Answer

(b) Electronic configuration of $X^{3+}$ is $[Ar] 3 d^{5}$

It repersents the total number of $e^{s}$ and oxidation state.

Therefore, atomic number of $X=18+5+3=26$

Hence, option (b) is correct.

• Option (a) 25: If the atomic number of element $X$ were 25, then in its +3 oxidation state, it would have lost 3 electrons. The electronic configuration of $X$ would be $[Ar] 3d^5 4s^2$. Removing 3 electrons would result in $[Ar] 3d^4$, not $[Ar] 3d^5$.

• Option (c) 27: If the atomic number of element $X$ were 27, then in its +3 oxidation state, it would have lost 3 electrons. The electronic configuration of $X$ would be $[Ar] 3d^7 4s^2$. Removing 3 electrons would result in $[Ar] 3d^6$, not $[Ar] 3d^5$.

• Option (d) 24: If the atomic number of element $X$ were 24, then in its +3 oxidation state, it would have lost 3 electrons. The electronic configuration of $X$ would be $[Ar] 3d^5 4s^1$. Removing 3 electrons would result in $[Ar] 3d^3$, not $[Ar] 3d^5$.

Answer

(a) $Cu(II)$ is more stable than $Cu(I)$. As it is known that, $Cu(I)$ has $3 d^{10}$ stable configuration while $Cu(II)$ has $3 d^{9}$ configuration. But $Cu(II)$ is more stable due to greater effective nuclear charge of $Cu(II)$ i.e., it hold 17 electrons instead of 18 in $Cu(I)$.

• (b) $Cu(II)$ is less stable: This is incorrect because $Cu(II)$ is actually more stable than $Cu(I)$ due to the greater effective nuclear charge in $Cu(II)$, which allows it to hold its electrons more tightly despite having one less electron than $Cu(I)$.

• (c) $Cu(I)$ and $Cu(II)$ are equally stable: This is incorrect because $Cu(II)$ is more stable than $Cu(I)$. The effective nuclear charge in $Cu(II)$ makes it more stable despite the $3d^9$ configuration, compared to the $3d^{10}$ configuration of $Cu(I)$.

• (d) Stability of $Cu(I)$ and $Cu(II)$ depends on nature of copper salts: This is incorrect because the intrinsic stability of $Cu(II)$ over $Cu(I)$ is due to the effective nuclear charge and not dependent on the nature of the copper salts.

Answer

(d) On moving left to right along period, metallic radius decreases while mass increases. Decrease in metallic radius coupled with increase in atomic mass results in increase in density of metal.

Hence, among the given four choices $Cu$ belongs to right side of Periodic Table in transition metal, and it has highest density $(89 g / cm^{3})$.

• (a) Fe: Iron (Fe) has a larger metallic radius (126 pm) compared to Co and Ni, and it is positioned to the left of Cu in the periodic table. This means it has a lower atomic mass and a larger atomic radius, resulting in a lower density compared to Cu.

• (b) Ni: Nickel (Ni) has a smaller metallic radius (125 pm) compared to Cu, but it is positioned to the left of Cu in the periodic table. Although it has a smaller radius, its atomic mass is lower than that of Cu, leading to a lower density.

• (c) Co: Cobalt (Co) has a metallic radius of 125 pm, similar to Ni, and is also positioned to the left of Cu in the periodic table. Like Ni, it has a lower atomic mass than Cu, resulting in a lower density.

Answer

(b) Transition elements form coloured salt due to the presence of unpaired electrons. In $CuF_2, Cu(II)$ contain one unpaired electron hence, $CuF_2$ is coloured in solid state.

• (a) $Ag_2SO_4$: Silver (Ag) in $Ag_2SO_4$ is in the +1 oxidation state, which has a completely filled d^10 configuration. There are no unpaired electrons, so it does not exhibit color.

• (c) $ZnF_2$: Zinc (Zn) in $ZnF_2$ is in the +2 oxidation state, which has a completely filled d^10 configuration. There are no unpaired electrons, so it does not exhibit color.

• (d) $Cu_2Cl_2$: Copper (Cu) in $Cu_2Cl_2$ is in the +1 oxidation state, which has a completely filled d^10 configuration. There are no unpaired electrons, so it does not exhibit color.

Answer

(a) On addition of $KMnO_4$ to concentrated $H_2 SO_4$, a green oily compound $Mn_2 O_7$ is obtained which is highly explosive in nature.

$$ 2 KMnO_4+2 H_2 SO_4 \text { (Conc.) } \longrightarrow Mn_2 O_7+2 KHSO_4+H_2 O $$

• (b) $MnO_2$: This compound is manganese dioxide, which is a solid and not a green oily liquid. It is also not highly explosive in nature.

• (c) $MnSO_4$: This compound is manganese(II) sulfate, which is a solid and typically forms a pink or pale red solution in water. It is not a green oily liquid and is not highly explosive.

• (d) $Mn_2 O_3$: This compound is manganese(III) oxide, which is a solid and not a green oily liquid. It is also not highly explosive in nature.

Thinking Process

This problem is based on calculation of magnetic moment can be done as Magnetic moment $(\mu)=\sqrt{n(n+2)} B M$.

Answer

(b) Greater the number of unpaired electron, higher will be its value of magnetic moment. Since, $3 d^{5}$ has 5 unpaired electrons hence highest magnetic moment.

$$ \begin{aligned} \mu & =\sqrt{5(5+2)} \\ & =\sqrt{35} \\ & =5.95 BM \end{aligned} $$

• (a) $3d^7$: This configuration has 3 unpaired electrons. The magnetic moment is calculated as: $$ \mu = \sqrt{3(3+2)} = \sqrt{15} \approx 3.87 , \text{BM} $$ Since it has fewer unpaired electrons than $3d^5$, its magnetic moment is lower.

• (c) $3d^8$: This configuration has 2 unpaired electrons. The magnetic moment is calculated as: $$ \mu = \sqrt{2(2+2)} = \sqrt{8} \approx 2.83 , \text{BM} $$ Since it has fewer unpaired electrons than $3d^5$, its magnetic moment is lower.

• (d) $3d^2$: This configuration has 2 unpaired electrons. The magnetic moment is calculated as: $$ \mu = \sqrt{2(2+2)} = \sqrt{8} \approx 2.83 , \text{BM} $$ Since it has fewer unpaired electrons than $3d^5$, its magnetic moment is lower.

