CBSE Class 12 Chemistry D And F Block Elements Worksheet Set 03

Short Answer Type Questions

Question. (i) Write the formula of an oxo-anion of Manganese (Mn) in which it shows the oxidation state equal to its group number. 
(ii) Write the formula of an oxo-anion of Chromium (Cr) in which it shows the oxidation state equal to its group number. 
Answer: (i) \( \text{MnO}_4^- / \text{KMnO}_4 \)
(ii) \( \text{Cr}_2\text{O}_7^{2-} / \text{CrO}_4^{2-} / \text{K}_2\text{Cr}_2\text{O}_7 / \text{K}_2\text{CrO}_4 \)

Question. (i) Why Lanthanoids cannot be easily separated?
(ii) Why does Zr (Z=40) and Hf (Z=72) exhibit almost identical atomic radii? 
Answer: (i) Due to lanthanoid contraction, the change in the atomic or ionic radii of these elements is very small. So, their chemical properties are similar. Hence, Lanthanoids cannot be easily separated.
(ii) Lanthanoid contraction is responsible for almost same atomic radii of 4d and 5d transition series elements.

Question. What are the transition elements? Write two characteristics of the transition elements.
Answer: These atoms whose d-orbitals are incomplete in ground state or in one of the most common oxidation state are called transition elements or d-block elements. The valence shell electronic configuration of transition elements is \( (n-1)d^{1-10}ns^{1-2} \).
Two characteristics of transition elements:
(i) Transition metals show variable oxidation states.
(ii) All transition metals act as catalyst.

Question. What is meant by ‘disproportionation’? Give an example of a disproportionation reaction in aqueous solution. 
OR
Suggest reasons for the following features of transition metal chemistry:
(i) The transition metals and their compounds are usually paramagnetic.
(ii) The transition metals exhibit variable oxidation states. 
Answer: Disproportionation is the reaction in which an element undergoes self-oxidation and self-reduction simultaneously. For example –
\( 2\text{Cu}^+ (\text{aq}) \rightarrow \text{Cu}^{2+} (\text{aq}) + \text{Cu(s)} \)
(Or any other correct equation)
OR
(i) Due to presence of unpaired electrons in d-orbitals.
(ii) Due to incomplete filling of d-orbitals. Due to very small energy difference between \( (n - 1) d \) and \( n s \)-orbitals.

Question. In the following ions:
\( \text{Mn}^{3+}, \text{V}^{3+}, \text{Cr}^{3+}, \text{Ti}^{4+} \)
(Atomic no: Mn = 25, V = 23, Cr = 24, Ti = 22)
(a) Which ion is most stable in an aqueous solution?
(b) Which ion is the strongest oxidizing agent?
(c) Which ion is colourless?
(d) Which ion has the highest number of unpaired electrons? 
Answer: (a) \( \text{Cr}^{3+} \)
(b) \( \text{Mn}^{3+} \)
(c) \( \text{Ti}^{4+} \)
(d) \( \text{Mn}^{3+} \)

Question. Explain the following observations:
(i) Copper atom has completely filled d orbitals \( (3d^{10}) \) in its ground state, yet it is regarded as a transition element.
(ii) \( \text{Cr}^{2+} \) is a stronger reducing agent than \( \text{Fe}^{2+} \) in aqueous solution. 
Answer: (i) Because it has incompletely filled d orbitals in one of its oxidation state \( (\text{Cu}^{2+}) \).
(ii) \( \text{Cr}^{2+}(d^4) \) changes to \( \text{Cr}^{3+}(d^3) \) while \( \text{Fe}^{2+}(d^6) \) changes to \( \text{Fe}^{3+} (d^5) \). In aqueous medium \( d^3 \) is more stable than \( d^5 \).

Question. Name the following:
(i) A transition metal which does not exhibit variation in oxidation state in its compounds.
(ii) A compound where the transition metal is in the +7 oxidation state.
(iii) A member of the lanthanoid series which is well known to exhibit +4 oxidation state.
(iv) Ore used in the preparation of Potassium dichromate.
Answer: (i) Scandium (Sc).
(ii) \( \text{KMnO}_4 \) or any other suitable example.
(iii) Cerium (Ce) or any other example.
(iv) Chromite ore.

