NCERT Exemplar Class 12 Chemistry Solutions Chapter 8 The d and f block elements

Question 1. Electronic configuration of a transition element X in +3 oxidation state is $[\mathrm{Ar}] 3 \mathrm{~d}^5$. What is its atomic number?
(i) 25
(ii) 26
(iii) 27
(iv) 24
Answer:

The answer is option (ii).
Electronic configuration of a transition element X in +3 oxidation state is $[\mathrm{Ar}] 3 \mathrm{~d}^5$.
Atomic number = 26

Question 2. The electronic configuration of Cu(II) is $3 d^{9}$ whereas that of Cu(I) is $3 d^{10}$. Which of the following is correct?
(i) Cu(II) is more stable
(ii) Cu(II) is less stable
(iii) Cu(I) and Cu(II) are equally stable
(iv) Stability of Cu(I) and Cu(II) depends on the nature of copper salts
Answer:

The answer is option (i).

The stability of Cu(II) is more due to greater effective nuclear charge of Cu(II)

Question 3. Metallic radii of some transition elements are given below. Which of these elements will have the highest density?

(i) Fe
(ii) CO
(iii) Ni
(iv) Cu

Answer:

The answer is option (iv). In a periodic table, moving across the period from left to right, the atomic radii of elements decrease. This is the reason behind the increase in density as we move rightwards in a period.
Amongst all the options, Cu lies to the rightmost side of the Periodic Table and has the highest density.

Question 4. Generally, transition elements form coloured salts due to the presence of unpaired electrons. Which of the following compounds will be coloured in solid-state?
(i) $\mathrm{Ag}_2 \mathrm{SO}_4$
(ii) $\mathrm{CuF}_2$
(iii) $\mathrm{ZnF}_2$
(iv) $\mathrm{Cu}_2 \mathrm{Cl}_2$
Answer:

The answer is that option (ii) $\mathrm{CuF}_2$ has one unpaired electron, due to which it forms coloured salts in the solid state.

Question 5. On addition of a small amount of $\mathrm{KMnO}_4$ to concentrated $\mathrm{H}_2 \mathrm{SO}_4$, a green oily compound is obtained which is highly explosive. Identify the compound from the following.
(i) $\mathrm{Mn}_2 \mathrm{O}_7$
(ii) $\mathrm{MnO}_2$
(iii) $\mathrm{MnSO}_4$
(iv) $\mathrm{Mn}_2 \mathrm{O}_3$
Answer:

The answer is option (i).

$2 \mathrm{KMnO}_4+2 \mathrm{H}_2 \mathrm{SO}_4($ Conc $) \rightarrow \mathrm{Mn}_2 \mathrm{O}_7+2 \mathrm{KHSO}_4+\mathrm{H}_2 \mathrm{O}$
Thus, we get $\mathrm{Mn}_2 \mathrm{O}_7$ which is highly explosive in nature.

Question 6. The magnetic nature of elements depends on the presence of unpaired electrons. Identify the configuration of the transition element which shows the highest magnetic moment.
(i) 3d⁷
(ii) 3d⁵
(iii) 3d⁸
(iv) 3d²
Answer:

The answer is option (ii).The value of the magnetic moment is directly proportional to the number of unpaired electrons. Therefore, 3d⁵ has the maximum number of unpaired electrons and the highest magnetic moment.
μ=5(5+2)

=35=5.95

Question 7. Which of the following oxidation state is common for all lanthanoids?
(i) +2
(ii) +3
(iii) +4
(iv) +5
Answer:

The answer is option (ii). +3 oxidation state is common for all lanthanoids. Although sometimes +2 and +4 ions are also obtained in solution or solid compounds.

Question 8. Which of the following reactions are disproportionation reactions?
(a) $\mathrm{Cu}^{+} \rightarrow \mathrm{Cu}^{2+}+\mathrm{Cu}$
(b) $3 \mathrm{MnO}_4^{-}+4 \mathrm{H}^{+} \rightarrow 2 \mathrm{MnO}_4^{-}+\mathrm{MnO}_2+2 \mathrm{H}_2 \mathrm{O}$
(c) $2 \mathrm{KMnO}_4 \rightarrow \mathrm{~K}_2 \mathrm{MnO}_4+\mathrm{MnO}_2+\mathrm{O}_2$
(d) $2 \mathrm{MnO}_4^{-}+3 \mathrm{Mn}^{+2}+2 \mathrm{H}_2 \mathrm{O} \rightarrow 5 \mathrm{MnO}_2+4 \mathrm{H}^{+}$
(i) a, b
(ii) a, b, c
(iii) b, c, d
(iv) a, d
Answer:

The answer is option (i). A disproportionation reaction is where an element is simultaneously oxidised and reduced.

(a) Cu⁺→Cu²⁺+Cu

It is a disproportionation reaction, in the product:

• One Cu⁺ is oxidised to Cu²⁺ (+1 → +2).
• The other Cu⁺ is reduced to metallic Cu (+1 → 0).

(b) 3MnO₄^-+4H⁺→2MnO^-₄+MnO₂+2H₂O

It is also a disproportionation reaction, in the product:

• Some Mn in MnO^-₄ is reduced from +7 → +4 (forming MnO₂).
• Some Mn is reduced from +7 → +6 (forming MnO₄^2-).

