Worksheet with Solutions: D and F - Block Elements | Chemistry Class 12 - NEET PDF Download

Answer: Actinoid contraction
Explanation: As we move across the actinoid series, the 5f electrons are poorly shielded, resulting in greater nuclear attraction. This causes a gradual decrease in atomic and ionic radii, known as actinoid contraction.

Answer: Coloured
Explanation: D-block elements often have partially filled d-orbitals, which allow d–d transitions. These transitions absorb certain wavelengths of light, leading to the appearance of coloured compounds.

Answer: Catalytic
Explanation: Transition metals can adopt multiple oxidation states and provide surface area for reactants to adsorb. This makes them efficient catalysts in many chemical reactions.

Answer: 3 to 12
Explanation: Groups 3 to 12 in the periodic table include all the transition metals, which belong to the d-block. These elements fill the d-orbitals progressively.

Answer: Yellow, Orange
Explanation:
Potassium chromate (K₂CrO₄) is yellow due to the presence of chromate ions, while potassium dichromate (K₂Cr₂O₇) is orange because of dichromate ions. Both are d-block compounds and show colour due to charge-transfer transitions.

Answer: (a) FeSO_4.7H₂O

Explanation:

• Green vitriol is the common name for ferrous sulfate heptahydrate, i.e., FeSO₄·7H₂O.

• CuSO₄·5H₂O is blue vitriol, and CaSO₄·2H₂O is gypsum

Answer: (c) d-block 

Explanation:

• Transition elements are located in the d-block of the periodic table.

• They span groups 3 to 12, and have partially filled d-orbitals either in the elemental state or in their ions.

Answer: (d) Zn²⁺

Explanation:

• Diamagnetic substances have no unpaired electrons.

• Electronic configurations:

  • Zn²⁺ → [Ar] 3d¹⁰ → all paired → diamagnetic

  • Co²⁺ → 3d⁷ → unpaired e⁻ present → paramagnetic

  • Ni²⁺ → 3d⁸ → unpaired e⁻ → paramagnetic

  • Cu²⁺ → 3d⁹ → 1 unpaired e⁻ → paramagnetic

Answer: (d) d-d transition

Explanation:

• In transition metal ions, the colour arises due to electronic transitions between split d-orbitals (in presence of ligands).

• This is called a d–d transition.

• Absorption of visible light promotes an electron from a lower d-orbital to a higher d-orbital.

Answer: (b) Cu >Au > Ag.

Actual melting points:

• Cu (Copper): 1085°C

• Au (Gold): 1064°C

• Ag (Silver): 962°C

• Hence, order: Cu > Au > Ag

Explanation: The assertion is incorrect because Cu⁺ has no unpaired electrons, Cu²⁺ has one.
The reason is also incorrect, Cuprous ion (Cu⁺) is colorless whereas cupric ion (Cu⁺⁺) is blue in the aqueous solution.

Explanation:

• Transition metals do show variable valency due to similar energy of (n-1)d and ns electrons.

• Reason is incorrect because the energy difference is small, not large, allowing both sets of electrons to participate in bonding.

Explanation:

• Transition metals provide a surface for reactions and can change oxidation states easily, making them good catalysts.

• V₂O₅ (in SO₂ to SO₃) and Pt are commonly used in catalytic processes.

• Assertion is incorrect: The magnetic moments of actinides often exceed spin-only values due to significant orbital contribution.

• Reason is also incorrect: They are paramagnetic, but the statement doesn’t relate properly to the assertion and lacks clarity.

Explanation:

• In acidic medium, KMnO₄ is reduced to Mn²⁺:
  MnO₄⁻ + 8H⁺ + 5e⁻ → Mn²⁺ + 4H₂O

• Assertion is correct.

• Reason is partially correct: The second part is confused—MnO₂ forms in neutral or slightly alkaline medium, not acidic.