Question. Explain the following observation:
(i) \( \text{Zn}^{2+} \) salt are colourless.
(ii) Copper has exceptionally positive \( E^\circ_{M^{2+}/M} \) value. 
Answer: (i) Due to absence of unpaired electrons. Detailed Answer: Zinc has no unpaired electrons in its d orbital and has a stable fully filled d orbital. Thus, due to absence of unpaired electrons, there will not be the excitation of electrons from lower energy level to higher level to exhibit complementary colour and hence \( \text{Zn}^{2+} \) salts are colourless.
(ii) Due to high \( \Delta_a H^\circ \) and low \( \Delta_{hyd}H^\circ \). Detailed Answer: As copper has high energy of atomisation \( \Delta_a H^\circ \) and low hydration energy \( \Delta_{hyd}H^\circ \), due to which \( E^\circ \) value is positive.

Question. Give reasons:
(i) Zn is not regarded as a transition element.
(ii) \( \text{Cr}^{2+} \) is a strong reducing agent. 
Answer: (i) In both, Zn and \( \text{Zn}^{2+} \) ions absence of incompletely filled d-orbital therefore, Zn is not regarded as a transition element.
(ii) \( \text{Cr}^{2+} \) has \( d^4 \) configuration while \( \text{Cr}^{3+} \) has more stable \( d^3 (t_{2g}^3) \) configuration. Thus, Cr has a tendency to acquire \( \text{Cr}^{3+} \) due to greater stability of +3 oxidation state. Therefore, \( \text{Cr}^{2+} \) acts as a strong a reducing agent.

Question. Explain the following observation:
(i) Silver atom has completely filled d-orbitals \( (4d^{10}) \) in its ground state, yet it is regarded as a transition element.
(ii) \( E^\circ \) value for \( \text{Mn}^{3+}/\text{Mn}^{2+} \) couple is much more positive than \( \text{Cr}^{3+}/\text{Cr}^{2+} \). 
Answer: (i) Silver can exhibit +2 oxidation state, wherein it will have incompletely filled d-orbital.
(ii) Much higher third ionisation energy of Mn where the required change is from \( d^5 \) to \( d^4 \).

Question. Write one similarity and one difference between the chemistry of lanthanoids and actinoids.
Answer: Similarity: (i) Both show contraction in size. (ii) Both show irregularity in their electronic configuration. (iii) Both are stable in +3 oxidation state.
Difference: (i) Actinoids are mainly radioactive but lanthanoids are not. (ii) Actinoids show wide range of oxidation states but lanthanoids do not. (iii) Actinoid contraction is greater than lanthanoid contraction. (Write any one of these or any other one similarity and one difference)

Question. Identify the following:
(i) Oxo anion of chromium which is stable in acidic medium.
(ii) The lanthanoid element that exhibits +4 oxidation state.
Answer: (i) \( \text{Cr}_2\text{O}_7^{2-} \)
(ii) Cerium

Question. Identify the following:
(i) Transition metal of 3d series that exhibits the maximum number of oxidation states.
(ii) An alloy consisting of approximately 95% lanthanoid metal used to produce bullet, shell and lighter flint.
Answer: (i) Mn
(ii) Mischmetal

Short Answer Type Questions

Question. The magnetic moment of few transition metal ions are given below:
Metal ion | Magnetic moment (BM)
\( \text{Sc}^{3+} \) | 0.00
\( \text{Cr}^{2+} \) | 4.90
\( \text{Ni}^{2+} \) | 2.84
\( \text{Ti}^{3+} \) | 1.73
(Atomic no. Sc = 21, Ti = 22, Cr = 24, Ni = 28)
Which of the given metal ions:
(i) has the maximum number of unpaired electrons?
(ii) gives colourless aqueous solution?
(iii) exhibits the most stable +3 oxidation state? 
Answer: (i) \( \text{Cr}^{2+} \)
(ii) \( \text{Sc}^{3+} \)
(iii) \( \text{Sc}^{3+} \)