Answer

(b) Lanthanoids show common oxidation state of +3 . Some of which also show +2 and +4 stable oxidation state alongwith +3 oxidation state. These are shown by those elements which by losing 2 or 4 electrons acquire a stable configuration of $f^{0}, f^{7}$ or $f^{14}$, e.g., $Eu^{2+}$ is $[Xe] 4 f^{7}, Yb^{2+}$ is $[Xe] 4 f^{14}, Ce^{4+}$ is $[Xe] 4 f^{0}$ and $Tb^{4 f}$ is $[Xe] 4 f^{7}$.

• (a) +2: While some lanthanoids can exhibit a +2 oxidation state, it is not common to all lanthanoids. The +2 state is typically less stable and is observed in only a few specific lanthanoids such as Eu and Yb.

• (c) +4: The +4 oxidation state is also not common to all lanthanoids. It is observed in a limited number of lanthanoids like Ce and Tb, where the elements achieve a stable electronic configuration by losing four electrons.

• (d) +5: The +5 oxidation state is extremely rare and not observed in lanthanoids. Lanthanoids typically do not achieve this high oxidation state due to the high energy required to remove five electrons.

Answer

(a) The reaction in which oxidation as well as reduction, occur upon same atom simultaneously is known as disproportionation reaction.

• Option (b) (i), (ii) and (iii):

  • Reaction (ii) is not a disproportionation reaction because it involves the reduction of $MnO_4^{-}$ to $MnO_2$ and the oxidation of $MnO_4^{-}$ to $MnO_4^{2-}$, but these changes do not occur on the same atom simultaneously.
  • Reaction (iii) is not a disproportionation reaction because it involves the decomposition of $KMnO_4$ into $K_2MnO_4$, $MnO_2$, and $O_2$, but there is no simultaneous oxidation and reduction of the same species.

• Option (c) (ii), (iii) and (iv):

  • Reaction (ii) is not a disproportionation reaction because it involves the reduction of $MnO_4^{-}$ to $MnO_2$ and the oxidation of $MnO_4^{-}$ to $MnO_4^{2-}$, but these changes do not occur on the same atom simultaneously.
  • Reaction (iii) is not a disproportionation reaction because it involves the decomposition of $KMnO_4$ into $K_2MnO_4$, $MnO_2$, and $O_2$, but there is no simultaneous oxidation and reduction of the same species.
  • Reaction (iv) is not a disproportionation reaction because it involves the reduction of $MnO_4^{-}$ to $MnO_2$ and the oxidation of $Mn^{2+}$ to $MnO_2$, but these changes do not occur on the same atom simultaneously.

• Option (d) (i) and (iv):

  • Reaction (iv) is not a disproportionation reaction because it involves the reduction of $MnO_4^{-}$ to $MnO_2$ and the oxidation of $Mn^{2+}$ to $MnO_2$, but these changes do not occur on the same atom simultaneously.

Answer

(d) When $KMnO_4$ solution is added to oxalic acid solution, the decolourisation is slow in the beginning but becomes instantaneous after some time because $Mn^{2+}$ acts as autocatalyst.

Reduction half $.MnO_4^{-}+8 H^{+}+5 e^{-} \longrightarrow Mn^{2+}+4 H_2 O] \times 2$

Oxidation half $.\quad C_2 O_4^{2-} \longrightarrow 2 CO_2+2 e^{-}] \times 5$

Overall equation

$$ 2 MnO_4^{-}+16 H^{+}+5 C_2 O_4^{2-} \longrightarrow 2 Mn^{2+}+10 CO_2+8 H_2 O $$

End point of this reaction Colourless to light pink

• (a) $CO_2$ is formed as the product: The formation of $CO_2$ as a product does not explain the change in reaction rate. The decolourisation becoming instantaneous is due to the catalytic effect of $Mn^{2+}$, not the production of $CO_2$.

• (b) reaction is exothermic: While the reaction may release heat, the exothermic nature of the reaction does not account for the initial slowness followed by a rapid decolourisation. The key factor is the autocatalytic role of $Mn^{2+}$.

• (c) $MnO_4^{-}$ catalyses the reaction: $MnO_4^{-}$ is actually a reactant in this reaction, not a catalyst. The catalyst in this reaction is $Mn^{2+}$, which is produced during the reaction and speeds up the process.

Answer

(c) $Tm(Z=69)$ do not belong to actinoid series. The actinoid series is with atomic numbers 90 to 103. Thulium (Tm) has atomic number 69 belongs to lanthanoids (4f series).

• (a) $U$ (Uranium) is incorrect because it belongs to the actinoid series with an atomic number of 92.
• (b) $Np$ (Neptunium) is incorrect because it belongs to the actinoid series with an atomic number of 93.
• (d) $Fm$ (Fermium) is incorrect because it belongs to the actinoid series with an atomic number of 100.

Answer

The reaction of $KMnO_4$ in which it acts as an oxidising agent in acidic medium is

$2 KMnO_4+3 H_2 SO_4 \longrightarrow K_2 SO_4+2 MnSO_4+3 H_2 O+5[O]$

$$ \frac{H_2 S+[O] \longrightarrow H_2 O+S] \times 5}{2 KMnO_4+3 H_2 SO_4+5 H_2 S \longrightarrow K_2 SO_4+2 MnSO_4+8 H_2 O+5 S} $$

5 moles of $S^{2-}$ ions react with 2 moles of $KMnO_4$. So, 1 mole of $S^{2-}$ ion will react with $\frac{2}{5}$ moles of $KMnO_4$

• Option (b) $\frac{3}{5}$ is incorrect because it suggests that 1 mole of $S^{2-}$ ions would react with $\frac{3}{5}$ moles of $KMnO_4$, which is not supported by the stoichiometry of the balanced chemical equation. The correct stoichiometry indicates that 1 mole of $S^{2-}$ ions reacts with $\frac{2}{5}$ moles of $KMnO_4$.

• Option (c) $\frac{4}{5}$ is incorrect because it overestimates the amount of $KMnO_4$ needed to react with 1 mole of $S^{2-}$ ions. According to the balanced equation, only $\frac{2}{5}$ moles of $KMnO_4$ are required for 1 mole of $S^{2-}$ ions.

• Option (d) $\frac{1}{5}$ is incorrect because it underestimates the amount of $KMnO_4$ needed to react with 1 mole of $S^{2-}$ ions. The balanced chemical equation shows that $\frac{2}{5}$ moles of $KMnO_4$ are necessary for 1 mole of $S^{2-}$ ions.

