Download The d- And f- Block Elements

The d- And f- Block Elements
Give reasons
1. Zn, Cd and Hg of group 12 are not regarded as transition metals.
Ans) A transition element is one which has incompletely filled d-orbital in its ground
state or in any one of its oxidation states. These metals have d10 configuration in
their ground state as well as in their common oxidation states.
2. Electronic configuration of Cr(Z=24) is [Ar] 3d54s1 instead of 3d44s2. Likewise
Cu(Z=29) is [Ar]3d104s1 instead of [Ar]3d94s2.
Ans) This is because half and completely filled sets of orbitals are relatively more stable
and this is possible because the energy gap between 3d and 4s sets of orbitals is
small.
3. Example 8.1 (pg 211)
4. Silver has outer configuration 4d105s1 in its ground state, but it is still a transition
metal.
Ans) Silver(Z=47) exhibits +2 state, wherein it will have incompletely filled 4d orbital.
Hence, it is a transition metal.
5. Transition metals have high melting and boiling points.
Ans) Due to the involvement of both ns and (n-1)d electrons in bonding, strong
interatomic metallic bonds are formed.
6. In any row of d-block elements the m.p of metals rise to a maximum at d5(except for
Mn and Tc) and then fall regularly.
Ans) This is because from d6 configuration onwards, due to pairing of electrons in d
orbitals, the no. of unpaired electrons are less. Therefore, interatomic bonds get
weaker. However, Mn and Tc show anomalous behaviour.
7. In many of the properties of the d-block elements, the maxima occur at about the
middle of each series.
Ans) This indicates that one unpaired electron per d-orbitals particularly a favourable
configuration for strong interatomic interaction.
8. Transition metals are hard, have high densities and high enthalpies of atomization.
Ans) Due to strong interatomic metallic bonds, because of large no. of unpaired electrons.
9. Metals of second and third transition series (4d and 5d series) have greater enthalpies
of atomization than the corresponding elements of first series (3d series).
Ans) Due to much more frequent metal-metal bonding in the heavier transition metals.
10. Successive ionization enthalpies for d-block elements do not increase as steeply as in
the main group elements with increasing atomic number.
Ans) Loss of variable numbers of electrons from (n-1)d orbitals is energetically
favourable.
11. In the series Sc(Z=21) to Zn(Z=30), the enthalpy of atomization of zinc in the lowest.
Ans) In the case of Zn, no electrons of 3d orbital are involved in the formation of metallic
bonds.
12. d-block elements are also called transition metals.
Ans) It is because of their position between the s-block elements (metallic)and p-block
elements(mostly non-metallic).
13. Atomic and ionic radii (ions of same charge) in a given series of d-block
progressively decrease with increase in atomic number.
Ans) This is because the new electron enters a d-orbital each time the nuclear charge
increases by unity. The shielding effect of d-electron is weak; hence the net
electrostatic attraction between the nucleus and the outermost electron increases and
radii decreases.
14. There is an increase in radii from 3d to 4d series but the radii of 5d series are nearly
the same as those of corresponding members of 4d series.
Ans) The 5d series of elements have fully filed 4f orbitals (i.e. they follow lanthanoids).
Due to very weak shielding effect of 4f electrons there is a regular decrease in
atomic radii. This effect is called Lanthanoid contraction, which compensates for
the expected increase in atomic size with increasing atomic number down the
group.
15. The 2nd and 3rd transition series e.g. Zr and Hf or Nb and Ta have similar physical and
chemical properties, much more than that expected from usual family relationship.
Ans) This is due to the similarity in their radii (due to Lanthanoid contraction).
16. From Ti(Z=22) to Cu(Z=29) there is a significant increase in density.
Ans) The decrease in metallic radius coupled with increase in atomic mass results in an
increase in density of these elements.
17. Ionisation enthalpy increase along each series.
Ans) Due to increase in nuclear charge which accompanies the filling of inner d-orbitals,
coupled by weak shielding effect of the d-electrons.
18. The trend in ionization enthalpy of 3d metals is irregular.
Ans) Removal of one electron alters the relative energies of 4s and 3d orbitals. So, the
unipositive ions have dn configurations with no 4s electrons. There is reorganization
energy accompanying ionization and exchange energy increases as no. of delectrons increase. Value for Cr is low because of no change in d-configuration but
value for Zn is higher because of ionization from 4s orbital.
