency to gain electrons and achieve a lower, more stable oxidation state.

EXPLANATION:

• The ions provided and their tendencies to get reduced:
  • Lu³⁺: Lutetium tends to have a stable +3 oxidation state, it does not have a high tendency to gain electrons and therefore is not a strong oxidizing agent.
  • Eu²⁺: Europium in the +2 oxidation state can be a moderately strong reducing agent, but not a strong oxidizing agent.
  • Tb⁴⁺: Terbium in the +4 oxidation state is highly unstable and has a strong tendency to accept electrons and revert to the more stable +3 oxidation state, making it a very strong oxidizing agent.
  • Ce³⁺: Cerium typically prefers the +3 oxidation state, but can be found in the +4 oxidation state as well. It does have some oxidizing power, but not as much as Tb⁴⁺.
• Among the ions given, Tb⁴⁺ has the highest oxidation state and is the most unstable. Therefore, it has the greatest tendency to accept electrons and be reduced to a more stable state, making it the strongest oxidizing agent.

Therefore, the strongest oxidizing agent among the given ions is Tb⁴⁺.

Concept:

The d-block elements, also known as transition metals, exhibit a variety of oxidation states due to the involvement of electrons from both the ns and (n-1)d orbitals in bonding. The stability of these oxidation states can vary significantly among different elements.

Explanation:

Transition metals have varied oxidation states, primarily because they can lose electrons from both the ns and (n-1)d orbitals. While different transition metals can exhibit different stable oxidation states, certain general trends are observed:

The most common and stable oxidation state for many d-block elements is +2. This stability arises due to the removal of the two ns electrons. Although many transition metals can exhibit higher oxidation states, the +2 state is particularly stable and commonly observed.

Conclusion:

The most general stable oxidation state exhibited by the d-block elements is: +2 due to the participation of ns electron.

CONCEPT:

Oxidation States of Transition Metals

• Transition metals are known for exhibiting a wide range of oxidation states.
• The number of oxidation states shown by a metal depends on the arrangement of its electrons, primarily in the d-orbitals.
• Higher oxidation states are typically found in the middle of the transition series, where there is a greater number of valence electrons available for bond formation.

Explanation:-

• 1) Iron (Fe):
  • Common oxidation states: +2, +3.
  • Maximum oxidation state: +6 (rare).
• 2) Manganese (Mn):
  • Common oxidation states: +2, +3, +4, +6, +7.
  • Maximum oxidation state: +7 (as seen in permanganates, MnO₄⁻).
  • Manganese exhibits the highest number of oxidation states among the 3d series metals, ranging from +2 to +7.
• 3) Titanium (Ti):
  • Common oxidation states: +2, +3, +4.
  • Maximum oxidation state: +4.
• 4) Cobalt (Co):
  • Common oxidation states: +2, +3.
  • Maximum oxidation state: +5 (very rare).

CONCLUSION:

The correct answer is (2) Manganese (Mn)

The correct answer is option 3, i.e. Show fixed oxidation state.

Key Points

• Transition elements are those elements whose two outermost shells are incomplete.
• These elements have partially filled d-subshell in the ground state or any of their common oxidation state and are commonly referred to as d-block transition elements.
• The d-block elements are categorized as 1st series transition elements, 2nd series transition elements, 3rd series transition elements, and 4th series transition elements.
• Examples are:- Cu, Zn, Ag, Cd, Au, Hg, etc.
• The f-block elements are termed inner transition elements.
• Some characteristics of Transition elements are;
  • They are hard and have high densities.
  • They always form colored ions and compounds.
  • They have high melting and boiling points.
  • They have more than one oxidation state.

Explanation:-

Cu has positive \(\mathrm{E}_{\mathrm{Cu}^{2+} / \mathrm{Cu}}^{\circ}\) value in 3d series.

In the first transition series, the standard electrode potential ((E0M2+/M) indicates the ease with which a metal can be reduced to its metallic form from its ion in aqueous solution. A positive value for (E⁰_M²⁺/M) means that the reduction process is favorable, and the metal is relatively less reactive compared to those with negative values.