Answer

(a) $V_2 O_5$ and $Cr_2 O_3$ are amphoteric oxide because both react with alkalies as well as acids.

Note In lower oxides, the basic character is predominant while in higher oxides, the acidic character is predominant.

• Option (b) $Mn_2 O_7, CrO_3$: Both $Mn_2 O_7$ and $CrO_3$ are not amphoteric oxides. They are acidic oxides because they react with bases to form salts and water but do not react with acids.

• Option (c) $CrO, V_2 O_5$: $CrO$ is a basic oxide, not amphoteric. It reacts with acids to form salts and water but does not react with bases. $V_2 O_5$ is amphoteric, but the presence of $CrO$ makes this option incorrect.

• Option (d) $V_2 O_5, V_2 O_4$: $V_2 O_4$ is a basic oxide, not amphoteric. It reacts with acids to form salts and water but does not react with bases. $V_2 O_5$ is amphoteric, but the presence of $V_2 O_4$ makes this option incorrect.

Answer

(a) Gadolinium belongs to $4 f$ series and has atomic number 64 . The correct electronic configuration of gadolinium is

$$ _{64} Gd= _{54}[Xe] 4 f^{7} 5 d^{1} 6 s^{2} $$

It has extra stability due to half-filled $4 f$ subshell.

• Option (b): $[Xe] 4 f^{6} 5 d^{2} 6 s^{2}$ is incorrect because it does not account for the extra stability provided by the half-filled $4f$ subshell. Gadolinium has 7 electrons in the $4f$ subshell, not 6.

• Option (c): $[Xe] 4 f^{8} 6 d^{2}$ is incorrect because it suggests that there are 8 electrons in the $4f$ subshell and 2 electrons in the $6d$ subshell, which does not match the actual electron count for gadolinium. Additionally, it omits the $6s$ electrons.

• Option (d): $[Xe] 4 f^{9} 5 s^{1}$ is incorrect because it suggests that there are 9 electrons in the $4f$ subshell and 1 electron in the $5s$ subshell, which is not the correct electron configuration for gadolinium. The $5s$ subshell should not be involved, and the $4f$ subshell should have 7 electrons.

Answer

(d) Interstitial compounds are formed when small atoms are trapped inside the crystal lattice of metals. Some of their important characteristics are as follows

(i) They are very hard and rigid.

(ii) They have high melting point which are higher than those of the pure metals.

(iii) They show conductivity like that of the pure metal.

(iv) They acquire chemical inertness.

• (a) They have high melting points in comparison to pure metals: This is incorrect because interstitial compounds indeed have high melting points, often higher than those of the pure metals due to the strong bonding between the metal atoms and the small atoms trapped in the lattice.

• (b) They are very hard: This is incorrect because interstitial compounds are known for their hardness and rigidity, which is a result of the small atoms fitting into the interstices of the metal lattice and strengthening the overall structure.

• (c) They retain metallic conductivity: This is incorrect because interstitial compounds retain the metallic conductivity of the pure metals, as the presence of small atoms in the lattice does not significantly disrupt the flow of electrons.

Answer

(b) The magnetic moment is associated with its spin angular momentum and orbital angular momentum.

Spin only magnetic moment value of $Cr^{3+}$ ion is $3 d^{3}$

Hence, magnetic moment

$$ \begin{aligned} (\mu) & =\sqrt{n(n+2)} BM \\ & =\sqrt{3(3+2)}=\sqrt{15} \\ & =3.87 BM \end{aligned} $$

• Option (a) $2.87 BM$: This value is incorrect because it corresponds to a different number of unpaired electrons. For a $Cr^{3+}$ ion, which has 3 unpaired electrons, the correct calculation yields $\sqrt{15} \approx 3.87 BM$, not $2.87 BM$.

• Option (c) $3.47 BM$: This value is incorrect because it does not match the result of the spin-only magnetic moment formula for 3 unpaired electrons. The correct calculation is $\sqrt{15} \approx 3.87 BM$, not $3.47 BM$.

• Option (d) $3.57 BM$: This value is incorrect because it also does not match the result of the spin-only magnetic moment formula for 3 unpaired electrons. The correct calculation is $\sqrt{15} \approx 3.87 BM$, not $3.57 BM$.

Answer

(c) $KMnO_4$ acts as an oxidising agent in alkaline medium. When alkaline $KMnO_4$ is treated with $KI$, iodide ion is oxidised to $IO_3^{-}$.

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

or, $\quad \quad v\quad \mathrm{I}^{-}+6 \mathrm{OH}^{-} \longrightarrow \mathrm{IO}_3^{-}+3 \mathrm{H}_2 \mathrm{O}+6 \mathrm{e}^{-}$

• (a) $I_2$: In an alkaline medium, $KMnO_4$ is a strong oxidizing agent and it oxidizes iodide ions ($I^-$) to higher oxidation states. Molecular iodine ($I_2$) is typically formed in acidic or neutral conditions, not in alkaline conditions.

• (b) $IO^{-}$: The oxidation state of iodine in $IO^{-}$ is +1. However, in the presence of a strong oxidizing agent like $KMnO_4$ in an alkaline medium, iodide ions are oxidized to a higher oxidation state than +1, specifically to +5 in $IO_3^{-}$.

• (d) $IO_4^{-}$: The oxidation state of iodine in $IO_4^{-}$ is +7. While $KMnO_4$ is a strong oxidizing agent, in an alkaline medium, it typically oxidizes iodide ions to the +5 oxidation state, forming $IO_3^{-}$, rather than the +7 oxidation state in $IO_4^{-}$.

Answer

(a) Copper lies below hydrogen in the electrochemical series and hence does not liberate $H_2$ from acids. Therefore, option (a) is not correct.

Other three options ( $b, c, d)$ are correct.

• (b) In its higher oxidation states, manganese forms stable compounds with oxygen and fluorine: This statement is correct. Manganese in higher oxidation states (like +4, +6, and +7) forms stable compounds such as ( \text{MnO}_2 ), ( \text{MnO}_3 ), and ( \text{MnF}_4 ).

• (c) ( \text{Mn}^{3+} ) and ( \text{Co}^{3+} ) are oxidising agents in aqueous solution: This statement is correct. Both ( \text{Mn}^{3+} ) and ( \text{Co}^{3+} ) have a tendency to gain electrons and get reduced, thus acting as oxidizing agents in aqueous solution.