Question 9. When KMnO₄ solution is added to the oxalic acid solution, the decolourisation is slow in the beginning but becomes instantaneous after some time because
(i) $\mathrm{CO}_2$ is formed as the product.
(ii) The reaction is exothermic.
(iii) $\mathrm{MnO}_4^{-}$ catalyses the reaction.
(iv) $\mathrm{Mn}^{2+}$ acts as an autocatalyst.
Answer:

The answer is option (iv). Acidified $\mathrm{KMnO}_4$ oxidises Oxalate ion to $\mathrm{CO}_2$ on addition and $\mathrm{Mn}^{2+}$ acts as an auto catalyst. This is why decolorisation is slow in the beginning but becomes instantaneous after some time.

Reduction half

$\left.\mathrm{MnO}_4^{-}+8 \mathrm{H}^{+}+5 \mathrm{e}^{-} \rightarrow \mathrm{Mn}^{2+}+\mathrm{H}_2 \mathrm{O}\right] \times 2$

Oxidation half

$\left.\mathrm{C}_2 \mathrm{O}_4^{2-} \rightarrow 2 \mathrm{CO}_2+2 \mathrm{e}^{-}\right] \times 5$

Overall equation

$2 \mathrm{MnO}_4^{-}+16 \mathrm{H}^{+}+5 \mathrm{C}_2 \mathrm{O}_4^{2-} \rightarrow 2 \mathrm{Mn}^{2+}+10 \mathrm{CO}_2+8 \mathrm{H}_2 \mathrm{O}$

End point of this reaction is colourless to light pink.

Question 10. There are 14 elements in the actinoid series. Which of the following elements does not belong to this series?
(i) U
(ii) Np
(iii) Tm
(iv) Fm
Answer:

The answer is option (iii). Tm (Thulium), which has an atomic number of 69, is a lanthanoid (4f) series.

Question 11. KMnO₄ acts as an oxidising agent in acidic medium. The number of moles of KMnO₄ that will be needed to react with one mole of sulphide ions in acidic solution is
(i) 25
(ii) 35
(iii) 45
(iv) 15
Answer:

The answer is the option (i).
$2 \mathrm{MnO}_4^{-}+5 \mathrm{~S}^{2-}+16 \mathrm{H}^{+} \rightarrow 2 \mathrm{Mn}^{2+}+5 \mathrm{~S}+8 \mathrm{H}_2 \mathrm{O}$
For 5 moles of S the number of moles of $\mathrm{KMnO}_4$=2
For 1 mole of S the number of moles of $\mathrm{KMnO}_4$=25

Question 12. Which of the following is amphoteric oxide?
Mn₂O₇, CrO₃, Cr₂O₃,CrO, V₂O₅,V₂O₄
(i)$\mathrm{V}_2 \mathrm{O}_5, \mathrm{Cr}_2 \mathrm{O}_3$
(ii) $\mathrm{Mn}_2 \mathrm{O}_7, \mathrm{CrO}_3$
(iii)$\mathrm{CrO}, \mathrm{V}_2 \mathrm{O}_5$
(iv) $\mathrm{V}_2 \mathrm{O}_5, \mathrm{~V}_2 \mathrm{O}_4$
Answer:

The answer is option (i). Since they react with both acids and bases, $\mathrm{V}_2 \mathrm{O}_5$ and $\mathrm{Cr}_2 \mathrm{O}_3$ are amphoteric oxides.
Also, the basic character is predominant in lower oxides, whereas the acidic character is predominant in higher oxides

Question 13. Gadolinium belongs to 4f series. Its atomic number is 64. Which of the following is the correct electronic configuration of gadolinium?
(i)$[\mathrm{Xe}] 4 \mathrm{f}^7 5 \mathrm{~d}^1 6 \mathrm{~s}^2$
(ii) $[\mathrm{Xe}] 4 \mathrm{f}^6 5 \mathrm{~d}^2 6 \mathrm{~s}^2$
(iii) $[\mathrm{Xe}] 4 \mathrm{f}^8 6 \mathrm{~d}^2$
(iv) $[\mathrm{Xe}] 44^9 5 \mathrm{~s}^1$
Answer:

The answer is the option (i) ${ }_{64} G d:[\mathrm{Xe}] 4 f^7 5 d^1 6 s^2$ is the correct electronic configuration of gadolinium that belongs to 4f series.

Question 14. Interstitial compounds are formed when small atoms are trapped inside the crystal lattice of metals. Which of the following is not a characteristic property of interstitial compounds?
(i) They have high melting points in comparison to pure metals.
(ii) They are very hard.
(iii) They retain metallic conductivity.
(iv) They are chemically very reactive.
Answer:

The answer is option (iv). Interstitial compounds are not chemically very active. Instead, they are inert in nature.

Question 15. The magnetic moment is associated with its spin angular momentum and orbital angular momentum. Spin only magnetic moment value of Cr³⁺ ion is___________.
(i) 2.87 B.M.
(ii) 3.87 B.M.
(iii) 3.47 B.M.
(iv) 3.57 B.M
Answer:

The answer is option (ii). The determination of the magnetic moment is done by the number of unpaired electrons. The μ is associated with its spin angular and orbital angular momentum.
Spin only magnetic moment value of Cr³⁺ ion is 3d³
Hence, magnetic moment (μ)=n(n+2)BM

=3(3+2)

=15=3.87BM

Question 16. KMnO₄ acts as an oxidising agent in alkaline medium. When alkaline KMnO₄ is treated with KI, iodide ion is oxidised to ____________.
(i) I₂
(ii)$\mathrm{IO}^{-}$
(iii) $\mathrm{IO}^{3-}$
(iv)$\mathrm{IO}_4^{-}$
Answer:

