d And f - Block Elements MCQ Quiz - Objective Question with Answer for d And f - Block Elements - Download Free PDF

CONCEPT:

Catalysts and their Applications

• A catalyst is a substance that speeds up a chemical reaction without being consumed in the process.
• Different reactions require specific catalysts to enhance the rate of reaction or achieve desired selectivity.

EXPLANATION:

• In the given problem, we need to match the catalysts in List-I with their corresponding reactions or components in List-II:
  • A. Haber Process: This is the industrial process to synthesize ammonia (NH₃) from nitrogen (N₂) and hydrogen (H₂). The catalyst used is iron (Fe). Hence, A matches with I.
  • B. Wacker Oxidation: This process involves the oxidation of alkenes (e.g., ethylene) to aldehydes or ketones using palladium chloride (PdCl₂) as the catalyst. Hence, B matches with II.
  • C. Wilkinson Catalyst: This is a homogeneous catalyst [(PPh₃)₃RhCl] used for hydrogenation reactions in organic chemistry. Hence, C matches with III.
  • D. Ziegler Catalyst: This catalyst (a combination of TiCl₄ and Al(CH₃)₃) is used for polymerization of alkenes to produce polyethylene. Hence, D matches with IV.

Correct Answer: Option 3) A-I, B-II, C-III, D-IV

CONCEPT:

Ferromagnetism and Electronic Configuration

• Ferromagnetism: It is a property of materials like iron, cobalt, and nickel, where magnetic moments of individual atoms align spontaneously in the same direction even in the absence of an external magnetic field. It is an extreme form of paramagnetism because the alignment of magnetic moments is much stronger and ordered in ferromagnetic substances compared to paramagnetic ones.
• Electronic Configuration of Ions:
  • For Cr²⁺ (Z = 24), the electronic configuration can be derived as:
    • Neutral Cr: [Ar] 3d⁵ 4s¹
    • After losing 2 electrons: Cr²⁺ = [Ar] 3d⁴
    • Number of unpaired electrons in 3d⁴ = 4
  • For Nd³⁺ (Z = 60), the electronic configuration is:
    • Neutral Nd: [Xe] 4f⁴ 6s²
    • After losing 3 electrons: Nd³⁺ = [Xe] 4f³
    • Number of unpaired electrons in 4f³ = 3

EXPLANATION:

• Statement I: "Ferromagnetism is considered as an extreme form of paramagnetism."
  • This statement is true because ferromagnetic materials exhibit a stronger and more ordered alignment of magnetic moments compared to paramagnetic materials. Ferromagnetism can be considered as an extreme or special case of paramagnetism.
• Statement II: "The number of unpaired electrons in a Cr²⁺ ion (Z = 24) is the same as that of a Nd³⁺ ion (Z = 60)."
  • This statement is false because:
    • Cr²⁺ has 4 unpaired electrons (from its 3d⁴ configuration).
    • Nd³⁺ has 3 unpaired electrons (from its 4f³ configuration).
    Hence, the number of unpaired electrons is not the same for Cr²⁺ and Nd³⁺.

Therefore, the correct answer is: Statement I is true but Statement II is false.

CONCEPT:

Main Group Elements

• The main group elements are the elements in groups 1, 2, and 13–18 of the periodic table.
• These elements have their valence electrons in the s or p orbitals.
• They include alkali metals, alkaline earth metals, and elements from the boron, carbon, nitrogen, oxygen, halogen, and noble gas groups.
• Transition metals and inner transition metals (lanthanides and actinides) are not considered main group elements because their valence electrons are in the d or f orbitals.

EXPLANATION:

• Analyze the given electronic configurations to determine whether they belong to main group elements:
  • A. [Ne]3s¹: This corresponds to sodium (Na), which is in Group 1. It is a main group element.
  • B. [Ar]3d³4s²: This corresponds to vanadium (V), which is a transition metal (Group 5). It is not a main group element.
  • C. [Kr]4d¹⁰5s²5p⁵: This corresponds to iodine (I), which is in Group 17. It is a main group element.
  • D. [Ar]3d¹⁰4s¹: This corresponds to copper (Cu), which is a transition metal (Group 11). It is not a main group element.
  • E. [Rn]5f⁰6d²7s²: This corresponds to thorium (Th), which is an actinide and not a main group element.

So, only A and C belong to main group elements.

