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 Calculation for Complexes

• The spin-only magnetic moment  of a complex is calculated using the formula:
  • n = Number of unpaired electrons in the complex.
• The formula applies to the complexes where the magnetic moment is influenced only by the spins of the unpaired electrons (i.e., without considering orbital contribution).

EXPLANATION:

• For [Mn(Br)6]3– Mn is in the +3 oxidation state with a 3d^4 electron configuration. The number of unpaired electrons is 4, so the magnetic moment is:
  
• For [Mn(CN)6]3– the strong field CN^- ligands cause electron pairing, leaving 2 unpaired electrons. The magnetic moment is:
  
• The sum of the magnetic moments is:
  

Therefore, the sum of the spin-only magnetic moments is 7.726 B.M.

 CONCEPT:

Diamagnetism in Ions

• Diamagnetism is the property of a substance to create an opposing magnetic field when subjected to an external magnetic field. This happens when all the electrons in an atom or ion are paired, which leads to no unpaired electrons in the electron configuration.
• Ions that are diamagnetic have only paired electrons, meaning there are no unpaired electrons in their molecular orbitals or atomic orbitals.
• In contrast, paramagnetic ions have at least one unpaired electron, leading to a net magnetic moment.

EXPLANATION:

• Option 1: La³⁺ and Ce⁴⁺
  • Both La³⁺ and Ce⁴⁺ have fully paired electrons in their electron configurations. La³⁺ has the electron configuration [Xe] 4f⁰, and Ce⁴⁺ has the configuration [Xe] 4f⁰ as well. Thus, they are both diamagnetic.
• Option 2: Yb²⁺ and Lu³⁺
  • Yb²⁺ has the electron configuration [Xe] 4f¹³ and Lu³⁺ has the configuration [Xe] 4f⁰. Yb²⁺ has all paired electrons, and Lu³⁺ has paired electrons as well. Therefore, both ions are diamagnetic.
• Option 3: La²⁺ and Ce³⁺
  • La²⁺ has an electron configuration of [Xe] 4f¹, and Ce³⁺ has [Xe] 4f¹ as well. Both ions have unpaired electrons, making them paramagnetic, not diamagnetic.
• Option 4: Yb³⁺ and Lu²⁺
  • The neutral ytterbium atom has the configuration [Xe] 4f¹⁴ 5d¹ 6s².
    The neutral lutetium atom has the configuration [Xe] 4f¹⁴ 5d¹ 6s². 

Therefore, the correct diamagnetic ions are found in Option 1, Option 2

CONCEPT:

Reaction of Permanganate Ion with Iodide Ion in Neutral Aqueous Medium

• The reaction of permanganate ion (MnO₄^–) with iodide ion (I^–) in a neutral aqueous medium leads to the reduction of the permanganate ion and the oxidation of iodide ion.
• In a neutral medium, the permanganate ion is reduced to MnO₂ (manganese dioxide), and iodide ions are oxidized to form iodate ions (IO₃^–).
• The balanced chemical equation for this reaction is as follows:
  • 2MnO₄^– + 2I^– + 2H₂O → 2MnO₂ + 2OH^– + I₂ (in acidic medium)
  • In a neutral medium, the final product of iodide oxidation is IO₃^– (iodate ion). Therefore, the permanganate ion is reduced to MnO₂ while iodide ion is oxidized to iodate (IO₃^–).

EXPLANATION:

• In neutral aqueous medium, the iodide ion (I^–) is oxidized by permanganate to form iodate (IO₃^–), which is the main product of the reaction.
• The reduction of permanganate results in the formation of manganese dioxide (MnO₂), which is a solid product.
• Thus, the correct product formed from the reaction of permanganate ion with iodide ion in neutral aqueous medium is IO₃^–.
• Hence, the correct answer is IO₃^– (Iodate ion).

Therefore, the correct answer is IO₃^–.

CONCEPT:

Terminal Oxygen Atoms in Chemical Compounds

• Terminal Oxygen Atoms: These are oxygen atoms that are at the end of a bond or functional group, typically attached to a central atom (like chromium in this case) and are not involved in further bonding.
• Chemical Reactions: The reactions provided involve chromium compounds, where the terminal oxygen atoms come from the formation of the dichromate ion (Cr₂O₇²⁻) in the final compound.

EXPLANATION:

No of terminal "O" = 6

Therefore, the correct answer is: The number of terminal "O" present in compound C is 6.

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:

​ 

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 = 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

Number of unpaired electrons = 5

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  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.

 = 0.34 V

 = –0.90 V

 = –1.18 V

 = = –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