The answer is the option (iii). On treating alkaline $\mathrm{KMnO}_4$ with KI , we get

$2 \mathrm{KMnO}_4+\mathrm{KI}+\mathrm{H}_2 \mathrm{O} \rightarrow 2 \mathrm{KOH}+2 \mathrm{MnO}_2+\mathrm{KIO}_3$

Question 17. Which of the following statements is not correct?
(i) Copper liberates hydrogen from acids.
(ii) In its higher oxidation states, manganese forms stable compounds with oxygen and fluorine.
(iii) Mn³⁺ andCo³⁺ are oxidising agents in aqueous solution.
(iv) Ti²⁺ andCr²⁺ are reducing agents in aqueous solution.
Answer:

The answer is the option (i). There is no liberation of hydrogen from acids when copper is added.
$\mathrm{Cu}+2 \mathrm{H}_2 \mathrm{SO}_4 \rightarrow \mathrm{CuSO}_4+\mathrm{SO}_2+2 \mathrm{H}_2 \mathrm{O}$

$3 \mathrm{Cu}+8 \mathrm{HNO}_3 \rightarrow 3 \mathrm{Cu}\left(\mathrm{NO}_3\right)_2+2 \mathrm{NO}+4 \mathrm{H}_2 \mathrm{O}$

Question 18. When acidified K₂Cr₂O₇ solution is added to Sn²⁺ salts then Sn²⁺ changes to
(i) Sn
(ii) Sn³⁺
(iii) Sn⁴⁺
(iv) Sn
Answer:

The answer is the option (iii). On addition of acidified K₂Cr₂O₇(Acidified potassium dichromate) solution to Sn²⁺ salt, Sn²⁺ changes to Sn⁴⁺.
The reaction is given below

Question 19. Highest oxidation state of manganese in fluoride is +4(MnF₄) but highest oxidation state in oxides is +7(Mn₂O₇) because ____________.
(i) Fluorine is more electronegative than oxygen.
(ii) Fluorine does not possess d-orbitals.
(iii) Fluorine stabilises a lower oxidation state.
(iv) in covalent compounds, fluorine can form a single bond only, while oxygen forms a double bond.
Answer:

The answer is option (iv). This is because in covalent compounds, fluorine can form single bonds only whereas oxygen has the ability to form multiple bonds and therefore it forms a double bond

Question 20. Although Zirconium belongs to the 4d transition series and Hafnium to 5d transition series even then they show similar physical and chemical properties because___________.
(i) both belong to d-block.
(ii) both have the same number of electrons.
(iii) both have a similar atomic radius.
(iv) Both belong to the same group of the periodic table
Answer:

The answer is option (iii). Zirconium and hafnium show similar physical and chemical properties due to the almost identical radii of Zr(160 pm) and Hf(159 pm). A similar atomic radius results from lanthanoid contraction.

Question 21. Why is HCl not used to make the medium acidic in oxidation reactions of KMnO₄ in an acidic medium?
(i) Both HCl and KMnO₄ act as oxidising agents.
(ii) KMnO₄ oxidises HCl into Cl₂ which is also an oxidising agent.
(iii) KMnO₄ is a weaker oxidising agent than HCl.
(iv) KMnO₄ acts as a reducing agent in the presence of HCl.
Answer:

The answer is the option (ii). HCl is oxidised by KMnO₄ into Cl₂ which is also an oxidising agent. Since hydrochloric acid is oxidised to chlorine, permanganate titrations in their presence are unsatisfactory.

Question 23. Transition elements show a magnetic moment due to spin and orbital motion of electrons. Which of the following metallic ions have almost the same spin only magnetic moment?
(i) Co²⁺
(ii) Cr²⁺
(iii) Mn²⁺
(iv) Cr³⁺
Answer:

The answer is the option (i, iv) The metallic ions $\mathrm{Co}^{2+}\left(3 d^7\right)$ and $\mathrm{Cr}^{3+}\left(3 d^3\right)$ have same number of unpaired electrons which is 3. Thus, they have similar spin-only magnetic moments.

Question 24. In the form of dichromate, Cr(VI) is a strong oxidising agent in acidic medium but Mo(VI) in MoO₃ and W(VI) in WO₃ are not because ___________.
(i) Cr(VI) is more stable than Mo(VI) and W(VI).
(ii) Mo(VI) and W(VI) are more stable than Cr(VI).
(iii) Higher oxidation states of heavier members of group 6 of the transition series are more stable.
(iv) Lower oxidation states of heavier members of group 6 of the transition series are more stable.
Answer:

The answer is option (ii), (iii). In the d-block elements, heavier elements exhibit higher oxidation states in their stable form. Like in group 6, Mo(VI) and W(VI) are more stable than Cr(VI). This is why, in the form of dichromate Cr(VI) is a stronger oxidizing agent in acidic medium whereas Mo(VI) in MoO₃ and W(VI) in WO₃ are not.

Question 25. Which of the following actinoids show oxidation states up to +7?
(i) Am
(ii) Pu
(iii) U
(iv) Np
Answer:

The answer is the option (ii, iv) oxidation state of Np and Pu is +7 as well.

