d AND f BLOCK ELEMENTS

CLASS – XII
CHEMISTRY
IMPORTANT QUESTIONS FOR PRE-BOARDS AND BOARD
1. Why do transition elements show variable oxidation states?
Due to incompletely filled d-orbitals and presence of unpaired electrons.
2. Why transition elements and their compounds are generally found to be good
catalysts?
Due to the presence of unpaired electrons in their incomplete d- orbitals and variable
oxidation states.
3. Explain the following observations :
(i) Transition elements generally form coloured compounds.
(ii) Zinc is not regarded as a transition element.
(iii) The enthalpies of atomization of transition metals are quite high.
(i) Because of presence of unpaired d electrons, which undergoes d-d transition by
absorption of energy from visible region and then emits complementary colours. This is
how transition elements form coloured compounds.

(ii) Zinc has completely filled d-orbitals in its ground state as well as its common
oxidation state +2. Hence considered as non-transition element.
(iii) Due to the presence of a large number of unpaired electrons in their atoms, they
have a stronger interatomic interaction and stronger bonding between the atoms.
4. Assign reasons for the following:
(i) Copper (I) ion is not known in aqueous solution.
(ii) Actinoids exhibit greater range of oxidation states than lanthanoids. OR The
chemistry of actinoids is not so smooth as that of lanthanoids.
(iii) Zn, Cd and Hg are soft metals.
(iv) HCl is not used to acidify KMnO4 solution.
(i) Cu2+(aq) is much more stable than Cu+(aq). This is because of a more negative
hydration enthalpy for Cu2+(aq) than that for Cu+(aq). Therefore, copper (I)
compounds are unstable in aqueous solution and undergo disproportionation as follows:
2Cu+ → Cu2+ + Cu
(ii) 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.
(iii) Due to completely filled d-orbitals.
(iv) HCl is not used to the acidify KMnO4 solution because KMnO4 is a very strong
oxidizing agent and it can oxidize HCl to liberate chlorine gas.
 5. Explain:
(i) Cr2+ is reducing in nature while with the same d-orbital configuration (d4) Mn3+ is
an oxidising agent.
(ii) In a transition series of metals, the metal which exhibits the greatest number of
oxidation states occurs in the middle of the series.
(iii) Metal-metal bonding is more frequent for the 4d and the 5d series of transition
metals than that for the 3d series.
(i) Cr2+ has the configuration 3d4 which easily changes to d3 due to stable half-filled t2g
orbitals. Therefore Cr2+ is reducing agent. While Mn2+ has stable half-filled d5
configuration. Hence Mn3+ easily changes to Mn2+ and acts as oxidising agent.

(ii) Due to presence of more unpaired electrons and use of all 4s and 3d electrons in the
middle of series.
(iii) Metal-metal bonding is more frequent for the 4d and the 5d series of transition
metals than that for the 3d series as these have their electrons of outer most shell at
greater distance from the nucleus, as compared to atoms of 3d transition metals.
6. Complete the following chemical equations:
(i) MnO4 (aq) + S2O32- (aq) + H2O (l) →
(ii) Cr2O72- (aq) + Fe2+ (aq) + H+ (aq) →

7. Complete the following chemical reaction equations:

8. (a) Which metal in the first transition 3d series exhibits +1 oxidation state most
frequently and why?
(b) Which of the following cations are coloured in aqueous solutions and why?
Sc3+, V3+, Ti4+, Mn2+.
(At. nos. Sc = 21, V = 23, Ti = 22, Mn = 25)
(a) Copper exhibits + 1 oxidation state more frequently i.e., Cu+1 because of its
electronic configuration 3d104s1.
 (b) V3+ and Mn2+ are coloured in their aqueous solution due to presence of unpaired
electron.
9. What is Lanthanoid contraction? What are its two consequences?
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, Zr and Hf have almost
similar sizes. 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. Hence basic character of hydroxides also decreases. Hence La(OH)3 is most basic
while Lu(OH)3 is least basic.
10. Explain:
(i) The transition metals (with the exception of Zn, Cd and Hg) are hard and have high
melting and boiling points.
(ii) The ionization enthalpies (first and second) in the first series of the transition
elements are found to vary irregularly.
(i) Because of stronger metallic bonding and high enthalpies of atomization.
(ii) Due to irregularities in the electronic configuration, there are irregularities in the
enthalpies of atomisation. Hence there is irregular variation in I.E.
11. Explain
(i) The highest oxidation state of a transition metal is usually exhibited in its oxide.
(ii) The oxidising power follows the order:
VO+2 < Cr2O72- < MnO4-
(i) The highest oxidation state of a metal is exhibited in its oxide or fluoride due to its
high electronegativity and small size.
(ii) Hint: Calculate the oxidation states of the central atoms V, Cr and Mn. Higher the
oxidation states, greater the oxidising potential.
12. Write one similarity and one difference between the chemistry of lanthanoids and that
of actinoids.
Similarity: Both lanthanoids and actinoids show contraction in size and irregularity in
their electronic configuration.
Difference: Actinoids show wide range of oxidation states but lanthanoids do not.
13. What is meant by ‘disproportionation’? Give an example of a disproportionation
reaction in aqueous solution.
Disproportionation: In a disproportionation reaction an element undergoes self-
oxidation as well as self-reduction forming two different compounds.
 14. Describe the oxidising action of potassium dichromate and write the ionic equations
for its reaction with
(i) iodine (ii) H2S.