• (d) ( \text{Ti}^{2+} ) and ( \text{Cr}^{2+} ) are reducing agents in aqueous solution: This statement is correct. Both ( \text{Ti}^{2+} ) and ( \text{Cr}^{2+} ) have a tendency to lose electrons and get oxidized, thus acting as reducing agents in aqueous solution.

Answer

(c) When acidified $K_2 Cr_2 O_7$ solution is added to $Sn^{2+}$ salts then $Sn^{2+}$ changes to $Sn^{4+}$. The reaction is given below

• Option (a) $Sn$: This option is incorrect because $Sn^{2+}$ is being oxidized in the reaction with acidified $K_2Cr_2O_7$. Oxidation involves an increase in the oxidation state, not a reduction to elemental tin ($Sn$), which would be a decrease in the oxidation state.

• Option (b) $Sn^{3+}$: This option is incorrect because the oxidation state of tin in this reaction changes from +2 to +4. There is no intermediate oxidation state of +3 involved in this specific redox reaction.

• Option (d) $Sn^{+}$: This option is incorrect because $Sn^{+}$ would represent a reduction in the oxidation state from +2 to +1. However, in the reaction with acidified $K_2Cr_2O_7$, $Sn^{2+}$ is oxidized, meaning its oxidation state increases, not decreases.

Answer

(d) Highest oxidation state of manganese in fluoride is $+4(MnF_4)$ but highest oxidation state in oxides is $+7(Mn_2 O_7)$. The reason is that in covalent compounds fluorine can form single bond while oxygen forms double bond.

• (a) Fluorine is more electronegative than oxygen, but this does not directly explain why manganese achieves a higher oxidation state in oxides compared to fluorides. The ability to form multiple bonds (as in the case of oxygen) is more relevant to the stabilization of higher oxidation states.

• (b) The fact that fluorine does not possess $d$ orbitals is not relevant to the oxidation state of manganese. The oxidation state is determined by the ability of the element to stabilize different charges, which is more influenced by bonding characteristics rather than the presence of $d$ orbitals in fluorine.

• (c) While fluorine can stabilize lower oxidation states due to its high electronegativity, this does not explain why manganese achieves a higher oxidation state in oxides. The key factor is the ability of oxygen to form multiple bonds, which stabilizes higher oxidation states of manganese.

Answer

(c) Due to lanthanoide contraction, $Zr$ and $Hf$ possess nearly same atomic and ionic radii i.e., $Zr=160 pm$ and $Hf=159 pm, Zr^{4+}=79 pm$ and $Hf^{4+}=78 pm$. Therefore, these two elements show similar properties (physical and chemical properties).

• (a) Both belong to $d$-block: While it is true that both zirconium and hafnium belong to the $d$-block, this is not the primary reason for their similar physical and chemical properties. Many elements in the $d$-block do not exhibit such close similarities in properties.

• (b) Both have same number of electrons: Zirconium and hafnium do not have the same number of electrons. Zirconium has 40 electrons, while hafnium has 72 electrons. The number of electrons is not the reason for their similar properties.

• (d) Both belong to the same group of the Periodic Table: Although zirconium and hafnium do belong to the same group (Group 4) of the Periodic Table, this alone does not account for their nearly identical physical and chemical properties. The primary reason is the lanthanoid contraction, which results in their nearly identical atomic and ionic radii.

Answer

(b) $HCl$ is not used to make the medium acidic in oxidation reactions of $KMnO_4$ in acidic medium. The reason is that if $HCl$ is used, the oxygen produced from $KMnO_4+HCl$ is partly utilised in oxidising $HCl$ to $Cl$, which itself acts as an oxidising agent and partly oxidises the reducing agent.

• (a) Both $HCl$ and $KMnO_4$ act as oxidising agents: This is incorrect because $HCl$ is not an oxidizing agent; it is a strong acid that can be oxidized by $KMnO_4$.

• (c) $KMnO_4$ is a weaker oxidising agent than $HCl$: This is incorrect because $KMnO_4$ is a much stronger oxidizing agent compared to $HCl$.

• (d) $KMnO_4$ acts as a reducing agent in the presence of $HCl$: This is incorrect because $KMnO_4$ is a strong oxidizing agent and does not act as a reducing agent in the presence of $HCl$.

Answer

$(a, b)$

$KMnO_4$ is coloured due to the charge transfer and not because of the presence of unpaired electrons. Similarly, oxidation state of $Ce$ in $Ce(SO_4)_2$ is +4 with $4 f^{\circ}$ electronic configuration. It is also coloured (yellow) due to charge transfer and not due to $f-f$ transition.

• $TiCl_4$ is not coloured because titanium in $TiCl_4$ is in the +4 oxidation state, which has a $3d^0$ electronic configuration, meaning there are no unpaired electrons to cause colour.
• $Cu_2Cl_2$ is not coloured because copper in $Cu_2Cl_2$ is in the +1 oxidation state, which has a $3d^{10}$ electronic configuration, meaning all electrons are paired and there are no unpaired electrons to cause colour.

Answer

$(a, d)$

Electronic configuration of $Co^{2+}=[Ar] 3 d^{7}$; Number of unpaired electrons $=3$

Electronic configuration of $Cr^{2+}=[Ar] 3 d^{4}$; Number of unpaired electrons $=4$

Electronic configuration of $Mn^{2+}=[Ar] 3 d^{5}$; Number of unpaired electrons $=5$

Electronic configuration of $Cr^{3+}=[Ar] 3 d^{3}$; Number of unpaired electrons $=3$

Hence, it is clearly seen that both $Co^{2+}$ and $Cr^{3+}$ have same number of unpaired electrons. i.e., 3 .

• $Cr^{2+}$: The electronic configuration of $Cr^{2+}$ is $[Ar] 3d^4$, which means it has 4 unpaired electrons. This is different from the 3 unpaired electrons in $Co^{2+}$ and $Cr^{3+}$.

• $Mn^{2+}$: The electronic configuration of $Mn^{2+}$ is $[Ar] 3d^5$, which means it has 5 unpaired electrons. This is different from the 3 unpaired electrons in $Co^{2+}$ and $Cr^{3+}$.

Answer

$(b, c)$

In d-block elements, for heavier elements, the higher oxidation states are more stable. Hence, $Mo(VI)$ and $W(VI)$ are more stable than $Cr(VI)$. Thats why, $Cr(VI)$ in the form of dichromate is a stronger oxidising agent in acidic medium whereas $MoO_3$ and $WO_3$ are not.