Question 26. General electronic configuration of actinoids is $(n-2) f^{1-14}(n-1) d^{0-2} n s^2$. Which of the following actinoids have one electron in 6d orbital?
(i) U (Atomic no. 92)
(ii) Np (Atomic no.93)
(iii) Pu (Atomic no. 94)
(iv) Am (Atomic no. 95)
Answer:

The answer is the option (i, ii) U and Np have one electron in 6d orbital.

$\begin{aligned} & 92 \mathrm{U}-5 f^3 6 d^1 7 s^2 \\ & 93 \mathrm{~Np}-5 f^4 6 d^1 7 s^2\end{aligned}$

Question 27. Which of the following lanthanoids show +2 oxidation state besides the characteristic oxidation state +3 of lanthanoids?
(i) Ce
(ii) Eu
(iii) Yb
(iv) Ho
Answer:

The answer is the option (ii, iii)
(a) Cerium (Z=57)
Electronic Configuration =$[X e] 4 f^5 5 d^0 6 s^2$
Oxidation state of Ce=+3,+4

(b) Europium (Z=63)
Electronic configuration = $[X e] 4 f^7 5 d^0 6 s^2$
Oxidation state of Eu=+2,+3

(c) Ytterbium (Z=70)
Electronic Configuration =$[X e] 4 f^{14} 5 d^0 6 s^2$
Oxidation state of Yb=+2,+3

(d) Holmium (Z=67)
Electronic Configuration =$[X e] 4 f^{11} 5 d^0 6 s^2$
The oxidation state of Ho=+3

Question 28. Which of the following ions show higher spin-only magnetic moment value?
(i) Ti³⁺
(ii) Mn²⁺
(iii) Fe²⁺
(iv) Co³⁺
Answer:

The answer is option (ii, iii).$\mathrm{Mn}^{2+}\left(3 d^5\right)$ and $\mathrm{Fe}^{2+}\left(3 d^6\right)$ will show higher values of spin only magnetic moment.

Question 29. Transition elements form binary compounds with halogens. Which of the following elements will form MF₃ type compounds?
(i) Cr
(ii) Co
(iii) Cu
(iv) Ni
Answer:

The answer is option (i, ii). Due to higher lattice energy in CoF₃ cobalt can form halides like MF₃ type of compounds. On the other hand, chromium can form CrF₆ due to higher bond enthalpy.

Question 30. Which of the following will not act as an oxidising agents?
(i) CrO₃
(ii) MoO₃
(iii) WO₃
(iv)$\mathrm{CrO}_4^{2-}$
Answer:

The answer is option (ii, iii). For a species to act as an oxidizing agent, the metal should be in a higher oxidation state, whereas stability is exhibited by its lower oxidation state. Since higher oxidation states of W and Mo are more stable; therefore, they do not act as oxidizing agents.

Question 31. Although +3 is the characteristic oxidation state for lanthanoids but cerium also shows +4 oxidation state because ___________.
(i) it has a variable ionisation enthalpy
(ii) it has a tendency to attain a noble gas configuration
(iii) it has a tendency to attain f0configuration
(iv) it resembles Pb⁴⁺
Answer:

The answer is option (ii, iii). The extra stability of empty, half-filled or completely filled f subshell gives rise to this irregularity. The noble gas configuration of Ce(IV) favours its formation.

Question 33. Why E⁰ values for Mn,Ni and Zn are more negative than expected?
Answer:

Eo values of Mn²⁺ and Zn²⁺ are more negative than expected because of the stability they have due to the half-filled d subshell $\left(3 d^5: \mathrm{Mn}^{2+}\right)$ and completely-filled $\left(3 d^{10}: Z n^{2+}\right)$ configuration respectively.

Question 34. Why first ionisation enthalpy of Cr is lower than that of Zn?
Answer:

Given below is the electronic configuration of chromium and zinc
Cr (24)=[Ar]3d⁵4s²

Zn (30)=[Ar]3d¹⁰4s²
Cr has a half-filled d orbital, which is very stable and results in lower Ionization enthalpy. Similarly, the value for Zn is higher because its electron comes out from completely filled 4s orbital.

Question 35. Transition elements show high melting points. Why?
Answer:

Transition metals have a high melting point because of a higher degree of metallic bonds formed. In addition to the ns electrons, the (n−1) d electrons also contribute to the metallic bonding.

Question 36. When Cu²⁺ ion is treated with KI, a white precipitate is formed. Explain the reaction with the help of the chemical equation.
Answer:

Iodide ions reduce the Cu²⁺ ions :

$2 \mathrm{Cu}^{2+}+4 \mathrm{I}^{-} \rightarrow \mathrm{Cu}_2 \mathrm{I}_2$ (White ppt.) $+\mathrm{I}_2$

Question 37. Out of Cu₂Cl₂ and CuCl₂, which is more stable and why?
Answer:

Due to a high hydration enthalpy, CuCl₂ exhibits higher stability Cu₂Cl₂.

Question 38. When a brown compound of manganese (A) is treated with HCl it gives a gas (B). The gas taken in excess reacts with NH₃ to give an explosive compound (C). Identify compounds A, B and C.
Answer:

A, B, C are shown below:
$A=\mathrm{MnO}_2 \quad B=\mathrm{Cl}_2 \quad \mathrm{C}=\mathrm{NCl}_3$
The reactions are explained as

$\mathrm{MnO}_2(\mathrm{~A})+4 \mathrm{HCl} \rightarrow \mathrm{MnCl}_2+\mathrm{Cl}_2(\mathrm{~B})+2 \mathrm{H}_2 \mathrm{O}$

$3 \mathrm{Cl}_2($ excess $)+\mathrm{NH}_3 \rightarrow \mathrm{NCl}_3(\mathrm{C})+3 \mathrm{HCl}$

Question 39. Although fluorine is more electronegative than oxygen, the ability of oxygen to stabilise higher oxidation states exceeds that of fluorine. Why?
Answer:

Fluorine $\left(1 s^2 2 s^2 2 p^5\right)$ can form a single bond as it has a single unpaired electron. However, Oxygen $\left(1 s^2 2 s^2 2 p^4\right)$ can form double bonds and can thus stabilize higher oxidation states.