It is also due to varying degree of stability of different 3d configurations, d0,d5 and
d10 are exceptionally stable.
19. Cr and Cu have unusually high second ionization enthalpy value and Zn
correspondingly low.
Ans) 24Cr : 1s22s22p63s23p63d54s1
2 2 6 2
6 10 1
29Cu : 1s 2s 2p 3s 3p 3d 4s
2 2 6 2 6 10 2
30Zn : 1s 2s 2p 3s 3p 3d 4s
In Cr and Cu the stable half filled and fully filled d5 and d10 configuration of the
ions are disrupted, while Zn2+ allows the production of stable d10 state.
20. Mn2+ compounds are more stable than Fe2+ towards oxidation to their +3 state.
Ans) 25Mn : 1s22s22p63s23p63d54s2
Mn2+ : 1s22s22p63s23p63d5 (a stable half-filled configuration). Therefore, Mn2+
resists oxidation to Mn3+.
2 2 6 2 6 6 2
26Fe : 1s 2s 2p 3s 3p 3d 4s
2+
Fe : 1s22s22p63s23p63d6
Fe3+ : 1s22s22p63s23p63d5 (stable half-filled configuration). Fe3+ has stable halffilled configuration. Therefore, Fe2+ tends to easily get oxidized to more stable +3
state.
21. The +2 state becomes more and more stable in the first half of the 3rd series with
increasing atomic number.
Ans) In unipositive ions M+, there is transfer of electrons from 4s to 3d and so there is
gain of exchange energy. Exchange energy is maximum where the d-orbitals are
singly occupied. So stability of +2 state increases to the middle of the series;
thereafter, pairing of d-electrons occur.
22. Transition metals show variable oxidation state (O.S).
Ans) Due to the participation of the ns as well as the (n-1)d electrons in bonding.
23. The elements which give the greatest number of oxidation state occur in or near the
middle of the series. (e.g. Mn, Cr) Mn = 7, Cr = 6.
Ans) Due to the availability of greater number of unpaired d-electrons for bonding.
24. The variation in oxidation state of transition elements is by unity while for nontransition elements (p-block) it is by 2.
Ans) In transition elements variability in O.S arises due to incompletely filled or partially
filled d-orbitals, while in non-transition elements it is due to inert pair effect and
excitation of elements.
Eg. O.S of V = +2, +3, +4, +5; Mn = +2 to +7; Cr = +2 to +6
While for Pb = +2 and +4; S = 2, 4, 6; P = 3, 5
25. Cr2+ is reducing and Mn3+ is oxidizing when both have d4 configuration.
Ans) Cr2+ is reducing (i.e. acts as reducing agent) as its configuration changes from d4 to
more stable d3. d3 is more stable because it has a half-filled t2g level. Whereas, Mn3+
is oxidizing (i.e. acts as an oxidizing agent) because it gets reduced to more stable
Mn2+ with half-filled d5 configuration.
26. In aqueous medium Cr2+ is a better reducing agent than Fe2+.
Ans) In case of Cr2+, d4  d3.
In case of Fe2+, d6  d5.
Water acts as ligands and in the presence of ligands, the 5d orbitals form two sets t2g
and eg. In Cr3+ there is half-filled configuration in t2g. So, Cr3+ is more stable.
Therefore, Cr2+ is a stronger reducing agent.
27. The highest oxidation state of a metal is exhibited in its oxide or fluoride.
Ans) The ability of fluorine to stabilize the highest oxidation state is due to either higher
lattice energy (for ionic compounds) or higher bond enthalpy (for covalent
compounds). The ability of oxygen to stabilize highest oxidation states is due to its
ability to form multiple bonds with the metal.
28. In the 3d series Mn exhibits largest number of oxidation states.
Ans) Because it has the maximum no.(7) unpaired electrons Mn [Ar] 3d54s2.
29. Co(II) is stable in aqueous solution but in the presence of complexing reagents it is
readily oxidized to (III) state.