• (a) $Cr(VI)$ is more stable than $Mo(VI)$ and $W(VI)$: This is incorrect because, in reality, $Mo(VI)$ and $W(VI)$ are more stable than $Cr(VI)$. The stability of higher oxidation states increases as we move down the group in the periodic table, making $Mo(VI)$ and $W(VI)$ more stable than $Cr(VI)$.

• (d) Lower oxidation states of heavier members of group-6 of transition series are more stable: This is incorrect because, for heavier elements in the d-block, the higher oxidation states are generally more stable. Therefore, the statement that lower oxidation states are more stable for heavier members of group-6 is not accurate.

Answer

$(b, d)$

The oxidation states of the following actinoids are

(a) Americium $(Z=95)$; Electronic configuration $=[R n] 5 f^{7} 6 d^{0} 7 s^{2}$ Oxidation states shown by $A m=+3,+4,+5,+6$.

(b) Plutonium $(Z=94)$; Electronic configuration $=[Rn] 5 f^{6} 6 d^{0} 7 s^{2}$ Oxidation states shown by $Pu=+3,+4,+5,+6,+7$.

(c) Uranium ( $Z=92$ ); Electronic configuration $=[Rn] 5 f^{3} 6 d^{1} 7 s^{2}$ Oxidation states shown by $U=+3,+4,+5,+6$.

(d) Neptunium ( $Z=93$ ); Electronic configuration $=[Rn] 5 f^{4} 6 d^{1} 7 s^{2}$ Oxidation states shown by $Np=+3,+4,+5,+6,+7$.

• Americium (Am): Americium does not show an oxidation state of +7. The highest oxidation state it exhibits is +6.

• Uranium (U): Uranium does not show an oxidation state of +7. The highest oxidation state it exhibits is +6.

Answer

( $a, b$ )

General electronic configuration of actinoids is $(n-1) f^{1-14}(n-1) d^{0-2} n s^{2} . U$ and $N p$ each have one electron in $6 d$ orbital. (Also, refer to Q. 25)

• Pu (Atomic number 94): Plutonium typically has the electronic configuration [Rn] 5f^6 7s^2. It does not have an electron in the 6d orbital.

• Am (Atomic number 95): Americium typically has the electronic configuration [Rn] 5f^7 7s^2. It does not have an electron in the 6d orbital.

Answer

$(b, c)$

(a) Cerium $(Z=57) \Rightarrow$ Electronic configuration $=[Xe] 4 f^{5} 5 d^{0} 6 s^{2}$ Oxidation state of $Ce=+3,+4$

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

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

(d) Holmium $(Z=67) \Rightarrow$ Electronic configuration $=[Xe] 4 f^{11} 5 d^{0} 6 s^{2}$ Oxidation state of $Ho=+3$

• Cerium (Ce): Cerium typically exhibits oxidation states of +3 and +4. It does not commonly show a +2 oxidation state.

• Holmium (Ho): Holmium predominantly exhibits an oxidation state of +3 and does not commonly show a +2 oxidation state.

Answer

$(b, c)$

As,

$$ \begin{aligned} Ti^{3+} & =[Ar] 3 d^{1} \\ Mn^{2+} & =[Ar] 3 d^{5},(t_{2 g}^{3} e_g^{2}) \\ Fe^{2+} & =[Ar] 3 d^{6}(t_{2 g}^{4} e_{2 g}^{2}) \\ Co^{3+} & =[Ar] 3 d^{6}(t_{2 g}^{6} e_g^{0}) \end{aligned} $$

Crystal field splitting energy (CFSE) is high in $Co^{3+}$, thus electrons pair up in $t_{2 g}$. Hence, only $Fe^{2+}$ and $Mn^{2+}$ show higher spin magnetic moment value.

• $Ti^{3+}$: This ion has a single unpaired electron in the 3d orbital ($3d^1$ configuration). The spin-only magnetic moment is directly related to the number of unpaired electrons. With only one unpaired electron, $Ti^{3+}$ has a lower spin-only magnetic moment compared to ions with more unpaired electrons.

• $Co^{3+}$: This ion has a $3d^6$ configuration. Due to the high crystal field splitting energy (CFSE) in $Co^{3+}$, the electrons pair up in the $t_{2g}$ orbitals, resulting in no unpaired electrons ($t_{2g}^6 e_g^0$). Therefore, $Co^{3+}$ has a very low or zero spin-only magnetic moment.

Answer

(a, b)

Transition elements such as $Cr$ and $Co$ form binary compounds with halogens, i.e., $CrF_3$ and $CoF_3$ whereas $Cu$ and $Ni$ do not form $CuF_3$ and $NiF_3$.

• Cu: Copper typically forms compounds in the +1 and +2 oxidation states, such as $CuF$ and $CuF_2$. The +3 oxidation state is less stable for copper, making $CuF_3$ unlikely to form.

• Ni: Nickel commonly forms compounds in the +2 oxidation state, such as $NiF_2$. The +3 oxidation state is not stable for nickel, hence $NiF_3$ is not typically formed.

Answer

$(b, c)$

A species can act as oxidising agent only when metal is present in high oxidation state but lower oxidation state show stability. As higher oxidations states of $W$ and $Mo$ are more stable, therefore they will not act as oxidising agents.

• (a) $CrO_3$: Chromium trioxide ($CrO_3$) contains chromium in the +6 oxidation state, which is a high oxidation state. Chromium in this state can readily accept electrons and be reduced to a lower oxidation state, making it a strong oxidizing agent.

• (d) $CrO_4^{2-}$: Chromate ion ($CrO_4^{2-}$) also contains chromium in the +6 oxidation state. Similar to $CrO_3$, chromium in this state can act as an oxidizing agent by accepting electrons and being reduced to a lower oxidation state.

Answer

$(b, c)$

Electronic configuration of $ _{58} Ce= _{54}[Xe] 4 f^{2} 5 d^{0} 6 s^{2}$.

Therefore, electronic configuration of $Ce^{4+}= _{54}[Xe] 4 f^{0}$.

Thus, it has a tendency to attain noble gas configuration and attain $f^{0}$ configuration.

• (a) it has variable ionisation enthalpy: This option is incorrect because the primary reason for cerium showing a +4 oxidation state is not due to variable ionisation enthalpy but rather its ability to achieve a stable electronic configuration.

• (d) it resembles $Pb^{4+}$: This option is incorrect because the resemblance to $Pb^{4+}$ is not a significant factor in cerium exhibiting a +4 oxidation state. The key reasons are related to its electronic configuration and stability.