Question 40. Although Cr³⁺ and Co²⁺ ions have the same number of unpaired electrons the magnetic moment of Cr³⁺ is 3.87 B.M. and that of Co²⁺ is 4.87 B.M. Why?
Answer:

Unlike Cr³⁺, Co²⁺ has a symmetrical electronic configuration, due to which the magnetic moment is higher for the Cobalt ion.

Question 41. Ionisation enthalpies of Ce, Pr and Nd are higher than Th, Pa and U. Why?
Answer:

6^th period elements like Ce, Pr and Nd have smaller sizes as compared to 7^th period elements like Th, Pa and U . As the size of the atom increases, the removal of electrons becomes easier (increased distance between outermost electrons and nucleus).

Question 42. Although Zr belongs to 4d and Hf belongs to 5d transition series but it is quite difficult to separate them. Why?
Answer:

Although Zr belongs to the 4f series and Hf to the 5f series, they have nearly the same size, because of which they exhibit very similar properties.

Question 43. Although +3 oxidation states are the characteristic oxidation state of lanthanoids, cerium also shows a +4 oxidation state. Why?
Answer:

Lanthanoids generally lose 3 electrons from $5 d^1 6 s^2$ to exhibit +3 oxidation state. However, Cerium has a configuration $4 f^1 5 d^1 6 s^2$, and gains additional stability by losing 4f¹ electrons as well. It is the case because on losing 4 electrons, it has only completely filled orbitals

Question 44. Explain why the colour of KMnO₄ disappears when oxalic acid is added to its solution in an acidic medium.
Answer:

In an acidic medium, KMnO₄ acts as an oxidising agent and itself converts to MnSO₄, which is colourless.

$\mathrm{KMnO}_4+3 \mathrm{H}_2 \mathrm{SO}_4 \rightarrow \mathrm{~K}_2 \mathrm{SO}_4+2 \mathrm{MnSO}_4+3 \mathrm{H}_2 \mathrm{O}+5[\mathrm{O}]$

$\begin{aligned} & {\left[\mathrm{COOH}+[\mathrm{O}] \rightarrow 2 \mathrm{CO}_2+\mathrm{H}_2 \mathrm{O}\right] \times 5} \\ & \mathrm{COOH}\end{aligned}$

$\mathrm{KMnO}_4+3 \mathrm{H}_2 \mathrm{SO}_4+5\binom{C O O H}{C O O H} \rightarrow \mathrm{~K}_2 \mathrm{SO}_4+2 \mathrm{MnSO}_4($ Colourless $)+$10CO₂+8H₂O

Question 45. When an orange solution containing $\mathrm{Cr}_2 \mathrm{O}_7^{2-}$ ion is treated with an alkali, a yellow solution is formed and when H⁺ ions are added to a yellow solution, an orange solution is obtained. Explain why does this happen.
Answer:

When $\mathrm{Cr}_2 \mathrm{O}_7^{2-}$ is treated with an alkali :
$\mathrm{Cr}_2 \mathrm{O}_7^{2-}$ (orange) $+\mathrm{OH}^{-} \rightarrow 2 \mathrm{CrO}_4^{2-}$ (yellow)
When the yellow solution is treated with an acid, we get back the orange solution

$2 \mathrm{CrO}_4^{2-}+2 \mathrm{H}^{+} \rightarrow \mathrm{Cr}_2 \mathrm{O}_7^{2-}$ (Orange) $+\mathrm{H}_2 \mathrm{O}$

Question 46. A solution of KMnO₄ on reduction yields either colourless solution or a brown precipitate or a green solution depending on the pH of the solution. What different stages of the reduction do these represent and how are they carried out?
Answer:

The oxidising nature of KMnO₄ depends on the pH of the solution. They change to colourless manganous ions in an acidic medium.
$\mathrm{MnO}_4^{-}+8 \mathrm{H}^{+}+5 \mathrm{e}^{-} \rightarrow \mathrm{Mn}^{2+}+4 \mathrm{H}_2 \mathrm{O}$
An alkaline solution will turn green because of the formation of manganate
$\mathrm{MnO}_4^{-}+\mathrm{e}^{-} \rightarrow \mathrm{MnO}_4^{2-}$
In a neutral solution, they will leave a brown precipitate, Manganese Oxide

$\mathrm{MnO}_4^{-}+2 \mathrm{H}_2 \mathrm{O}+3 \mathrm{e}^{-} \rightarrow \mathrm{MnO}_2+4 \mathrm{OH}^{-}$

Question 47. The second and third rows of transition elements resemble each other much more than they resemble the first row. Explain why?
Answer:

The f-orbital contributes extraordinarily little to shielding, because of which the effective nuclear charge increases and the size decreases. Due to this the second and third row transition elements have similar atomic radii and resemble each other much more than the third-row elements.