Ans) Due to crystal field splitting energy which more than compensates the 3rd ionization
enthalpy.
30. The d1 configuration is very unstable for ions.
Ans) The hydration or lattice energy more than compensates the ionization enthalpy
involved in removing electron from d1. Moreover, the species also acquires the
nearest noble gas configuration.
31. In the 3d series, only Cu exhibits +1 state .
Ans) Because Cu+ has stable 3d10 configuration.
32. Cu+ ion is not stable in aqueous solution.
Ans) 2Cu+(aq)  Cu2+(aq) + Cu(s)
Cu+ undergoes disproportionation. The Eo value for this reaction is favourable. Cu2+
also has higher hydration energy than Cu+ which compensates for the high 2nd I.E of
Cu.
33. The highest oxidation state is exhibited in oxoanions of a metal.
Ans) Small size and high electro-negativity of Oxygen oxidizes metal to highest O.S.
Oxygen can also form multiple bonds with metal.
34. The lowest oxide of transition metal is basic, the highest amphoteric or acidic.
Ans) In lower oxidation state of metal, the oxides are ionic and ionic oxides are basic. In
higher oxidation states, they are covalent and covalent oxides are either acidic or
amphoteric.
35. The Eo (M2+/M) values for first row series are irregular.
Ans) Eo values depend on I.E1 + I.E2 as well as hydration and sublimation energies which
show irregular variation.
36. The Eo (Cu2+/Cu) value for copper is positive (0.34V) while for all other metals of 3d
series it is negative.
Ans) Due to its high sublimation energy and low hydration energy. The energy required
to change Cu  Cu2+ is not balanced.
37. Eo (M3+/M2+) values of Sc and Fe are low but that of Zn and Mn are high.
Ans) Low values indicate stability of +3 more than +2 while high values indicate +2 state
is more stable than +3.
Sc : +3 state is stable because it has a noble gas configuration of Ar.
Fe3+ is more stable due to half-filled 3d5 configuration. Zn2+ is more stable due to
3d10 configuration. Mn2+ is stable because of 3d5 configuration.
[Eo indicates reduction potential, high Eo indicates high tendency to get reduced i.e.
from M3+ to M2+].
38. Increasing oxidation power in VO2+ < Cr2O72 < MnO4.
Ans) This is due to increasing stability of the lower species to which they are reduced i.e.
Mn2+, Cr3+ and V2+.
39. Eo value for Mn3+/Mn2+ couple is much more positive than that for Cr3+/Cr2+ or
Fe3+/Fe2+.
Ans) Mn2+ is more stable than Mn3+. Mn2+ has half-filled 3d5 configuration, therefore
Mn3+ is easily reduced to Mn2+. Whereas, Cr3+ and Fe3+ are more stable, so they are
less readily reduced.
40. Transition metals and many of their compounds show paramagnetic behaviour.
Ans) Due to the presence of one more unpaired electron in the orbitals.
(Use spin formula to calculate magnetic moment:
  n(n  2)
Where n = no. of unpaired electrons. Unit of  is BM. Higher the n value, greater
the degree of paramagnetism.)
Sc3+, Cu+, Zn2+ are diamagnetic due to the absence of unpaired electrons.
41. The transition metals generally form coloured compounds or coloured ions.
Ans) In compounds or complexes, the d-orbital of the transition metal split into two sets
of orbitals of slightly different energies. In transition elements with partly filled dorbitals electrons can move from one set of d-orbitals to another. The energy
required for this is available in the visible region. The colour observed is
complementary to the colour of the light absorbed. Ions with empty d-orbitals (d0
configurations) or fully filled d-orbitals (d10 configuration) will be colourless. E.g.
Sc3+, Zn2+, Cu+ are colourless. Cr3+, Mn2+, Ni2+ are coloured (partially filled dorbitals).
42. Transition metals form a large no. of complex compounds.
Ans) Due to small size of metal ions, their high ionic charges and availability of empty dorbitals for bond formation.
43. Transition metals and their many compounds act as good catalyst.
Ans) This is due to their ability to adopt multiple oxidation state. They provide a suitable
surface for adsorption of reactant molecules. They can form unstable intermediate
complexes with the reactants and provide a new path for the reaction with lower
activation energy.