Question. Consider the standard electrode potential values \( (M^{2+}/M) \) of the elements of the first transition series.
Ti: –1.63, V: –1.18, Cr: –0.90, Mn: –1.18, Fe: –0.44, Co: –0.28, Ni: –0.25, Cu: +0.34, Zn: –0.76
Explain:
(i) \( E^\circ \) value for copper is positive.
(ii) \( E^\circ \) value of Mn is more negative as expected from the trend.
(iii) \( \text{Cr}^{3+} \) is a stronger reducing agent than \( \text{Fe}^{2+} \). 
Answer: (i) The high energy to transform Cu(s) to \( \text{Cu}^{2+}(\text{aq}) \) is not balanced by its hydration enthalpy.
(ii) \( \text{Mn}^{2+} \) has \( d^5 \) configuration (stable half-filled configuration).
(iii) \( d^5 \) to \( d^3 \) occurs in case of \( \text{Cr}^{2+} \) to \( \text{Cr}^{3+} \). (More stable \( t_{2g}^3 \)) while it changes from \( d^6 \) to \( d^5 \) in case of \( \text{Fe}^{2+} \) to \( \text{Fe}^{3+} \).

Question. (i) (a) Transition metals form alloys. Why?
(b) Chromium is typically hard metal but mercury is liquid. Why?
(ii) Based on the data, arrange \( \text{Fe}^{2+}, \text{Mn}^{2+} \) and \( \text{Cr}^{2+} \) in the increasing order of stability of +2 oxidation state.
\( E^\circ_{\text{Cr}^{3+}/\text{Cr}^{2+}} = – 0.4 \text{ V} \)
\( E^\circ_{\text{Mn}^{3+}/\text{Mn}^{2+}} = + 1.5 \text{ V} \)
\( E^\circ_{\text{Fe}^{3+}/\text{Fe}^{2+}} = + 0.8 \text{ V} \)
Answer: (i) (a) Due to the almost similar atomic sizes of elements, atoms can easily occupy the position in the crystal lattice of the other atom of another element (metal). Thus, formation of alloys is the property of transition metals.
(b) Due to the presence of large number of unpaired electrons, metal-metal interaction is strong whereas mercury does not have unpaired electrons and forms weak metallic bonds.
(ii) \( \text{Cr}^{2+} < \text{Fe}^{2+} < \text{Mn}^{2+} \)

Question. Account for the following:
(i) \( \text{CuCl}_2 \) is more stable than \( \text{Cu}_2\text{Cl}_2 \).
(ii) Atomic radii of 4d and 5d series elements are nearly same.
(iii) Hydrochloric acid is not used in permanganate titration. 
Answer: (i) In \( \text{CuCl}_2 \), Cu is in +2 oxidation state which is more stable due to high hydration enthalpy as compared to \( \text{Cu}_2\text{Cl}_2 \) in which Cu is in +1 oxidation state.
(ii) Due to lanthanoid contraction.
(iii) Because HCl is oxidised to chlorine.

Question. (i) Give reasons for the following:
(a) Compounds of transition elements are generally coloured.
(b) \( \text{MnO} \) is basic while \( \text{Mn}_2\text{O}_7 \) is acidic.
(ii) Calculate the magnetic moment of a divalent ion in aqueous medium if its atomic number is 26. 
Answer: (i) (a) Due to d-d transition.
(b) Due to higher oxidation state of \( \text{Mn}_2\text{O}_7 \) / Due to high polarizing power of Mn(VII).
(ii) \( \mu = \sqrt{4(4 + 2)} = 4.90 \text{ B.M.} \)

Question. Give reasons:
(i) Mn shows the highest oxidation state of +7 with oxygen but with fluorine it shows the highest oxidation state of +4.
(ii) Transition metals show variable oxidation states.
(iii) Actinoids show irregularities in their electronic configurations. 
Answer: (i) Mn can form \( p\pi - d\pi \) bond with oxygen by using 2p orbital of oxygen and 3d-orbital of Mn because of which it shows highest oxidation state of +7. With fluorine, Mn cannot form \( p\pi - d\pi \) bond thus shows the highest oxidation state of +4.
(ii) Transition metal show variable oxidation state due to comparable energies of \( ns \) and \( (n – 1)d \) orbitals and partially filled d orbitals. So, both these orbitals take part in the reactions.
(iii) Due to comparable energies of 5f, 6d and 7s orbitals and the relative stabilities of \( f^0, f^7 \) and \( f^{14} \) occupancies of the 5f orbitals.

