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

CONCEPT:

Formation of Prussian Blue Using Ferricyanide and Ferrous Ions

• Potassium ferricyanide (K₃[Fe(CN)₆]) is a coordination complex containing iron(III) and six cyanide (CN⁻) ligands.
• When ferrous ions (Fe²⁺) are added to this solution, they react with the cyanide ligands to form an insoluble blue complex called Prussian blue.
• The deep blue color is due to the formation of a mixed-valence iron complex where Fe²⁺ and Fe³⁺ are bridged by CN⁻ ligands.

EXPLANATION:

• In this reaction:

  Fe²⁺ + [Fe(CN)₆]³⁻ → Fe³⁺[Fe²⁺(CN)₆]⁻ (Prussian blue)

  • The [Fe(CN)₆]³⁻ ion provides CN⁻ ligands that bridge the Fe²⁺ and Fe³⁺ centers.
  • The iron(III) complex acts as a **source of cyanide ligands (CN⁻)** which coordinate to Fe²⁺.
• Other ligands like NO₂⁻, CO, and SCN⁻ are not present in potassium ferricyanide.

Therefore, the correct answer is: (A) CN⁻

CONCEPT:

Magnetic Moment

• The magnetic moment of an ion is determined by the number of unpaired electrons it has.
• It is calculated using the formula:

  µ = √(n(n + 2)) BM

• Here, µ is the magnetic moment (in Bohr Magneton, BM), and "n" is the number of unpaired electrons in the ion.
• The magnetic moment increases with the number of unpaired electrons present in the ion.

EXPLANATION:

• Yb²⁺: This ion belongs to the f-block elements, specifically lanthanides. 
  [Xe]4f¹⁴
  It has fewer unpaired electrons, resulting in a lower magnetic moment.
• Eu³⁺:
   [Xe]4f⁶
  This ion has 6 unpaired electrons, leading to a magnetic moment less than 7.9 BM.
• Gd³⁺: 
  [Xe] 4f⁷.
  This ion has 7 unpaired electrons. Using the formula:

  µ = √(7(7 + 2)) = √63 ≈ 7.9 BM

  Hence, Gd³⁺ has a magnetic moment of 7.9 BM.
• Ce⁴⁺: This ion has no unpaired electrons, resulting in a magnetic moment of 0 BM.
• Therefore, the correct answer is option 3 (Gd³⁺).

Thus, the ion with a magnetic moment of 7.9 BM is Gd³⁺.

CONCEPT:

Half-filled f-shell of lanthanide ion

• The electronic configuration of lanthanides involves the filling of 4f orbitals.
• A half-filled f-shell corresponds to an electronic configuration where the 4f subshell contains 7 electrons (since the maximum capacity of the f-subshell is 14 electrons).
• The presence of exactly 7 electrons in the 4f subshell provides extra stability due to symmetrical distribution and exchange energy.

EXPLANATION:

• Ce⁴⁺: Cerium in the +4 oxidation state has an electronic configuration of [Xe].
  s² 2s² 2p⁶ 3s² 3p⁶ 3d¹⁰ 4s² 4p⁶ 4d¹⁰ 5s² 5p⁶.
  Its 4f subshell is empty, so it does not have a half-filled f-shell.
• Tb⁴⁺: Terbium in the +4 oxidation state has an electronic configuration of [Xe]4f⁷. It has exactly 7 electrons in the 4f subshell, which corresponds to a half-filled f-shell.
• Yb²⁺: Ytterbium in the +2 oxidation state has an electronic configuration of [Xe]4f¹⁴. Its 4f subshell is completely filled, not half-filled.
• Eu³⁺: Europium in the +3 oxidation state has an electronic configuration of [Xe]4f⁶. Its 4f subshell is not half-filled (it has fewer than 7 electrons).
• Among the given options, Tb⁴⁺ has a half-filled 4f subshell with 7 electrons.

Therefore, the correct answer is Tb⁴⁺.

CONCEPT:

Lanthanide Contraction and Ionic Radius

• The lanthanides are elements with atomic numbers 57 (La) to 71 (Lu), where 4f orbitals are progressively filled.
• Lanthanide contraction refers to the gradual decrease in ionic radii of the Ln³⁺ ions with increasing atomic number.
• This occurs because:
  • The poor shielding effect of 4f electrons increases the effective nuclear charge.
  • This increased nuclear attraction pulls the electrons closer, reducing the ionic size.
• Thus, as we move from left to right in the lanthanide series, the ionic radius decreases.

EXPLANATION:

Nd³⁺ > Gd³⁺ > Ho³⁺ > Tm³⁺

• Nd³⁺ (atomic number 60) has the largest radius among the given ions.
• Gd³⁺ (atomic number 64) is smaller than Nd³⁺.
• Ho³⁺ (atomic number 67) is smaller than Gd³⁺.
• Tm³⁺ (atomic number 69) is even smaller.
• Hence, the correct order based on decreasing ionic radius is:

Therefore, the correct answer is Option 3: Nd³⁺ > Gd³⁺ > Ho³⁺ > Tm³⁺

CONCEPT:

Reaction of Potassium Chlorate (KClO₃) with Moist Oxalic Acid

• Potassium chlorate (KClO₃) undergoes a redox reaction in the presence of moist oxalic acid (C₂H₂O₄).
• Oxalic acid acts as a reducing agent, while KClO₃ acts as an oxidizing agent.
• During the reaction, KClO₃ is reduced to chlorine dioxide (ClO₂), and oxalic acid is oxidized to carbon dioxide (CO₂).
• Potassium ions from KClO₃ combine with oxalate ions to form potassium oxalate (K₂C₂O₄).

EXPLANATION:

KClO₃ + C₂H₂O₄ → K₂C₂O₄ + ClO₂ + CO₂

• Given reaction:
• Steps in the reaction:
  • Oxalic acid (C₂H₂O₄) is oxidized to carbon dioxide (CO₂).
  • KClO₃ is reduced to chlorine dioxide (ClO₂).
  • Potassium ions (K⁺) from KClO₃ react with oxalate ions (C₂O₄²⁻) to form potassium oxalate (K₂C₂O₄).

Therefore, the correct answer is Option 3: K₂C₂O₄ + ClO₂ + CO₂.

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