44. Transition metals form interstitial compounds easily.
Ans) Because small atoms like H, C, N, B etc. are trapped in the interstitial sites in the
crystal lattice of these metals.
45. Transition metals form alloys easily.
Ans) Alloys are homogenous solid solutions. Alloys are formed by atoms whose radii are
within 15% of each other. Since variations of radii of transition metals are very less,
they form alloys easily.
46. K2Cr2O7 is preferred over Na2Cr2O7 as oxidizing agent in volumetric analysis.
Ans) Na2Cr2O7 is hygroscopic, i.e. it absorbs moisture from air, so it cannot be used as a
primary standard in volumetric analysis.
47. Orange colour of K2Cr2O7 turns to yellow on adding alkali (e.g. NaOH).
Ans) Due to the formation of yellow chromate ions (CrO42)
Cr2O72 + 2OH  2CrO42 + H2O
orange
yellow
48. In permanganate and dichromate ions (MnO4 and Cr2O72), all the bonds between
the metal and oxygen are covalent.
Ans) Because Mn and Cr are in +7 and +6 state respectively. In higher oxidation states,
bonds are covalent.
49. HCl is not used to acidify KMnO4 or K2Cr2O7 during volumetric analysis.
Ans) KMnO4 and K2Cr2O7 are strong oxidizing agents and they oxidize HCl to Cl2. This
will affect volumetric estimation. Therefore, H2SO4 is preferred.
50. Though stable O.S of lanthanoids is +3, Ce and Tb exhibit +4 state. Eu and Yb
exhibit +2 state as well.
Ans) Ce in +4 state has stable noble gas configuration of Xe. Tb4+ has stable half-filled
4f7 configuration.
Eu2+ and Yb2+ have stable 4f7 and 4f14 configurations respectively.
51. Eu2+ and Yb2+ are reducing agents while Ce4+ and Tb4+ are oxidants.
Ans) Eu2+ and Yb2+ are oxidized to +3 state, the most stable state of lanthanoids.
Whereas, Ce4+ and Tb4+ are reduced to the +3 state.(+3 state is the most stable state
of the lanthanoids)
52. Most of the lanthanoids and actinoids are paramagnetic and coloured.
Ans) Due to the presence of unpaired electrons in 4f and 5f orbitals of these metals.
53. The basic strength of hydroxides of lanthanoids M(OH)3 decreases from La to Lu.
Ans) Due to lanthanoid contraction, there is a steady decrease in size of M3+ ions from La
to Lu. Therefore, MOH bond strength increases and basic strength decreases.
54. Actinoid contraction is greater from element to element than lanthanoid contraction.
Ans) Due to weaker shielding effect of 5f electrons from element to element in the series.
55. Separation of lanthanoid elements from one another is difficult.
Ans) Since the change in the ionic radius is very small, their chemical properties are
similar. This makes their separation difficult.
56. The chemistry of actinoids is more complex than lanthanoids.
Ans) Due to the ability of actinoids to exist in more no. of oxidation states than
lanthanoids. Lanthanoids : +2, +3, +4 ; Actinoids : +3, +4, +5, +6, +7.
57. There is a greater range in O.S in actinoids than lanthanoids.
Ans) Because 5f, 6d and 7s levels are of comparable energies, electrons of all these
orbitals are involved in bonding.
58. Ionisation enthalpy of actinoids is less than that of corresponding lanthanoids.
Ans) Penetration of 5f electrons are less than 4f electrons, so 5f electrons are better
shielded from the nucleus, than 4f electrons. So, it is easier to remove outer
electrons in actinoids than in lanthanoids.
59. in group 6, MoO3 and WO3 are not oxidizing agents but Cr(VI) in the form of
dichromate ions in acid medium is astrong oxidant.
Ans) Higher oxidation states(+6) is more stable for the heavier members of the group.
60. MnO42─ ions undergo disproportionation in acidic or neutral medium.
Ans) In MnO42─ Mn is in the +6 state which is not a stable state of the metal so it
undergoes disproportionation to the more stable MnO4─ (+7) and MnO2(+4)
3 MnO42─ + 4 H+ → 2 MnO4─ + MnO2 + 2 H2O
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