Among the given options:

Chromium (Cr) has a negative standard reduction potential.
Vanadium (V) also has a negative value.
Copper (Cu) has a positive standard reduction potential ((E⁰_Cu²⁺/Cu = +0.34 V)), meaning it is readily reduced to its metallic state.
Nickel (Ni) has a negative standard reduction potential.

\(\mathrm{E}_{\mathrm{Cu}^{2+} / \mathrm{Cu}}^{\circ}\) = 0.34 V

\(\mathrm{E}_{\mathrm{Cr}^{2+} / \mathrm{Cr}}^{\circ}\) = –0.90 V

\(\mathrm{E}_{\mathrm{V}^{2+} / \mathrm{V}}^{\circ}\) = –1.18 V

\(\mathrm{E}_{\mathrm{Ni}^{2+} / \mathrm{Ni}}^{\circ}\) = = –0.25 V

Explanation:-

Outer electron configuration of Ti = 3d²4s²

So, Maximum O.S. of Ti = +4 

Outer E.C. of V = 3d³4s²

So, Maximum O.S. of V = +5 Outer E.C. of Mn = 3d⁵4s²

So, Maximum O.S. of Mn = +7

Outer E.C. of Cu = 3d¹⁰4s¹

So, Maximum O.S. of Cu = +2

So, correct option is : A-II, B-III, C-I, D-IV

CONCEPT:

Gun Metal Composition

• Gun metal, also known as red brass in the United States, is a type of bronze – an alloy of copper, tin, and zinc.
• The typical composition of gun metal is:
  • Copper (Cu): Approximately 88%
  • Tin (Sn): Approximately 8%
  • Zinc (Zn): Approximately 4%

EXPLANATION:

• The primary components of gun metal are copper (Cu), tin (Sn), and zinc (Zn).
• Other options include different metals which are not typical components of gun metal:
  • Option 2: Al, Mg, Mn, Cu – This is not gun metal.
  • Option 3: Cu, Ni, Fe – This is not gun metal.
  • Option 4: Cu, Sn, Fe – This is not gun metal.

CONCLUSION:

The correct composition of gun metal includes copper (Cu), tin (Sn), and zinc (Zn).

CONCEPT:

Oxidizing Agents

• An oxidizing agent gains electrons and, therefore, gets reduced in a chemical reaction.
• Ions with stable and energetically favorable configurations after gaining electrons tend to be good oxidizing agents.
• Elements can act as good oxidizing agents if they can readily change to a lower, more stable oxidation state.

EXPLANATION:

• The electronic configurations of the given ions are:
  • Sm²⁺: [Xe] 4f⁶
  • Ce²⁺: [Xe] 4f²
  • Ce⁴⁺: [Xe] 4f⁰
  • Tb⁴⁺: [Xe] 4f⁷
• Good oxidizing agents among these ions:
  • Ce⁴⁺ and Tb⁴⁺ are good oxidizing agents because they can readily be reduced to the +3 oxidation state, which is very stable.

CONCLUSION:

The correct answer is Option 3: C and D only.

Explanation:-

• Mischmetal is an alloy of rare-earth elements. It is also called cerium mischmetal, or rare-earth mischmetal.
• Misch metal contains Lanthanide metals (94-95)% + Iron (5) % + Traces of S, C, Si, Ca, and Al.
• A typical composition includes approximately 55% cerium, 25% lanthanum, and 15~18% neodymium, with traces of other rare earth metals.
• Hence, It is an alloy of rare-earth elements. So, statement A is correct.
• Mischmetal is prepared from monazite, an anhydrous phosphate of the light lanthanides and thorium.
• The ore was cracked by reaction at high temperatures with either concentrated sulfuric acid or sodium hydroxide.
• Monazite-derived mischmetal typically was about 48% cerium, 25% lanthanum, 17% neodymium, and 5% praseodymium, with the balance being the other lanthanides.
• Hence, the composition of cerium is highest in mischmetal. So, statement B is incorrect.
• Misch metal alloyed with iron is the flint (spark-producing agent) in cigarette lighters and similar devices. Misch metal is also used as a deoxidizer in various alloys and to remove oxygen in vacuum tubes. So, statement C is correct.
• As, Misch metal contains Lanthanide metals (94-95)% + Iron (5) % + Traces of S, C, Si, Ca, and Al. Statement D is also correct.

Conclusion:-

• Hence, statements A, C, and D are correct.