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Important Questions from - d and f Block Elements | XII CBSE/ISC | Board Exam Preparation

Q1. Silver atom has completely filled d-orbitals (4d10) in its ground state. How can you say that it is a transition element?

Ans. Silver (Z = 47) can exhibit +2 oxidation state wherein it will have incompletely filled d-orbitals (4d), hence a transition element.


Q2. In the series Sc (Z = 21) to Zn (Z = 30), the enthalpy of atomisation of zinc is the lowest, i.e., 126 kJ mol–1. Why?

Ans - In the formation of metallic bonds, no electrons from 3d-orbitals are involved in case of zinc, while in all other metals of the 3d series, electrons from the d-orbitals are always involved in the formation of metallic bonds. That is why, the enthalpy of atomisation of zinc is the lowest in the series.


Q3. Which of the 3d series of the transition metals exhibits the largest number of oxidation states and why?

Ans - Manganese (Z = 25), as it has the maximum number of unpaired electrons in d-subshell. Thus, it shows oxidation states from +2 to +7 (+2, +3, +4, +5, +6 and +7) which is the maximum number.


Q4. How would you account for the irregular variation of ionisation enthalpies (first and second) in first series of the transition elements?

Ans - Irregular variation of ionisation enthalpies is mainly attributed to varying degree of stability of different 3d configuration (e.g., d0 , d 5 , d10 are exceptionally stable).


Q5. Why is the highest oxidation state of a metal exhibited in its oxide or fluoride only?

Ans - Due to small size and high electronegativity, oxygen or fluorine can oxidise a metal to its highest oxidation state. As a result of this they can oxidise a metal to its highest oxidation state.

Q6. Which is a stronger reducing agent— Cr2+ or Fe2+ and why?

Ans - Cr2+ is a stronger reducing agent than Fe2+ because after the loss of one e electron Cr2+ becomes Cr3+ which has more stable t2g-3 (half-filled) configuration in a medium like water.


Q7. Calculate the spin-only magnetic moment of Co2+(Z=27) by writing the electronic configuration of Co and Co2+.

Ans - Electronic configuration of M atom with Z = 27 is [Ar] 3d7 4s2

Electronic configuration of M2+ = [Ar] 3d7 , i.e., it has 3 unpaired electrons

Therefore, Magnetic Moment = √n(n+2) = √3(3+2) = 3.87 BM


Q8. Explain why Cu+ ion is not stable in aqueous solutions?

Ans - The higher stability of Cu2+ in aqueous solution may be attributed to its greater negative ∆hyd. Ho than that of Cu+ . It compensates the second ionisation enthalpy of Cu involved in the formation of Cu2+ ions.


Q9. Actinoid contraction is greater from element to element than lanthanoid contraction. Why?

Ans - This is because the 5f electrons themselves provide poor shielding from element to element in the series.


Q10. Why are Mn2+ compounds more stable than Fe2+ compounds towards oxidation to their +3 state?

Ans - Electronic configuration of Mn2+ is 3d5 which is half-filled and hence stable. So, 3rd ionisation enthalpy is very high, i.e., 3rd electron cannot be easily lost. In case of Fe2+, electronic configuration is 3d6 . Thus, it can lose one electron easily to give the stable configuration 3d5 .

Q11. What is lanthanoid contraction? What are the consequences of lanthanoid contraction?

Ans - The gradual decrease in the size of elements of lanthanoids because of poor shielding of 4f orbitals is called Lanthanoid Contraction.

The consequences of lanthanoid contraction are as follows:

(i) The properties of second and third transition series are similar. 312 Xam idea Chemistry–XII

(ii) Basic strength decreases from La(OH)3 to Lu(OH)3.

(iii) Lanthanide contraction makes separation of lanthanoids possible.


Q12. Which of the d-block elements may not be regarded as the transition elements?

Ans - The electronic configuration of Zn, Cd and Hg are represented by the general formula (n – 1) d 10 ns 2. The orbitals in these elements are completely filled in the ground state as well as in their common oxidation states. Therefore, they are not regarded as transition elements.


Q13. What are the different oxidation states exhibited by lanthanoids?

Ans - +2, +3 and +4 (+3 being most common).


Q14. The enthalpies of atomisation of the transition metals are high, why?

Ans - The transition elements exhibit high enthalpy of atomisation because they have large number of unpaired electrons in their atoms. Due to this they have stronger interatomic interaction.