Answer

Copper not replace hydrogen from acids because $Cu$ has positive $E^{\circ}$ value, i.e., less reactive than hydrogen which has electrode potential $0.00 V$.

Answer

Negative values of $Mn^{2+}$ and $Zn^{2+}$ are related to stabilities of half-filled and completely filled configuration respectively. But for $Ni^{2+}, E^{\circ}$ value is related to the highest negative enthalpy of hydration.

Hence, $E^{s}$ values for $Mn, Ni$ and $Zn$ are more negative than expected.

Answer

Ionisation enthalpy of $Cr$ is less than that of $Zn$ because $Cr$ has stable $d^{5}$ configuration. In case of zinc, electron comes out from completely filled $4 s$-orbital. So, removal of electron from zinc requires more energy as compared to the chromium.

Answer

Transition elements show high melting point, due to involvement of greater number of electrons in the interatomic bonding from $(n-1) d$-orbitals in addition to $n s$ electrons in forming metallic bond. Thus, large number of electrons participate forming large number of metallic bond.

Answer

When $Cu^{2+}$ ion is treated with $KI$, it produces $Cu_2 I_2$ white precipitate in the final product.

$$ 2 Cu^{2+}+4 I^{-} \longrightarrow \underset{\text { (White ppt.) }}{Cu_2 I_2}+I_2 $$

(In this reaction, $CuI_2$ is formed which being unstable, dissociates into $Cu_2 I_2$ and $I_2$ ).

Answer

Among $Cu_2 Cl_2$ and $CuCl_2, CuCl_2$ is more stable. Stability of complex can be explained on the basis of stability of oxidation state of copper. Stability of $Cu^{2+}(a q)$ rather than $Cu^{+}(a q)$ is due to much more negative value of $[\Delta_{hyd} H^{s}.$ of $.Cu^{2+}(a q)]$ than $Cu^{+}$which more than compensates for the second ionisation enthalpy of $Cu$.

Thinking Process

This problem is based on the properties of $MnO_2$ and preparation of $NCl_3$.

Answer

$MnO_2$ is the brown compound of $Mn$ which reacts with $HCl$ to give $Cl_2$ gas. This gas forms an explosive compound $NCl_3$ when treated with $NH_3$. Thus, $A=MnO_2 ; B=Cl_2 ; C=NCl_3$ and reactions are as follows

(i) $\underset{[A]}{MnO_2+4 HCl} \longrightarrow MnCl_2+\underset{[B]}{Cl_2}+2 H_2 O$

(ii) $NH_3+\underset{\text { (Excess) }}{3 Cl_2} \longrightarrow \underset{\text { [C] }}{NCl_3}+3 HCl$

Answer

Oxygen can form multiple bonds with metals, while fluorine can’t form multiple bond with metals. Hence, oxygen has more ability to stabilize higher oxidation state rather than fluorine

Answer

Magnetic moment of any metal ion can be decided on the basis of spin as well as orbital contribution of electron. Due to symmetrical electronic configuration, there is no orbital contribution in $Cr^{3+}$ ion. However, appreciable orbital contribution takes place in $Co^{2+}$ ion.

Answer

$Ce, Pr$ and $Nd$ are lanthanoids and have incomplete $4 f$ shell while $Th, Pa$ and $U$ are actinoids and have $5 f$ shell incomplete.

In the beginning, when 5 -orbitals begin to be occupied, they will penetrate less into the inner core of electrons. The $5 f$-electrons will therefore, be more effectively shielded from the nuclear charge than $4 f$ electrons of the corresponding lanthanoids.

Therefore, outer electrons are less firmly held and they are available for bonding in the actinoids

Answer

Separation of $Zr$ and $Hf$ are quite difficult because of lanthanoid contraction. Due to lanthanoid contraction, they have almost same size $(Zr=160 pm$ and $Hf=159 pm)$ and thus, similar chemical properties. That’s why it is very difficult to separate them by chemical methods.

Answer

It is due to the fact that after losing one more electron Ce acquires stable $4 f^{\circ}$ electronic configuration. So, Ce shows +4 oxidation state also alongwith +3 oxidation state.

Answer

When oxalic acid is added to acidic solution of $KMnO_4$, its colour disappear due to reduction of $MnO_4^{-}$ion to $Mn^{2+}$. Chemical reaction occurring during this neutralisation reaction is as follows

$$ 5 C_2 O_4^{2-}+\underset{\text { (Coloured) }}{2 MnO_4^{-}}+16 H^{+} \longrightarrow \underset{\text { (Colourless) }}{2 Mn^{2+}}+8 H_2 O+10 CO_2 $$

Answer

When orange solution containing $Cr_2 O_7^{2-}$ ion is treated with an alkali, a yellow solution of $\mathrm{CrO}_4^{2-}$ is obtained. On the same way, $\underset{\substack{\text { Dischromate } \\ \text { (orange) }}}{\mathrm{Cr}_2 \mathrm{O}_7^{2-}} \xrightarrow[\mathrm{H}^{+}]{\mathrm{OH}^{-}} \underset{\begin{array}{l}\text { Chromate } \\ \text { (yellow) }\end{array}}{\mathrm{CrO}_4{ }^{2-}}$

when $H^{+}$ions are added to yellow solution, an orange solution is obtained due to interconversion.

Answer

Oxidising behaviour of $KMnO_4$ depends on $pH$ of the solution.

In acidic medium ( $pH<7$ )

$$ MnO_4^{-}+8 H^{+}+5 e^{-} \longrightarrow \underset{\text { Colourless }}{Mn^{2+}}+4 H_2 O $$

In alkaline medium $(pH>7)$

$$ MnO_4^{-}+e^{-} \longrightarrow \underset{(\text { Green })}{MnO_4^{2-}} $$

In neutral medium $(pH=7)$

$$ MnO_4^{-}+2 H_2 O+3 e^{-} \longrightarrow \underset{\text { (Brown ppt) }}{MnO_2}+4 OH^{-} $$

Answer

Due to lanthanoid contraction, the atomic radii of the second and third row transition elements is almost same. So, they resemble each other much more as compared to first row elements and show similar character.

Answer

$E^{S}$ value of $Cu$ is positive because of the fact that sum of sublimation enthalpy and ionisation enthalpy to convert $Cu(s)$ to $Cu^{2+}(a q)$ is so high that it is not compensate by its hydration enthalpy. $E^{\circ}$ value for $Zn$ is negative because of the fact that after removal of electrons from $4 s$ orbital, stable $3 d^{10}$ configuration is obtained.