Question 48. $E^{\Theta}$ of Cu is + 0.34V while that of Zn is – 0.76V. Explain.
Answer:

Cu has to lose an electron from a fully filled 3d¹⁰ orbital to become Cu²⁺. This isn’t balanced by the hydration enthalpy giving Cu a positive $E^{\Theta}$ value. On the other hand, Zn forms a relatively more stable Zn²⁺ with an electronic configuration of 3d¹⁰, which is completely filled, it has a lower ionization enthalpy than Cu²⁺, but nearly the same hydration energy and a negative $E^{\Theta}$ value.

Question 49. The halides of transition elements become more covalent with increasing oxidation state of the metal. Why?
Answer:

With an increase in the oxidation state, the element’s charge increases and size decreases. Fajan’s rule states that the smaller the ion, the more the bond will exhibit a covalent nature

Question 50. While filling up of electrons in the atomic orbitals, the 4s orbital is filled before the 3d orbital but the reverse happens during the ionisation of the atom. Explain why?
Answer:

While electrons are filled according to n+l rule (which is lower for 4s(4+0=4) than 3d(3+2=5), they leave according to the ionisation enthalpy. 4s electrons, being further from the nucleus, are more loosely held and are removed.

Question 51. Reactivity of transition elements decreases almost regularly from Sc to Cu. Explain.
Answer:

Effective nuclear charge increases as we move along the period from left to right. Due to this, there is a decrease in size as well. Therefore, the electrons will be held more tightly and removing them from the outermost shell will be difficult. Thus, the ionization enthalpy also increases and reactivity decreases. Sc is more reactive than Cu.

Question 53. Match the compounds/elements given in Column I with uses given in Column II.

Answer:

(i → b), (ii → a), (iii →d), (iv → e), (v → c)

(i) Uses of Lanthanoid oxide: Colour TV screens, Fluorescent lamps, LEDs, etc.

(ii) Lanthanoids improve the properties of metals, so they are used in making alloys, the production of iron alloys and aluminium alloys.

(iii) Mischmetals are a mix of lanthanides, mainly Ce and La, which are used in high-strength steel and superalloys.

(iv) Magnesium-based alloy is a constituent of bullets.

(v) Mixed oxides of lanthanoids are indeed employed in petroleum cracking processes, especially in catalysis. Lanthanoid oxides have unique redox properties, acid-base characteristics, and thermal stability that make them effective as catalysts.

Question 54. Match the properties given in Column I with the metals given in Column II.

Answer:

(i → c), (ii → a), (iii → b)

(i) The element Osmium (Os) exhibits multiple oxidation states, ranging from −2 to +8, which depend on the compound it forms.

(ii) Manganese (Mn) is a transition metal known for having the widest range of oxidation states, ranging from −3 to +7, which depends on the compound it forms.

(iii) Chromium (Cr) has the highest melting point due to strong metallic bonding and its electron configuration (five unpaired 3d electrons).

Question 55. Match the statements given in Column I with the oxidation states given in Column II.

Answer:

(i → c), (ii → a), (iii → e), (iv → b)

(i) Oxidation satate of Mn in MnO₂

Let the oxidation state of Mn be x,

x + (-4) = 0

X = +4

(ii) Electronic configuration of Mn: [Ar] 3d⁵ 4s²

In the +2 oxidation state, Mn loses the two 4s electrons. This gives a half-filled 3d subshell (3d⁵), which is very stable.

(iii) The most unstable oxidation state of Mn in oxides is +7. The +7 oxidation state of manganese occurs in Mn₂O₇. It is highly oxidizing, volatile, and explosive even in small quantities. It also decomposes easily, releasing oxygen and forming lower oxides (like MnO₂).

(iv) Characteristic oxidation state of lanthanoids is +3. It arises from the loss of two 6s and one 4f/5d electron.

Question 57. Match the property given in Column I with the element given in Column II.

Answer:

(i) → (b) (ii) → (d) (iii) → (a) (iv) → (e) (v) → (c)

(i) Cerium has oxidation state +4

$\mathrm{Ce}=[\mathrm{Xe}] 4 \mathrm{~F}^2 5 \mathrm{~d}^0 6 \mathrm{~s}^2 ;$ oxidation state $=+3,+4$

(ii) Europium has an oxidation state of +2

$\mathrm{Eu}=[\mathrm{Xe}] 4 \mathrm{~F}^7 5 \mathrm{~d}^0 6 \mathrm{~s}^2 ;$ oxidation state $=+2,+3$

(iii) Promethium is a man-made radioactive lanthanoid.

(iv) Gadolinium has electronic configuration 4F⁷ in +3 oxidation state is.

${ }_{64} \mathrm{Gd}=[\mathrm{Xe}] 4 \mathrm{~F}^7 5 \mathrm{~d}^1 6 \mathrm{~s}^2 ;$ oxidation state $=+3$

Question 58. Match the properties given in Column I with the metals given in Column II.