Question. (i) Account for the following:
(a) \( \text{Cu}^+ \) is unstable in an aqueous solution.
(b) Transition metals form complex compounds.
(ii) What is mischmetal? give one use.
\( \text{Cr}_2\text{O}_7^{2-} + 8\text{H}^+ + 3\text{NO}_2^- \rightarrow \) 
Answer: (i) (a) Because \( \text{Cu}^+ \) undergoes disproportionation as \( 2\text{Cu}^+ \rightarrow \text{Cu} + \text{Cu}^{2+} \). Hydration enthalpy of \( \text{Cu}^{2+} \) is higher than that of \( \text{Cu}^+ \) which compensates the \( \text{I.E.}_2 \) of Cu involved in the formation of \( \text{Cu}^{2+} \) ions.
(b) Because of small size of metal, high ionic charge and availability of vacant d-orbital.
(ii) Misch metal is an alloy with composition (95%) lanthanide metal, (5%) iron and traces of S, C, Ca and Al. It's alloy of magnesium is used to produce bullets, shells, flints.

Question. Give reasons for the following:
(a) Transition metals show variable oxidation states.
(b) \( E^\circ \) value for \( (\text{Zn}^{2+}/\text{Zn}) \) is negative while that of \( (\text{Cu}^{2+}/\text{Cu}) \) is positive.
(c) Higher oxidation state of Mn with fluorine is +4 whereas with oxygen is +7. 
Answer: (a) Because of availability of partially filled orbitals and comparable energies of \( ns \) and \( (n – 1)d \) orbitals.
(b) \( E^\circ \) value for \( (\text{Zn}^{2+}/\text{Zn}) \) is negative due to stable completely filled \( d^{10} \) configuration in \( \text{Zn}^{2+} \). The positive value of \( (\text{Cu}^{2+}/\text{Cu}) \) accounts for its ability to liberate \( \text{H}_2 \) from acids due to its high enthalpy of atomization and low hydration energy.
(c) Mn can form multiple bonds with oxygen by using 2p orbital of oxygen and 3d orbital of Mn because of which it shows highest oxidation state of +7 with fluorine, Mn cannot form multiple bonds thus shows an oxidation state of +4.

Question. Give reasons for the following:
(i) Transition elements act as catalysts.
(ii) It is difficult to obtain oxidation state greater than two for copper.
(iii) \( \text{Cr}_2\text{O}_7^{2-} \) is a strong oxidising agent in acidic medium whereas \( \text{WO}_3 \) and \( \text{MoO}_3 \) are not. 
Answer: (i) Due to large surface area and ability to show variable oxidation states.
(ii) Due to high value of third ionisation enthalpy.
(iii) \( \text{Mo(VI)} \) and \( \text{W(VI)} \) are more stable than \( \text{Cr(VI)} \).

Question. Following ions are given:
\( \text{Cr}^{2+}, \text{Cu}^{2+}, \text{Cu}^+, \text{Fe}^{2+}, \text{Fe}^{3+}, \text{Mn}^{3+} \)
Identify the ion which is
(i) a strong reducing agent.
(ii) unstable in aqueous solution.
(iii) a strong oxidising agent.
Give suitable reason in each. 
Answer: (i) \( \text{Cr}^{2+} \), because its configuration changes from \( d^4 \) to \( d^3 \) and having a half-filled \( t_{2g} \) level.
(ii) \( \text{Cu}^+ \) in an aqueous medium energy is required to remove one electron from \( \text{Cu}^+ \) to \( \text{Cu}^{2+} \), high hydration energy of \( \text{Cu}^{2+} \) compensates for it. Therefore \( \text{Cu}^+ \) ion in an aqueous solution is unstable.
\( 2\text{Cu}^+ \rightarrow \text{Cu}^{2+}(\text{aq}) + \text{Cu(s)} \)
(iii) \( \text{Mn}^{3+} \), because its configuration changes from \( \text{Mn}^{3+} \) to \( \text{Mn}^{2+} \) results in the half filled \( d^5 \) configuration, which has extra stability.