Q15. Transition metals and their many compounds act as good catalyst.

Ans - The transition metals and their compounds are known for their catalytic activity. This activity is due to their ability to adopt multiple oxidation states and to form complexes.


Q16. What are interstitial compounds? Why are such compounds well known for transition metals?

Ans - Interstitial compounds are those in which small atoms occupy the interstitial sites in the crystal lattice. Interstitial compounds are well known for transition metals because small-sized atoms of H, B, C, N, etc., can easily occupy positions in the voids present in the crystal lattices of transition metals.


Q17. How is the variability in oxidation states of transition metals different from that of the non transition metals? Illustrate with examples.

Ans - The oxidation states of transition elements differ from each other by unity e.g., Fe2+ and Fe3+, Cu+ and Cu2+ (due to incomplete filling of d-orbitals) whereas oxidation states of non-transition elements normally differ by two units e.g., Pb2+ and Pb4+, Sn2+ and Sn4+, etc.


Q18. The d4 species, Cr2+ is strongly reducing while manganese (III) is strongly oxidising.

Ans - E° value for Cr3+/Cr2+ is negative (–0.41 V) whereas E° value for Mn3+/Mn2+ is positive (+1.57 V). Thus, Cr2+ ions can easily undergo oxidation to give Cr3+ ions and, therefore, act as strong reducing agent. On the other hand, Mn3+ can easily undergo reduction to give Mn2+ and hence act as oxidising agent.


Q19. Cobalt (II) is stable in aqueous solution but in the presence of complexing reagents it is easily oxidised.

Ans - This is because in presence of complexing reagents the CFSE value compensates more than the third ionisation energy of cobalt.


Q20. Which metal in the first transition series exhibits +1 oxidation state most frequently and why?

Ans - Cu has the electronic configuration 3d10 4s1 . It can easily lose 4s1 electron to give the stable 3d10 configuration. Hence, it shows +1 oxidation state.


Q21. The lowest oxide of transition metal is basic, the highest is amphoteric/acidic.

Ans - The lowest oxide of transition metal is basic because the metal atom has low oxidation state. This means that it can donate valence electrons which are not involved in bonding to act like a base. Whereas the highest oxide is acidic due to the highest oxidation state as the valence electrons are involved in bonding and are unavailable. For example, MnO is basic whereas Mn2O7 is acidic.


Q22. A transition metal exhibits higher oxidation states in oxides and fluorides.

Ans - A transition metal exhibits higher oxidation states in oxides and fluorides because oxygen and fluorine are highly electronegative elements, small in size (and strongest oxidising agents). For example, osmium shows an oxidation states of +6 in OsF6 and vanadium shows an oxidation state of +5 in V2O5.


Q23. The highest oxidation state is exhibited in oxo-anions of a metal.

Ans - Oxometal anions have the highest oxidation state, e.g., Cr in Cr2 7 O2– has an oxidation state of +6 whereas Mn in MnO4 – has an oxidation state of +7. This is again due to the combination of the metal with oxygen, which is highly electronegative and oxidising element.


Q24. What are alloys? Name an important alloy which contains some of the lanthanoid metals. Mention its uses.

Ans - An alloy is a homogeneous mixture of two or more metals, or metals and non-metals. An important alloy containing lanthanoid metals is misch metal which contains 95% lanthanoid metals and 5% iron alongwith traces of S, C, Ca and Al. It is used in Mg-based alloy to produce bullets, shells and lighter flints.


Q25. What are inner-transition elements? Decide which of the following atomic numbers are the numbers of the inner-transition elements: 29, 59, 74, 95, 102, 104.

Ans - The f-block elements, i.e., in which the last electron enters into f-subshell are called inner-transition elements. These include lanthanoids (58–71) and actinoids (90–103). Thus, elements with atomic numbers 59, 95 and 102 are inner-transition elements.


Q26. Why is zinc not regarded as a transition clement?

Ans - As zinc atom has completely filled d-orbitals (3d10) in its ground state as well as in oxidised state, therefore, it is not regarded as a transition element.


Q27. Why does a transition series contain 10 elements?

Ans - There are five d-orbitals in an energy level and each orbital can contain two electrons. As we move from one element to the next, an electron is added and for complete filling of the five d-orbitals, 10 electrons are required.


Q28. What are interstitial compounds?

Ans - Interstitial compounds are those in which small atoms occupy the interstitial sites in the crystal lattice.