Answer:

(i → c), (ii → d), (iii → b), (iv → a)
i) Cu+ has a configuration 3d10 , making removal of the next electron a challenge
ii) Similar to Cu⁺ , Zn²⁺ has a configuration 3d¹⁰
iii) M=Cr
iv) Energy of atomization is highest for Nickel

Question 60. In the following question, a statement of assertion followed by a statement of reason is given. Choose the correct answer out of the following choices.
Assertion: Separation of Zr and Hf is difficult.
Reason: Because Zr and Hf lie in the same group of the periodic table.
(i) Both assertion and reason are true, and reason is the correct explanation of the assertion.
(ii) Both assertion and reason are true, but the reason is not the correct explanation of the assertion.
(iii) The assertion is not true, but the reason is true.
(iv) Both assertion and reason are false.
Answer:

The answer is the option (ii). Due to their similar sizes, it is difficult to separate Zr and Hf

Question 61. In the following question, a statement of assertion followed by a statement of reason is given. Choose the correct answer out of the following choices.
Assertion: Actinides form relatively less stable complexes as compared to lanthanoids.
Reason: Actinides can utilise their 5f orbitals along with 6d orbitals in bonding but lanthanoids do not use their 4f orbital for bonding.
(i) Both assertion and reason are true, and reason is the correct explanation of the assertion.
(ii) Both assertion and reason are true, but the reason is not the correct explanation of the assertion.
(iii) The assertion is not true, but the reason is true.
(iv) Both assertion and reason are false.
Answer:

The answer is option (iii). Actinoids form more stable complex compounds than Lanthanoids.

Question 62. In the following question, a statement of assertion followed by a statement of reason is given. Choose the correct answer out of the following choices.
Assertion : Cu cannot liberate hydrogen from acids.
Reason: Because it has positive electrode potential.
(i) Both assertion and reason are true, and reason is the correct explanation of the assertion.
(ii) Both assertion and reason are true, but the reason is not the correct explanation of the assertion.
(iii) The assertion is not true, but the reason is true.
(iv) Both assertion and reason are false.
Answer:

(i) Cu has a higher electrode potential than hydrogen, because of which it can’t liberate Hydrogen in acidic solutions.

Question 63. In the following question, a statement of assertion is followed by a statement of reason. Choose the correct answer out of the following choices.
Assertion: The highest oxidation state of osmium is +8.
Reason: Osmium is a 5d-block element
(i) Both assertion and reason are true, and reason is the correct explanation of the assertion.
(ii) Both assertion and reason are true, but the reason is not the correct explanation of the assertion.
(iii) The assertion is not true, but the reason is true.
(iv) Both assertion and reason are false.
Answer:

The answer is the option (ii). Osmium can use all its 8 electrons to expand its octet and show the oxidation state of +8.

Question 65. When a chromite ore (A) is fused with sodium carbonate in free excess of air and the product is dissolved in water, a yellow solution of compound (B) is obtained. After treatment of this yellow solution with sulphuric acid, compound (C) can be crystallized from the solution. When compound (C) is treated with KCl, orange crystals of compound (D) crystallise out. Identify A to D and also explain the reactions.
Answer:

The compounds A, B, C and D are given as under:
$\mathrm{A}=\mathrm{FeCr}_2 \mathrm{O}_4 \quad \mathrm{~B}=\mathrm{Na}_2 \mathrm{CrO}_4 \quad \mathrm{C}=\mathrm{Na}_2 \mathrm{Cr}_2 \mathrm{O}_7 \cdot 2 \mathrm{H}_2 \mathrm{O} \quad \mathrm{D}=\mathrm{K}_2 \mathrm{Cr}_2 \mathrm{O}_7$
The following are the reactions :

$4 \mathrm{FeCr}_2 \mathrm{O}_4(\mathrm{~A})+8 \mathrm{NaCO}_3+7 \mathrm{O}_2 \rightarrow 8 \mathrm{Na}_2 \mathrm{CrO}_4(\mathrm{~B})+2 \mathrm{Fe}_2 \mathrm{O}_3+8 \mathrm{CO}_2$

$2 \mathrm{NaCrO}_4+2 \mathrm{H}^{+} \rightarrow \mathrm{Na}_2 \mathrm{Cr}_2 \mathrm{O}_7+2 \mathrm{Na}^{+}+\mathrm{H}_2 \mathrm{O}$
$\mathrm{Na}_2 \mathrm{Cr}_2 \mathrm{O}_7(\mathrm{C})+2 \mathrm{KCl} \rightarrow \mathrm{K}_2 \mathrm{Cr}_2 \mathrm{O}_7+2 \mathrm{NaCl}$

$\mathrm{Na}_2 \mathrm{Cr}_2 \mathrm{O}_7+2 \mathrm{KCl} \rightarrow \mathrm{K}_2 \mathrm{Cr}_2 \mathrm{O}_7+2 \mathrm{NaCl}$

Question 66. When an oxide of manganese (A) is fused with KOH in the presence of an oxidizing agent and dissolved in water, it gives a dark green solution of compound (B). Compound (B) disproportionates in neutral or acidic solution to give purple compound (C). An alkaline solution of compound (C) oxidises potassium iodide solution to a compound (D) and compound (A) is also formed. Identify compounds A to D and also explain the reactions involved.
Answer:

The compounds (A), (B), (C) and (D) are given as under:

$A=\mathrm{MnO}_2 \quad B=K_2 \mathrm{MnO}_4 \quad C=K \mathrm{MnO}_4 \quad D=K I O_3$

$2 \mathrm{MnO}_2(A)+4 \mathrm{KOH}+\mathrm{O}_2 \rightarrow 2 \mathrm{~K}_2 \mathrm{MnO}_4(B)+2 \mathrm{H}_2 \mathrm{O}$

$3 \mathrm{MnO}_4^{2-}+4 \mathrm{H}^{+} \rightarrow 2 \mathrm{MnO}_4^{-}+\mathrm{MnO}_2+2 \mathrm{H}_2 \mathrm{O}$

$2 \mathrm{MnO}_4^{-}+\mathrm{H}_2 \mathrm{O}+\mathrm{KI} \rightarrow 2 \mathrm{MnO}_2(i)+2 \mathrm{OH}^{-}+\mathrm{KIO}_3(i v)$.