Thinking Process

This problem is based on concept of Fajan’s rule and its application.

Answer

As the oxidation state increases, size of the ion of transition element decreases. As per Fajan’s rule, as the size of metal ion decreases, covalent character of the bond formed increases.

Therefore, the halide of transition elements become more covalent with increasing oxidation state of the metal.

Answer

During filling up of electrons follow $(n+l)$ rule. Here $4 s$ has lower energy than $3 d$ orbital. After the orbitals are filled $4 s$ goes beyond $3 d$, i.e., $4 s$ is farther from nucleus than $3 d$. So, electron from $4 s$ is removed earlier than from $3 d$.

Answer

Reactivity of transition elements depends mostly upon their ionisation enthalpies. As we move from left to right in the periodic table ( $Se$ to $Cu$ ), ionisation enthalpies increase almost regularly.

Hence, their reactivity decreases almost regularly from $Se$ to $Cu$.

Answer

A. $\rightarrow(3)$

B. $\rightarrow(4)$

C. $\rightarrow$ (2)

E. $\rightarrow(1)$

Answer

A. $\rightarrow(2)$

B. $\rightarrow(1)$

C. $\rightarrow(4)$

D. $\rightarrow(5)$

E. $\rightarrow(3)$

Answer

A. $\rightarrow(3)$

B. $\rightarrow(1)$

C. $\rightarrow(2)$

A. Osmium is an element which can show +8 oxidation state.

B. $3 d$ block element that can show upto +7 oxidation state is manganese.

C. $3 d$ block element with highest melting point is chromium.

Answer

A. $\rightarrow(3)$

B. $\rightarrow(1)$

C. $\rightarrow(5)$

D. $\rightarrow(2)$

A. Oxidation state of $Mn$ in $MnO_2$ is +4 .

B. Most stable oxidation state of $Mn$ is +2 .

C. Most stable oxidation state of $Mn$ in oxides is +7 .

D. Characteristic oxidation state of lanthanoids is +3 .

Answer

A. $\rightarrow(4)$

B. $\rightarrow(1) \quad$

C. $\rightarrow(2) \quad$

D. $\rightarrow(5)$

E. $\rightarrow(6)$

Answer

A. $\rightarrow(2)$

B. $\rightarrow(4)$

C. $\rightarrow(1)$

D. $\rightarrow(5)$

E. $\rightarrow(3)$

A. Lanthanoid which shows +4 oxidation state is cerium.

$ _{58} Ce=[Xe] 4 f^{2} 5 d^{0} 6 s^{2} ;$ Oxidation state $=+3,+4$

B. Lanthanoid which can show +2 oxidation state is europium.

$ _{63} Eu=[Xe] 4 f^{7} 5 d^{0} 6 s^{2} ;$ Oxidation state $=+2,+3$

C. Radioactive lanthanoid is promethium. It is the only synthetic (man-made) radioactive lanthanoid.

D. Lanthnoid which has $4 f^{7}$ electronic configuration in +3 oxidation state is gadolinium.

$ _{64} Gd=[Xe] 4 f^{7} 5 d^{1} 6 s^{2} ;$ Oxidation state $=+3$

E. Lanthanoid which has $4 f^{14}$ electronic configuration in +3 oxidation state is lutetium

$ _{71} Lu=[Xe] 4 f^{14} 5 d^{1} 6 s^{2} ;$ Oxidation state $=+3$

Answer

A. $\rightarrow(3)$

B. $\rightarrow(4)$

C. $\rightarrow(2)$

D. $\rightarrow(1)$

A. $Cu^{+}=3 d^{10}$ which is very stable configuration due to full-filled orbitals. Hence, removal of second electron requires very high energy.

B. $Zn^{2+}=3 d^{10}$ which is very stable configuration. Hence, removal of third electron requires very high energy.

C. Metal carbonyl with formula $M(CO)_6$ is $Cr(CO)_6$.

D. Nickel is the element with highest heat of atomisation.

Answer

(a) Assertion and reason both are correct and reason is the correct explanation of assertion.

Copper (II) iodide $(CuI_2)$ is not known because $Cu^{2+}$ oxidises $I^{-}$to lodine.

Answer

(b) Assertion and reason are true but reason is not correct explanation of assertion.

Separation of $Zr$ and $Hf$ is difficult; it is not because of they lie in the same group of Periodic Table. This is due to lanthanoid contraction which causes almost similar radii of both of them.

Answer

(c) Assertion is not true but reason is true.

Actinoids form relatively more stable complexes as compared to lanthanoids because of actinoids can utilise their $5 f$ orbitals along with $6 d$ orbitals in bonding but lanthanoids do not use their $4 f$ orbitals for bonding.

Answer

(a) Assertion and reason both are correct and reason is correct explanation of assertion. Cu can not liberate hydrogen from acids because it has positive electrode potential. Metals having negative value of electrode potential liberate $H_2$ gas.

Answer

(b) Assertion and reason both are correct but reason is not the correct explanation of assertion.

The highest oxidation state of osmium is +8 due to its ability to expand their octet by using its all 8 electrons ( 2 from 6 s and 6 from $5 d$ ).

Answer

The substances from $A$ to $E$ are

$$ A=Cu ; B=Cu(NO_3)_2 ; C=[Cu(NH_3)_4]^{2+} ; D=CO_2 ; E=CaCO_3 $$

Reactions:

(i) $\mathrm{{CuCO_3 \xrightarrow{\Delta} CuO+CO_2] \times 2}}$

(ii) $\mathrm{2 CuO+CuS \longrightarrow \underset{\text{[A]}}{3 Cu}+SO_2}$

(iii) $\mathrm{\underset{\text{[A]}}{Cu}+4 HNO_3 (conc.) \longrightarrow \underset{\text{[B]}}{Cu(NO_3)_2}+2 NO+2 H_2 O}$

(iv) $\mathrm{\underset{\text{[B]}}{Cu^{2+}}+NH_3 \longrightarrow \underset{\substack{ \text{[C]} \\ \text { (Blue Solution) }}}{[Cu(NH_3)_4]}}$

(v) $\mathrm{Ca}(\mathrm{OH})_2+ \underset{\text{[D]}}{\mathrm{CO}_2} \longrightarrow \underset{\substack{\text{[E]} \\ \text{(Milky)}}}{\mathrm{CaCO}_3}+\mathrm{H}_2 \mathrm{O}$

(vi) $\mathrm{CaCO_3+H_2 O+CO_2 \longrightarrow Ca(HCO_3)_2}$

Answer

$K_2 Cr_2 O_7$ is an orange compound. It is formed when $Na_2 Cr_2 O_7$ reacts with $KCl$. In acidic medium, yellow coloured $CrO_4^{2-}$ (chromate ion) changes into dichromate.

