NCERT Questions for Class 12 Chemistry Chapter 4 – The D and F Block Elements

Important Questions with Solutions of Class 12 Chemistry Chapter 4 – The D and F Block Elements

1) CuSO₄.5H₂O is blue while ZnSO₄ and CuSO₄ are colourless.

Ans – H₂O is a ligand in CuSO₄.5H₂O exhibiting crystal field splitting. So, the d-d transition is attainable and possible in CuSO₄.5H₂O. As water is absent in anhydrous CuSO₄, and ZnSO₄, ligand crystal field splitting is unattainable, and no colour is formed.

2) Silver (Ag) has filled d-orbital (4d¹⁰) during its ground state. Then, why has it been denoted as a transition element?

Ans – During its ground state silver has completely occupied 4d -orbital (4d¹⁰, 5s¹). The oxidation states of silver are at +1 and +2. An electron is eliminated from the s- orbital in the +1 oxidation state. However, the d-orbital gets one-electron reduction in the +2 oxidation state. So, the d-orbital remains incomplete (4d⁹). Hence, it is a transition element.

3) Cu²⁺ (aq) is more stable than Cu⁺ (aq). Why?

Ans – The stability of the ions in aqueous solutions depends upon the hydration energy (enthalpy) of different ions when they connect to the water molecules. Cu²⁺ ions have more charge density than Cu⁺ ions and form much stronger contact interactions, which require higher energy.

4) State the reason for the following:

 a) All the bonds between manganese and oxygen in permanganate ions are covalent.

Ans – MnO₄^– in its highest oxidation state will remain +7. The bond formed by transitional metals at high oxidation states is covalent (According to Fajan’s principle, as the oxidation state increases, the ionic character decreases).

b) Permanganate titrations in the hydrochloric acid are unacceptable.

Ans – Since hydrochloric acid is oxidized to chlorine, permanganate titrations in the acid cannot be trusted. In preparative organic chemistry, it is derived as a favored oxidant.

5) Answer the below questions:

a) Transition metals have such an extreme melting point?

Ans – Transition elements have a large density with melting and boiling points. These characteristics arise since metallic bonding through delocalized d-electrons can increase cohesion with the total shared electrons.

b) Which one has a higher melting point Fe or Cu?

Ans – The melting point of Fe is higher than Cu. Since iron has 4 unpaired electrons in the 3d-subshell whereas copper has only one electron in the 4s-subshell. Hence, the metallic bonds of iron are considerably stronger than copper’s.

6) Transition metals reveal the low oxidation state of carbon monoxide.

Ans – Synergic bonds provide the transition elements to develop complex in a zero oxidation state. In metal carbonyl, the M-C bond contains s and p characters. The M-C sigma bond develops when a lone pair of electrons of carbonyl carbon gets donated towards a vacant orbital of the metal.

The M-C bond is formed by donating an electron pair from a metal’s full d-orbital to the unoccupied anti-bonding pi* orbital on carbon monoxide. The synergic effect of metal-to-ligand bonding strengthens the carbon monoxide-metal bond.

7) Explain the following:

a) La(OH)₃ is stronger base than Lu(OH)₃

Ans – The most basic is La(OH)₃, and the least basic is Lu(OH)₃. The covalent nature of the hydroxides increases as the size of lanthanide ions changes from La³⁺ to Lu³⁺ and the basic strength reduces.

b) zn²⁺ salts are white

Ans – No unpaired electron in the electronic configuration of Zn²⁺ (3d¹⁰) salts. Generally, unpaired electrons cause a transition in the visible regions. Hence, salts of Zn²⁺ ion are white.

c) Cu (1) compounds aren’t stable in aqueous solutions and get disproportion.

Ans – Cu⁺ is more unstable in aqueous solution than Cu²⁺ as copper’s 2^nd I.E. is considerable, Cu²⁺ hydrations enthalpy is lower than Cu⁺ & overcompensates for copper’s 2nd I.E. Also, many Cu⁺ complexes in aqueous solution are unstable and disproportionate.

8) Answer the following:

a) Describe the preparation of potassium dichromate (K₂Cr₂O₇) Write the chemical equations of the reactions involved.

Ans – Potassium dichromate is prepared from chromite ore (FeO.Cr₂O₃) as follows:

Step 1: Chromite ore is roasted into sodium chromate:

4FeO.Cr₂O₃ +8Na₂CO₃ +7O₂ → 2Fe₂O₃ +8Na₂CrO₄ +8CO₂

Step 2: Sodium chromate is converted into sodium dichromate:

2Na₂CrO₄ +H₂SO₄ → Na₂Cr₂O₇ + Na₂SO₄ +H₂O

Step 3: Sodium dichromate converted to potassium dichromate:

Na₂Cr₂O₇ +2KCI → K₂Cr₂O₇ +2NaCl

(b) The chromates and dichromates are interconvertible by the change in pH of the medium. Why? Give chemical equations in favor of your answer.

Ans – The equilibrium reaction of dichromate and chromate ions is given as

Cr₂O₇^2- + H₂O → 2CrO₄^2- + 2H⁺

In the above reaction, dichromate ions ( Cr₂O₇^2-) give an orange colour, and chromate ions (CrO₄^2-) offer a yellow colour.

Hence, if the concentration of H⁺ ions is increased in the solution then pH is decreased, which shows the pH of the solution is acidic & the reaction shifts towards the left side and subsequently gives orange-coloured dichromate ions. The concentration of the OH ions will increase in the solution and hence, the pH of the solution increases and the equilibrium of the reaction will shift towards the right side and give yellow-coloured chromate ions.

9) Explain giving reasons:

a) Transition metals are less reactive than the alkali metals and alkaline earth metals.

Ans – Transition metals have higher ionization potential and melting point than alkali metals due to which transition metals are less reactive than alkali metals. As we move from the left to the right side of the periodic table, electrons will occupy the same shell or the same orbital.

The outermost shell is the least full for alkali metals with one electron hence they have fewer protons and have fewer nucleus attractions. All transition metals from the same period contain massive protons interacting at the same distance from the electrons in alkali metals.

b) The value of E° (Cu²⁺/Cu) is positive

Ans – Compared to H⁺/H₂, a positive value for Cu²⁺/Cu reveals that this reduction happens rapidly. Thus, it is a more powerful oxidizing agent, with redox couple Cu²⁺ /Cu more effective than H⁺/H2, Additionally, this means that Cu cannot convert H⁺ ions in acid to H₂.

c) Elements in the middle of the transition series have higher melting points.

Ans – The higher melting point of transition metals is due to the interatomic metallic bonding involving massive (n-1)d electrons using the ns electrons. Except for unusual values of manganese and technetium, the melting points of such metals rise to the maximum at d⁵ as the atomic number increases for the 3d series.

d) The atomic size of the elements of the transition series decreases slightly.

Ans – The outer electrons carry greater nuclear charges owing to the poor shielding effect of d-electrons. Since they have a higher effective nuclear charge, their size decreases slightly, and there is less fluctuation in size among transition elements.