Question 67. On the basis of lanthanoid contraction, explain the following:
(a) Nature of bonding in La₂O₃ and Lu₂O₃.
(b) Trends in the stability of oxo-salts of lanthanoids from LatoLu.
(c) Stability of the complexes of lanthanoids.
(d) Radii of 4d and 5d-block elements.
(e) Trends in acidic character of lanthanoid oxides.
Answer:

(a). The covalent character of the bond increases as the size of the atom decreases. La is larger than Lu (Lanthanide contraction), hence, La₂O₃ displays more ionic behaviour, but Lu₂O₃ displays more covalent behaviour.
(b). Decrease in size of atoms also leads to da ecrease in stability of the oxo-salts.
(c). A decrease in the size of atoms corresponds with the production of lower stable compounds.
(d). 4d and 5d block elements in the same column have nearly equal radii.
(e) There is an increase in the acidic nature of oxides as the size decreases.

Question 68. (a) Answer the following questions:
(i) Which element of the first transition series has the highest second ionization enthalpy?
(ii) Which element of the first transition series has the highest third ionization enthalpy?
(iii) Which element of the first transition series has the lowest enthalpy of atomization?
(b) Identify the metal and justify your answer:
(i) Carbonyl M(CO)₅ (ii) MO₃F
Answer:

(a)
(i)Copper has an electronic configuration $3 d^{10} 4 s^1$ & the second electron, which has to be removed from a fully filled d-orbital, leads to a very high ionization enthalpy.
(ii) Zinc has an electronic configuration of $3 d^{10} 4 s^2$ & the third electron, which has to be removed from a fully filled d-orbital, leads to a very high ionization enthalpy.
(iii) With no unpaired electrons, Zinc has the lowest energy of atomization
(b)
(i) Fe(EANrule)
(ii) Mn (oxidation state of +7)

Question 69. Mention the type of compounds formed when small atoms like H, C and N get trapped inside the crystal lattice of transition metals. Also give physical and Chemical characteristics of these compounds.
Answer:

Interstitial compounds are formed when the crystal lattice of larger transition metals traps smaller atoms like Hydrogen, Carbon and Nitrogen. These compounds have a higher melting point than pure metals, are very hard, conduct electricity and are chemically inert.

Question 70. (a) Transition metals can act as catalysts because these can change their oxidation state.
How does Fe(III) catalyse the reaction between iodide and persulphate ions?
(b) Mention any three processes where transition metals act as catalysts.
Answer:

(a) The reaction between iodide and persulfate ions is :
$2 \mathrm{I}^{-}+\mathrm{S}_2 \mathrm{O}_8^{2-} \xrightarrow{F e(I I I)} \mathrm{I}_2+2 \mathrm{SO}_4^{2-}$
Role of $\mathrm{Fe}(I I I)$ ions : $2 \mathrm{Fe}^{3+}+2 \mathrm{I}^{-} \rightarrow 2 \mathrm{Fe}^{2+}+I_2$$2 \mathrm{Fe}^{2+}+\mathrm{S}_2 \mathrm{O}_8^{2-} \rightarrow 2 \mathrm{Fe}^{3+}+2 \mathrm{SO}_4^{2-}$

(b) (i)MnO₂ acts as a catalyst : $\mathrm{KClO}_3 \rightarrow \mathrm{O}_2$
(ii) Vanadium Oxide acts as a catalyst in the contact process: $\mathrm{SO}_2 \rightarrow \mathrm{SO}_3$
(iii) Finely divided iron acts as a catalyst in Haber's process: $\mathrm{N}_2+\mathrm{H}_2 \rightarrow \mathrm{NH}_3$

Question 71. A violet compound of manganese (A) decomposes on heating to liberate oxygen and compounds (B) and (C) of manganese are formed. Compound (C) reacts with KOH in the presence of potassium nitrate to give compound (B). On heating compound (C) with cone. H₂SO₄ and NaCl, chlorine gas is liberated and a compound (D) of manganese along with other products is formed. Identify compounds A to D and also explain the reactions involved.
Answer:

The compounds A, B, C and D are given as :
$A=\mathrm{KMnO}_4 \quad B=\mathrm{K}_2 \mathrm{MnO}_4 \quad \mathrm{C}=\mathrm{MnO}_2 \quad D=\mathrm{MnCl}_2$
The reactions are as

$\mathrm{KMnO}_4(A) \xrightarrow{\Delta} \mathrm{K}_2 \mathrm{MnO}_4(B)+\mathrm{MnO}_2(C)+\mathrm{O}_2$

$\mathrm{MnO}_2+\mathrm{KOH}+\mathrm{O}_2 \rightarrow 2 \mathrm{~K}_2 \mathrm{MnO}_4+2 \mathrm{H}_2 \mathrm{O}$

$\mathrm{MnO}_2+4 \mathrm{NaCl}+4 \mathrm{H}_2 \mathrm{SO}_4 \rightarrow \mathrm{MnCl}_2(\mathrm{D})+2 \mathrm{NaHSO}_4+2 \mathrm{H}_2 \mathrm{O}+\mathrm{Cl}_2$