EXPLANATION:

• Given:
  • β2 = 1.2 × 10⁷
  • K1 = 4.0 × 10⁴
• Formula:

  β2 = K1 × K2 ⇒ K2 = β2 / K1

  = (1.2 × 10⁷) / (4.0 × 10⁴)

  = 0.3 × 10³

  = 3.0 × 10²

Therefore, the second stepwise constant K2 is 3.0 × 102.

EXPLANATION:

• Given:
  • K1 = 1.5 × 10⁴
  • K2 = 6.0 × 10³
  • K3 = 2.0 × 10²
• Formula:

  β3 = K1 × K2 × K3

  = (1.5 × 10⁴) × (6.0 × 10³) × (2.0 × 10²)

  = 1.5 × 6.0 × 2.0 × 10⁴⁺³⁺²

  = 18.0 × 10⁹

Therefore, the overall stability constant β3 is 1.8 × 1010.

Concept:

• The D-block elements are known as transition elements.
• There total of 4 blocks in the modern periodic table. They are as follows: s block, p block, d block, f block.
• There are 18 groups and 7 periods in the modern periodic table.
• Total elements are 118 out of which 91 are metals, 7 are metalloids, and 20 are non-metals.

Explanation:

• 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 generalised electronic configuration of these elements is (n-1) d1–10 ns1–2.
• The d-block elements are categorised as 1st series transition elements, 2nd series transition elements, 3rd series transition elements and 4th series transition elements.
• The examples are:- Cu, Zn, Ag, Cd, Au, Hg, etc.

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.

Concept:

Paramagnetic materials:

• Small, positive susceptibility to magnetic fields.
• These materials are slightly attracted by a magnetic field.
• Paramagnetic properties are due to the presence of some unpaired electrons, and from the realignment of the electron paths caused by the external magnetic field
• Paramagnetic materials include magnesium, molybdenum, lithium, and tantalum.

​

Magnetic moment:

• The strength of a magnet and its orientation in presence of a magnetic field is called its magnetic moment.
• The complexes have lone electrons in them, unpaired which contribute to the magnetic moment. It is given by the formula:

​ \(μ = {\sqrt{n(n+2)} }BM\)

Explanation:

• Chromium belongs to the d block. Its electronic configuration is: [Ar]3d⁵4s¹.
• In the Cr⁺² oxidation state, it loses two electrons and has the configuration [Ar]3d⁴.
• The number of unpaired electrons n = 4.

​

• Hence, the magnetic moment is:

\(μ = {\sqrt{n(n+2)} }BM\)

\(μ = {\sqrt{4(4+2)} }BM\)

μ = 4.90 BM

Hence, the magnetic moment of Cr²⁺ is 4.90 BM.

Additional Information

Diamagnetic materials

• Weak, negative susceptibility to magnetic fields
• Diamagnetic materials are slightly repelled by a magnetic field.
• All the electrons are paired so there is no permanent net magnetic moment per atom.
• Most elements in the periodic table, including copper, silver, and gold, are diamagnetic.

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. 

Explanation:

The magnetic moment of a divalent ion in an aqueous solution depends on its electronic configuration, which in turn is determined by the element's atomic number.

Atomic number (z = 25) belongs to Mn atom

E.C. of Mn = [Ar]3d⁵4s²

E.C. of Mn²⁺ ion = [Ar]3d⁵4s⁰

\(μ=\sqrt{n(n+2)} B M\)

Number of unpaired electrons = 5

\(μ=\sqrt{5(5+2)}=\sqrt{35} \mathrm{BM}\)

μ = 5.92 BM

Explanation:

• When compounds containing chloride are heated with potassium dichromate in presence of sulphuric acid, a red liquid of chromyl chloride is formed.
• It is a test for compounds containing chlorine in them.
• Compounds containing chlorine will produce red-colored vapours of chromylchloride when heated with potassium dichromate in presence of sulphuric acid.
• Sulphuric acid acts as a dehydrating agent.

The net reaction is:

K₂Cr₂O₇ + 6KCl + H₂SO₄ → Cr₂O₂Cl₂

Hence, the formula of the deep red liquid formed on heating potassium dichromate with KCI in conc. H2SO4 is CrO2Cl2.

Key Points

• Chromyl chloride is Cr₂O₂Cl₂, which is a transition metal complex.
• The structure is tetrahedral.​
• It is deep red liquid which is unusual because transition complexes are mostly solids in nature.
• It is volatile at room temperature and the IUPAC name is Chromium (VI) dichloride dioxide.
• It can also be prepared by the reaction of chromium oxide and anhydrous HCl.
• The reaction is:

CrO₃ + HCl →  Cr₂O₂Cl₂.

It is used in Etard's reaction.

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:

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)

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