Answer:

(i) Due to lanthanide contraction, size reduces. With the size reduction, the covalent character increases. Therefore, Lu₂O₃ is more covalent than La₂O₃ .

(ii) Oxosalts contain oxygen as an anion. As the size of the cation reduces from La to Lu, according to Fajan’s rules, the polarising power of the cation will increase, and it will distort the cloud of oxygen(anion) significantly. Thus the bond weakens, and the stability also reduces.

(iii) As the size of the central atom reduces, the stability of the complex increases. A small metal ion with a greater charge attracts the ligands better.

(iv) In 5d block elements, the effective nuclear charge increases due to poor shielding of f orbitals, thereby reducing the size. This is called lanthanide contraction. So, the radii of 4d and 5d block elements end up being very similar.

(v) From La to Lu, the acidic character increases. As the size reduces from La to Lu, the ability to lose electrons(Lewis base character) reduces, so the acidity increases.

Answer:

• The following reaction takes place when Cr₂O₇^2- is treated with an alkali, 
  (orange) Cr₂O₇^2- + OH^– → 2CrO₄^2- (yellow)
• When the yellow solution is treated with an acid, we get back the orange solution 
  (yellow) 2CrO₄^2- + 2H⁺ → Cr₂O₇^2– (orange) + H₂O
• This reaction is reversible under proper conditions.

Answer: Due to lanthanoid contraction, the atomic radii of the second and third-row transition elements are almost identical. So, they resemble each other much more than first-row elements.

Answer: Atomic orbitals are filled in order of increasing energies. Since the energy of 3d orbital is more than 4s orbital, based on the (n+l) rule, it is filled after 4s orbital. But during ionisation, electrons in the outermost orbital are lost. Since 4s will be the outermost orbital, in this case, electrons from this orbital will be ionised first.

Answer: As the oxidation state increases, the size of the ion of the transition element decreases. As per Fajan’s rule, as the size of metal ions decreases, the covalent character of the bond formed increases. Therefore, the halides of transition elements become more covalent with the increasing oxidation state of the metal.

Answer: (i) From titanium to copper the atomic size of elements decreases and mass increases as a result of which density increases.
(ii) The catalytic properties of the transition elements are due to the presence of unpaired electrons in their incomplete d- orbitals and variable oxidation states.

Answer: Lanthanoid contraction: The overall decrease in atomic and ionic radii with increasing atomic number is known as lanthanoid contraction. In going from La⁺³ to Lu⁺³ in lanthanoid series, the size of ion decreases. This decrease in size in the lanthanoid series is known as lanthanoid contraction. The lanthanoid contraction arises due to imperfect shielding of one 4f electron by another present in the same subshell.

Consequences :
(i) Similarity in properties: Due to lanthanoid contraction, the size of elements which follow (Hf – Hg) are almost similar to the size of the elements , of previous row (Zr – Cd) and hence these are difficult to separate. Due to small change in atomic radii, the chemical properties of lanthanoids are very similar due to which separation of lanthanoid becomes very difficult.
(ii) Basicity difference : Due to lanthanoid contraction, the size decreases from La⁺³ to Lu⁺³. Thus covalent character increases. Hence basic character of hydroxides also decreases i.e. why La(OH)₃ is most basic while Lu(OH)₃ is least basic.

Answer: Potassium dichromate (K₂Cr₂O₇) acts as a strong oxidising agent in acidic medium using H₂SO₄.
K₂Cr₂O₇ + 4H₂SO₄ → K₂SO₄ + Cr₂(SO₄)₃ + 4H₂O + 3[O]
Ionic reactions :

Answer: (i) The catalytic properties of the transition elements are due to the presence of unpaired electrons in their incomplete d-orbitals and variable oxidation states.
(ii) Due to lanthanoid contraction in second series after lanthanum, the atomic radii of elements of second and third series become almost same and hence show similarities in properties.
(iii) Because of very small energy gap between 5f, 6d and 7s subshells all their electrons can take part in bonding and shows variable oxidation states.

Answer: