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

\(\mu_{\text{spin}} = \sqrt{n(n + 2)}\)

• The spin-only magnetic moment \(( \mu_{\text{spin}} ) \) 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:
  \(\mu_{\text{spin}} = \sqrt{4(4 + 2)} = \sqrt{24} = 4.898 \, \text{B.M.} \\ \)
• For [Mn(CN)6]3– the strong field CN^- ligands cause electron pairing, leaving 2 unpaired electrons. The magnetic moment is:
  \( \mu_{\text{spin}} = \sqrt{2(2 + 2)} = \sqrt{8} = 2.828 \, \text{B.M.} \\ \)
• The sum of the magnetic moments is:
  \(4.898 \, \text{B.M.} + 2.828 \, \text{B.M.} = 7.726 \, \text{B.M.}\)

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

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:

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

CONCEPT:

Magnetic Moment and Unpaired Electrons

μ = √n(n + 2) B.M.

• The magnetic moment (μ) of a metal ion is related to the number of unpaired electrons (n) present in the ion, and it can be calculated using the formula:
• Where 'n' represents the number of unpaired electrons in the metal ion, and 'μ' is the magnetic moment in Bohr Magneton (B.M.).
• Given that the magnetic moment is 4.9 B.M., we can substitute this value into the formula and solve for 'n'.

EXPLANATION:

4.9 = √n(n + 2)

24.01 = n(n + 2)

• Given the magnetic moment μ = 4.9 B.M., we can substitute this into the formula:
• Squaring both sides:
• Solving the quadratic equation: n² + 2n - 24.01 = 0, we get n = 4, which indicates 4 unpaired electrons.
• Now, we examine the electron configurations for each metal ion:
  • A. Cr²⁺: [Ar] 3d⁴ → 4 unpaired electrons (Correct)
  • B. Fe²⁺: [Ar] 3d⁶ → 4 unpaired electrons (Correct)
  • C. Fe³⁺: [Ar] 3d⁵ → 5 unpaired electrons (Incorrect)
  • D. Co²⁺: [Ar] 3d⁷ → 3 unpaired electrons (Incorrect)
  • E. Mn³⁺: [Ar] 3d⁴ → 4 unpaired electrons (Correct)

Therefore, the correct answer is: A, B, and E only.

CONCEPT:

• Potassium ferrocyanide, K₄[Fe(CN)₆], reacts with various metal salts to form characteristic precipitates based on the metal ion involved.
• The nature and color of the precipitate help in identifying metal ions qualitatively.
• Specific reactions:
  • CuSO₄ (acidified with acetic acid): Reacts with K₄[Fe(CN)₆] to form a chocolate brown precipitate of Cu₂[Fe(CN)₆].
  • FeCl₃: Reacts with K₄[Fe(CN)₆] to form Prussian blue precipitate, Fe₄[Fe(CN)₆]₃.
  • ZnCl₂ (neutralized with NH₄OH): Forms white or bluish white precipitate K₂Zn₃[Fe(CN)₆]₂.
  • MgCl₂ and BaCl₂: Do not give such characteristic precipitates with K₄[Fe(CN)₆].

EXPLANATION:

\(2\mathrm{CuSO}_4 + \mathrm{K}_4[\mathrm{Fe(CN)}_6] \xrightarrow{\mathrm{CH}_3\mathrm{COOH}} \mathrm{Cu}_2[\mathrm{Fe(CN)}_6] \downarrow + 2\mathrm{K}_2\mathrm{SO}_4 \\ \text{(Chocolate brown ppt.)} \\ 4\mathrm{FeCl}_3 + 3\mathrm{K}_4[\mathrm{Fe(CN)}_6] \rightarrow \mathrm{Fe}_4[\mathrm{Fe(CN)}_6]_3 \downarrow + 12\mathrm{KCl} \\ \text{(Prussian Blue ppt.)} \\ 3\mathrm{ZnCl}_2 + 2\mathrm{K}_4[\mathrm{Fe(CN)}_6] \xrightarrow{\mathrm{NH}_4\mathrm{OH}} \mathrm{K}_2\mathrm{Zn}_3[\mathrm{Fe(CN)}_6]_2 \downarrow + 6\mathrm{KCl} \\ \text{(White or bluish white ppt.)}\)

• Option A: Correct. CuSO₄ with K₄[Fe(CN)₆] gives chocolate brown precipitate.
• Option B: Correct. FeCl₃ with K₄[Fe(CN)₆] gives Prussian blue precipitate.
• Option C: Correct. ZnCl₂ neutralized with NH₄OH reacts with K₄[Fe(CN)₆] to give white/bluish white precipitate.
• Option D: Incorrect. MgCl₂ does not give blue precipitate with K₄[Fe(CN)₆].
• Option E: Incorrect. BaCl₂ neutralized with NaOH does not give white precipitate with K₄[Fe(CN)₆].

Therefore, the correct tests with respective observations are A, B, and C only.

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