15. When chromite ore FeCr2O4 is fused with NaOH in presence of air, a yellow-coloured
compound (A) is obtained which on acidification with dilute sulphuric acid gives a
compound (B). Compound (B) on reaction with KCl forms an orange-coloured
crystalline compound (C).
(i) Write the formulae of the compounds (A), (B) and (C).
(ii) Write one use of compound (C).

Na2Cr2O7 + 2KCl → 2NaCl + K2Cr2O7 (C)

Use of C – Used as a strong oxidising agent.
16. Explain:
a. E0 value for Mn3+ /Mn2+ couple is much more positive than Cr3+/Cr2+
b. The atomic radii of the metals of the third (5d) series of transition elements are
virtually the same as those of the corresponding members of the second (4d) series.
c. Many of the transition elements are known to form interstitial compounds.
d. The E°M2+/M for copper is positive (0.34 V).
e. Transition metals are much harder than the alkali metals.
f. MnO is basic while Mn2O7 is acidic.
g. Transition metals form a large number of complexes.

a. The large positive E° value for Mn3+/Mn2+ shows that Mn2+ is much more stable
than Mn+3 due to stable half filled configuration (3d5). Therefore the 3rd
ionisation energy of Mn will be very high and Mn3+ is unstable and can be easily
reduced to Mn2+. E° value for Cr3+ | Cr2+ is positive but small i.e. Cr3+ can also be
reduced to Cr2+ but less easily. Thus Cr3+ is more stable than Mn3+.
b. Due to lanthanoid contraction the atomic radii of elements of second and third
series become almost same and hence show similarities in properties.
 c. The transition metals form a large number of interstitial compounds in which
small atoms such as hydrogen, carbon, boron and nitrogen occupy the empty
spaces in the crystal lattices of transition metals.
d. Copper has high enthalpy of atomisation and low enthalpy of hydration.
e. Strong metallic bonding due to presence of unpaired electrons in their ‘d’
orbitals.
f. MnO is basic while Mn2O7 is acidic because the basic nature decreases as the
oxidation state or number of oxygen atoms increases i.e. MnO (+4) and Mn 2O7
(+7)
g. Because of smaller size of their ions and availability of vacant d-orbitals,
transition metals from a large number of complexes.

17. Describe the preparation of potassium dichromate from chromite ore. What is the
effect of change of pH on dichromate ion?
Preparation given above Q15.
Effect of pH
2CrO4-2 + 2H+ → Cr2O7-2 + H2O
Cr2O7-2 + 2OH–→ 2CrO4-2 + H2O
18. Complete the following equations:

19. The elements of 3d transition series are given as:

Sc, Ti, V, Cr, Mn, Fe, Co, Ni, Cu, Zn
Answer the following:
(i) Write the element which shows maximum number of oxidation states. Give reason.
(ii) Which element has the highest m.p.?
(iii) Which element shows only +3 oxidation state?
(iv) Which element is a strong oxidizing agent in +3 oxidation state and why?
(i) Mn
(ii) Cr
(iii) Sc
(iv) Mn is a strong oxidizing agent in +3 oxidation state because after reduction it
attains +2 oxidation state in which it has the most stable half-filled (d5) configuration.
 20. Name a member of the lanthanoid series which is well known to exhibit +4 oxidation
state.
Ce
21. Why are Cu2+ salts are coloured while Zn2+ salts are white?
Cu2+ has the configuration 3d9 with one unpaired electron which gets excited in the
visible region to impart its colour while Zn2+ has 3d10 configuration without any
unpaired electron so no d – d transition possible and hence colourless.
22. Orange colour of Cr2O72- ion changes to yellow when treated with an alkali. Why?
Potassium dichromate when treated with NaOH solution turns yellow due to formation
of chromate.
23. Following are the transition metal ions of 3d series:
Ti4+, V2+, Mn3+, Cr3+
(Atomic numbers: Ti = 22, V= 23, Mn = 25, Cr = 24)
Answer the following:
(i) Which ion is most stable in an aqueous solution and why?
(ii) Which ion is a strong oxidising agent and why?
(iii) Which ion is colourless and why?
(i) Cr3+ is most stable because of its small size and t32g configuration.
(ii) Mn3+ is a strong oxidising agent because after gaining one electron it is converted to
Mn2+ which has stable d5 configuration.
(iii) Ti4+ is colourless due to d° configuration, i.e., no unpaired electrons.