Question. Account for the following:
(i) \( \text{Eu}^{2+} \) is a strong reducing agent.
(ii) Orange colour of dichromate ion changes to yellow in alkaline medium.
(iii) \( E^\circ(M^{2+}/M) \) values for transition metals show irregular variation.
Answer: (i) \( \text{Eu}^{2+} \) is a strong reducing agent because \( \text{Eu}^{3+} \) is more stable than \( \text{Eu}^{2+} \).
(ii) Dichromate ion changes to chromate ion/OH-
\( \text{Cr}_2\text{O}_7^{2-} \text{ (orange)} \rightarrow \text{CrO}_4^{2-} \text{ (yellow)} \)
(iii) Due to the irregular variation in ionisation enthalpies (sum of 1st and 2nd ionisation enthalpies), heat of sublimation and enthalpy of hydration/due to irregular electronic configurations from left to right in a period, which changes the ionisation potential.

Question. Explain the following:
(a) Out of \( \text{Sc}^{3+}, \text{Co}^{2+} \) and \( \text{Cr}^{3+} \) ions, only \( \text{Sc}^{3+} \) is colourless in aqueous solutions. (Atomic no. Co = 27; Sc = 21 and Cr = 24)
(b) The \( E^\circ_{\text{Cu}^{2+}/\text{Cu}} \) for copper metal is positive (+0.34), unlike the remaining members of the first transition series.
(c) \( \text{La(OH)}_3 \) is more basic than \( \text{Lu(OH)}_3 \). 
Answer: (a) \( \text{Co}^{2+} : [\text{Ar}]3d^7, \text{Sc}^{3+} : [\text{Ar}]3d^0, \text{Cr}^{3+} : [\text{Ar}]3d^3 \). \( \text{Co}^{2+} \) and \( \text{Cr}^{3+} \) have unpaired electrons. Thus, they are coloured in aqueous solution. \( \text{Sc}^{3+} \) has no unpaired electron. Thus it is colourless.
(b) Metal copper has high enthalpy of atomisation and enthalpy of ionisation. Therefore the high energy required to convert Cu(s) to \( \text{Cu}^{2+}(\text{aq}) \) is not balanced by its hydration enthalpy.
(c) Due to lanthanoid contraction the size of lanthanoid ion decreases regularly with increase in atomic size. Thus covalent character between lanthanoid ion and \( \text{OH}^– \) increases from \( \text{La}^{3+} \) to \( \text{Lu}^{3+} \). Thus the basic character of hydroxides decreases from \( \text{La(OH)}_3 \) to \( \text{Lu(OH)}_3 \).

Question. Give the reasons for following:
(i) Transition elements and their compounds act as catalysts.
(ii) \( E^\circ \) value for \( (\text{Mn}^{+2}|\text{Mn}) \) is negative whereas for \( (\text{Cu}^{+2}|\text{Cu}) \) is positive.
(iii) Actinoids show irregularities in their electronic configuration. 
Answer: (i) Due to variable oxidation state. Transition metals have the ability to adsorb many other substances on their surface and activate them as a result of chemisorption.
(ii) \( \text{Mn}^{2+} \) is stable due to exactly half filled \( 3d^5 \) configuration/ Due to high \( \Delta_a H^\circ \) and low \( \Delta_{hyd}H^\circ \) for \( \text{Cu}^{2+}/\text{Cu} \) is positive.
(iii) Due to comparable energies of 5f , 6d and 7s orbitals. Electrons can easily move between these subshells.