Q29. Copper atom has completely filled d-orbitals in its ground state but it is a transition element. Why?

Ans - Copper exhibits +2 oxidation state wherein it has incompletely filled d orbitals (3d 9 4s 0 ) hence, a transition element.


Q30. Give reason: Zn is soft whereas Cr is hard.

Ans - Cr (3d5 4s1 ) has five unpaired electrons in its d-orbitals whereas Zn (3d10 4s2 ) has no unpaired electrons in its d-orbitals. As a result of this weak metallic bonds exist in Zn whereas strong metallic bonds exist in Cr. Hence, Zn is soft whereas Cr is hard.

Q31. Reactivity oftransition elements decreases almost regularly from Sc to Cu.Explain.

Ans - It is due to regular increase in ionisation enthalpy.


Q32. Zn2+ salts are white while Cu2+ salts are coloured. Why?

Ans - Cu2+ (3d9 4s0 ) has one unpaired electron in d-subshell which absorbs radiation in visible region resulting in d-d transition and hence Cu2+ salts are coloured. Zn2+ (3d10 4s0 ) has completely filled d-orbitals. No radiation is absorbed for d-d transition and hence Zn2+ salts are colourless.


Q33. In the following ions: Mn3+, V3+, Cr3+, Ti4+

(Atomic no. : Mn = 25, V = 23, Cr = 24, Ti = 22)

(i) Which ion is most stable in an aqueous solution?

(ii) Which ion is the strongest oxidizing agent?

(iii) Which ion is colourless?

(iv) Which ion has the highest number of unpaired electrons?

Ans - (i) Cr3+ because of half filled t2g level.

(ii) Mn3+, as the change from Mn3+ to Mn2+ results in stable half filled (d5 ) configuration.

(iii) Ti4+, as Ti4+ has empty d-orbitals therefore d-d transition cannot occur in Ti4+.

(iv) Mn3+ (3d4 4s 0 ). It has 4 unpaired electrons.


Q34. Why Ti3+ is coloured whereas Sc3+ is colourless in aqueous solution.

Ans - Ti3+ is coloured because of one electron in d orbtal d-d transition is possible incase of Ti+3 whereas no d electrons are present in Sc+3.


Q35. Why the enthalpies of atomisation of transition metals are quite high?

Ans - This is because transition metals have strong metallic bonds as they have large number of unpaired electrons.

Q36. There is a close similarity in physical and chemical properties of the 4d and 5d series of the transition elements, much more than expected on the basis of usual family relationship.

Ans - This is because 5d and 4d-series elements have virtually the same atomic and ionic radii due to lanthanoid contraction. Due to equality in size of Zr and Hf, Nb and Ta, Mo and W, etc., the two elements of each pair have the same properties.


Q37. The members in the actinoid series exhibit larger number of oxidation states than the corresponding members in the lanthanoid series.

Ans - The members in the actinoid series exhibit larger number of oxidation states than the corresponding members in the lanthanoid series due to the fact that the 5f, 6d and 7s levels are of comparable energies.


Q38. Transition metals form large number of complex compounds.

Ans - The tendency to form complex compounds is due to:

—Small size and high charge on metal ion.

—The availability of d orbitals for accommodating electrons donated by the ligand.


Q39. Write one similarity and one difference between the chemistry of lanthanoid and actinoid elements.

Ans - Similarities

— Both show mainly an oxidation state of +3.

— Actinoids show actinoid contraction like lanthanoid contraction is shown by lanthanoids.

— Both are electropositive and very reactive. (Any one)

Differences

— Except promethium (Pm) lanthanoids are non-radioactive whereas actinoids are radioactive.

— Lanthanoids do not form oxocation whereas actinoids form oxocation.

— Lanthanoids have less tendency towards complex formation whereas actinoids have greater tendency towards complex formation.


Q40. Scandium (Z = 21) does not exhibit variable oxidation states and yet it is regarded as a transition element.

Ans - This is because scandium has partially filled d-orbitals in the ground state (3d1 4s 2 ).


Q41. The enthalpies of atomisation of transition elements are high.

Ans - This is because transition metals have strong metallic bonds as they have a large number of unpaired electrons.


Q42. The E0 value for the Mn3+/Mn2+ couple is much more positive than that of Cr3+/Cr2+.

Ans - This is due to much larger third ionisation energy of Mn as Mn2+ is very stable on account of stable d5 configuration.


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