The given process is the preparation method of potassium dichromate from chromite ore.

$$ \mathrm{A=FeCr_2 O_4 ; B=Na_2 CrO_4 ; C=Na_2 Cr_2 O_7 ; D=K_2 Cr_2 O_7 .} $$

(i) $\mathrm{\underset{\text{[A]}}{4 FeCr_2 O_4} + 8 Na_2 CO_3+7 O_2 \longrightarrow \underset{\text{[B]}}{8 Na_2 CrO_4}+2 Fe_2 O_3+8 CO_2}$

(ii) $\mathrm{2 Na_2 CrO_4+2 H^{+} \longrightarrow Na_2 Cr_2 O_7+2 Na^{+}+H_2 O}$

(iii) $\mathrm{\underset{\text{[C]}}{Na_2 Cr_2 O_7}+2 KCl \longrightarrow \underset{\text{[D]}}{K_2 Cr_2 O_7}+2 NaCl}$

Thinking Process

This problem based on the concept of preparation and properties of potassium permanganate.

Answer

It is the method of preparation of potassium permanganate (purple). Thus, $(A)=MnO_2$

$(C)=KMnO_4$

$(B)=K_2 MnO_4$

$(D)=KIO_3$

$$ \begin{aligned} & \underset{[A]}{2 MnO_2}+4 KOH+O_2 \longrightarrow \underset{[B]}{2 K_2 MnO_4}+2 H_2 O \\ & 3 MnO_4^{2-}+4 H^{+} \longrightarrow \underset{[C]}{2 MnO_4^{-}}+MnO_2+2 H_2 O \\ & 2 MnO_4^{-}+H_2 O+KI \longrightarrow \underset{[A]}{2 MnO_2}+\underset{[D]}{2 OH^{-}}+KIO_3 \end{aligned} $$

Answer

(i) As the size decreases covalent character increases. Therefore, $La_2 O_3$ is more ionic and $Lu_2 O_3$ is more covalent.

(ii) As the size decreases from La to Lu, stability of oxo salts also decreases.

(iii) Stability of complexes increases as the size of lanthanoids decreases.

(iv) Radii of $4 d$ and $5 d$-block elements will be almost same.

(v) Acidic character of oxides increases from La to Lu.

Answer

(a) (i) $Cu$, because the electronic configuration of $Cu$ is $3 d^{10} 4 s^{1}$. So, second electron needs to be removed from completely filled $d$-orbital which is very difficult.

(ii) Zinc, because of electronic configuration of $Zn=3 d^{10} 4 s^{2}$ and $Zn^{2+}=3 d^{10}$ which is fully filled and hence is very stable. Removal of third electron requires very high energy.

(iii) Zinc, because of it has completely filled $3 d$ subshell and no unpaired electron is available for metallic bonding.

(b) (i) Carbonyl $M(CO)_5$ is $Fe(CO)_5$

According to EAN rule, the effective number of a metal in a metal carbonyl is equal to the atomic number of nearest inert gas EAN is calculated as

EAN $=$ number of electrons in metal $+2 \times(CO)$ $=$ atomic number of nearest inert gas

$\ln M(CO)_5=x+2 \times(5)=36$ ( $Kr$ is the nearest inert gas)

$x=26$ (atomic number of metal)

So, the metal is $Fe$ (iron).

(ii) $MO_3 F$ is $MnO_3 F$.

In $MO_3 F$

Let us assume that oxidation state of $M$ is $x$

$$ x+3 \times(-2)+(-1)=0 $$

or, $x=+7$ i.e., $M$ is in +7 oxidation state of +7 . Hence, the given compound is $MnO_3 F$.

Answer

When small atoms like $H, C$ and $N$ get trapped inside the crystal lattice of transition metals.

(a) Such compounds are called interstitial compounds.

(b) Their characteristic properties are;

(i) They have high melting points, higher than those of pure metals.

(ii) They are very hard.

(iii) They retain metallic conductivity.

(iv) They are chemically inert.

Answer

(a) Reaction between iodide and persulphate ions is

$ \begin{array}{cccc} & 2 \mathrm{I}^{-}+\mathrm{S}_2 \mathrm{O}_8^{2-} & \xrightarrow{\mathrm{Fe}(\mathrm{III})} & \mathrm{ I_2 + 2SO_4^{2-}} \\ \text { Role of } \mathrm{Fe}(\mathrm{III}) \text { ions } & \mathrm{Fe}^{3+}+2 \mathrm{I}^{-} & \longrightarrow & 2 \mathrm{Fe}^{2+}+\mathrm{I}_2 \\ &2 \mathrm{Fe}^{2+}+\mathrm{S}_2 \mathrm{O}_8^{2-} & \longrightarrow & 2 \mathrm{Fe}^{3+}+2 \mathrm{SO}_4^{2-} \end{array} $

(b) (i) Vanadium ( $V$ ) oxide used in contact process for oxidation of $SO_2$ to $SO_3$.

(ii) Finely divided iron in Haber’s process in conversion of $N_2$ and $H_2$ to $NH_3$.

(iii) $MnO_2$ in preparation of oxygen from $KClO_3$.

Thinking Process

This problem is based on preparation and properties of $KMnO_4 K_2 MnO_4$ and $MnO_2$.

Answer

Since, compound $(C)$ on treating with conc. $H_2 SO_4$ and $NaCl$ gives $Cl_2$ gas, so it is manganese dioxide $(MnO_2)$. It is obtained alongwith $MnO_4^{2-}$ when $KMnO_4$ (violet) is heated.

Thus,

$(A)=KMnO_4$

$(B)=K_2 MnO_4 $

$(C)=MnO_2$

$(D)=MnCl_2$

$$ \begin{matrix} \underset{[A]}{KMnO_4} \stackrel{\Delta}{\longrightarrow} \underset{[B]}{K_2 MnO_{4}}+\underset{[C]}{MnO_2}+O_2 \\ \\ 2 MnO_2+4 KOH+O_2 \longrightarrow 2 K_2 MnO_4+2 H_2 O \\ \\ MnO_2+4 NaCl+4 H_2 SO_4 \longrightarrow MnCl_2+4 NaHSO_4+2 H_2 O+Cl_2 \end{matrix} $$