Question. Give reasons for the following:
(a) Transition metals have high enthalpies of atomization.
(b) Manganese has lower melting point even though it has a higher number of unpaired electrons for bonding.
(c) \( \text{Ce}^{4+} \) is a strong oxidising agent. 
Answer: (a) Because of strong interatomic interactions / Strong metallic bonding between atoms. Transition metals have large number of unpaired electrons in their atoms. This results in stronger interatomic interaction.
(b) Due to stable \( 3d^5 \) configuration, interatomic interaction is poor between unpaired electrons. Mn has low melting point bounded half-filled d orbital electrons with nucleus results in strong interatomic interaction.
(c) Because Ce is more stable in +3 oxidation state. The formation of \( \text{Ce}^{4+} \) is promoted by its noble gas configuration reverting to the common +3 state. \( E^\circ \) value for \( \text{Ce}^{4+}/\text{Ce}^{3+} \) is + 1.74 V thus readily gains an electron and acts as a strong oxidising agent.

Question. (a) What happens when
(i) Manganate ions \( (\text{MnO}_4^{2–}) \) undergoes disproportionation reaction in acidic medium?
(ii) Lanthanum is heated with sulphur?
(b) Explain the following trends in the properties of the members of the First series of transition elements:
(i) \( E^\circ (M^{2+}/M) \) value for copper is positive (+0.34 V) in contrast to the other members of the series.
(ii) \( \text{Cr}^{2+} \) is reducing while \( \text{Mn}^{3+} \) is oxidising, though both have \( d^4 \) configuration.
(iii) The oxidising power in the series increases in the order \( \text{VO}_2^+ < \text{Cr}_2\text{O}_7^{2-} < \text{MnO}_4^- \).
Answer: (a) (i) \( 3\text{MnO}_4^{2-} + 4\text{H}^+ \rightarrow 2\text{MnO}_4^- + \text{MnO}_2 + 2\text{H}_2\text{O} \)
(ii) \( 2\text{La} + 3\text{S} \xrightarrow{\text{heat}} \text{La}_2\text{S}_3 \)
(b) (i) Copper has high enthalpy of atomisation and low enthalpy of hydration, thus the high energy is required to transform Cu(s) to \( \text{Cu}^{2+}(\text{aq}) \) which is not balanced by hydration enthalpy.
(ii) \( \text{Cr}^{2+} \) is reducing as its configuration changes from \( d^4 \) to \( d^3 \), the latter having more stable half filled \( t_{2g} \) level. On the other hand, the change from \( \text{Mn}^{3+} \) to \( \text{Mn}^{2+} \) results an extra stable \( d^5 \) configuration.
(iii) This is due to the increasing stability of the species of lower oxidation state to which they are reduced.
 

Long Answer Type Questions

Question. (i) Complete the following equations:
(a) \( \text{Cr}_2\text{O}_7^{2–} + 2\text{OH}^– \rightarrow \)
(b) \( \text{MnO}_4^– + 4\text{H}^+ + 3e^– \rightarrow \)
(ii) Account for the following:
(a) Zn is not considered as a transition element.
(b) Transition metals form a large number of complexes.
(c) The \( E^\circ \) value for the \( \text{Mn}^{3+}/\text{Mn}^{2+} \) couple is much more positive than that for \( \text{Cr}^{3+}/\text{Cr}^{2+} \) couple.
Answer: (i) (a) \( \text{Cr}_2\text{O}_7^{2–} + 2\text{OH}^– \rightarrow 2\text{CrO}_4^{2-} + \text{H}_2\text{O} \)
(b) \( \text{MnO}_4^- + 4\text{H}^+ + 3e^– \rightarrow \text{MnO}_2 + 2\text{H}_2\text{O} \)
(ii) (a) Because Zn/\( \text{Zn}^{2+} \) has fully filled d-orbitals.
(b) This is due to smaller ionic sizes, higher ionic charge and availability of d-orbitals.
(c) Because \( \text{Mn}^{2+} \) is more stable (\( 3d^5 \)) than \( \text{Mn}^{3+} \)(\( 3d^4 \)), while \( \text{Cr}^{+3} \) is more stable due to \( t_{2g}^3/d^3 \) configuration.

Question. 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 is not regarded as a transition element. Give reason.
(ii) Which element has the highest m.p?
(iii) Write the element which can show an oxidation state of +1.
(iv) Which element is a strong oxidizing agent in +3 oxidation state and why? 
Answer: (i) Zn, because it does not have partially filled d-orbital in its ground state or ionic state.
(ii) Cr has the highest melting point. As the number of unpaired electrons increases upto \( d^5 \) configuration, it results in the increase in the strength of metallic bonds. To break the metallic bond, significant energy is required thus Cr with highest number of unpaired electrons i.e., 6 has the highest melting point.
(iii) Cu can show +1 oxidation state as it can loose one electron present in 4s orbital.
(iv) Mn is a strong oxidising agent in +3 oxidation state because change of \( \text{Mn}^{3+} \) to \( \text{Mn}^{2+} \) give stable half filled (\( d^5 \)) configuration.

Question. (i) (a) How is the variability in oxidation states of transition metals different from that of the p-block elements?
(b) Out of \( \text{Cu}^+ \) and \( \text{Cu}^{2+} \), which ion is unstable in aqueous solution and why?
(c) Orange colour of \( \text{Cr}_2\text{O}_7^{2-} \) ion changes to yellow when treated with an alkali. Why?
(ii) Chemistry of actinoids is complicated as compared to lanthanoids. Give two reasons. 
Answer: (i) (a) In p-block elements the difference in oxidation state is 2 and in transition metals the difference is 1.
(b) \( \text{Cu}^+ \), due to disproportionation reaction and low hydration enthalpy.
(c) Due to formation of chromate ion/\( \text{CrO}_4^{2-} \) ion, which is yellow in colour.
(ii) Actinoids are radioactive, actinoids show wide range of oxidation states.

Question. (i) (a) Which transition element in 3d series has positive \( E^\circ_{M^{2+}/M} \) value and why?
(b) Name a member of lanthanoid series which is well known to exhibit +4 oxidation state and why?
(ii) Account for the following
(a) The highest oxidation state is exhibited in oxoanions of transition metals.
(b) HCl is not used to acidify \( \text{KMnO}_4 \) solution.
(c) Transition metals have high enthalpy of atomisation. 
Answer: (i) (a) Copper; Due to high \( \Delta_a H \) and low \( \Delta_{hyd}H \).
(b) Cerium; Due to stable \( 4f^0 \) configuration/Tb; Due to stable \( 4f^7 \) configuration.
(ii) (a) Due to ability of oxygen to form multiple bonds to metal. Due to high electronegativity and small size, oxygen acts as a strong oxidising agent.
(b) HCl is oxidized to chlorine. As \( \text{KMnO}_4 \) is a very strong oxidising agent, it oxidizes HCl resulting in evolution of chlorine gas.
(c) Due to strong interatomic bonding.

Question. (a) Give reasons:
(i) Transition metals and their compounds show catalytic activities.
(ii) Separation of a mixture of Lanthanoid elements is difficult.
(iii) Zn, Cd and Hg are soft and have low melting point.
(b) Account for the following:
(i) \( \text{Ti}^{3+} \) is coloured whereas \( \text{Sc}^{3+} \) is colourless in aqueous solution.
(ii) \( \text{Cr}^{2+} \) is a strong reducing agent. 
Answer: (a) (i) The catalytic activities of transition metals and their compounds is due to the ability of adopt variable oxidation states and to form complexes. It can also provide a large surface area for the reactants to be adsorbed.
(ii) Separation of lanthanoid elements is difficult because all lanthanoid elements have almost similar physical as well as chemical properties. Due to the lanthanoid contraction the change in the atomic or ionic radii is very small.
(iii) Zn, Cd and Hg are soft and have low melting point because no d-orbitals are available for metallic bond formation and bonds formed are very weak.
(b) (i) \( \text{Ti}^{3+} \) has incomplete d (\( 3d^1 \)) orbital whereas \( \text{Sc}^{3+} \) has empty (\( 3d^0 \)) d-orbital.
(ii) \( \text{Cr}^{2+} \) ion can lose electron to form \( \text{Cr}^{3+} \), so acts as a strong reducing agent.