Chemical properties Questions in English

(D) Iron becomes passive when treated with concentrated nitric acid $(HNO_3)$.
This happens because a thin,protective,and non-porous layer of iron oxide $(Fe_2O_3)$ forms on the surface of the metal,which prevents further reaction with the acid.

(A) The oxidising nature of the trioxides of group $6$ elements depends on the stability of the lower oxidation states.
As we move down the group from $Cr$ to $W$,the stability of the $+6$ oxidation state increases,while the stability of lower oxidation states decreases.
Therefore,$CrO_3$ is a strong oxidising agent because $Cr$ prefers to be in the $+3$ oxidation state.
In contrast,$MoO_3$ and $WO_3$ are much less oxidising because $Mo$ and $W$ are more stable in the $+6$ oxidation state.
The correct order of oxidising nature is $CrO_3 > MoO_3 > WO_3$.

(D) The ability of an element to exhibit a higher oxidation state depends on its ability to form multiple bonds with the central metal atom.
Oxygen $(O)$ is capable of forming $p\pi-d\pi$ multiple bonds with transition metals,which stabilizes higher oxidation states.
In contrast,Fluorine $(F)$ can only form single covalent bonds $(F-M)$ and cannot form multiple bonds.
Therefore,oxygen can stabilize the $ 7$ oxidation state of manganese in $Mn_2O_7$,whereas fluorine can only stabilize up to the $ 4$ oxidation state in $MnF_4$.

(D)

Ion	$E^o \ (M^{3+}/M^{2+}) \ (V)$
$Ti^{2+}$	$-0.37$
$V^{2+}$	$-0.26$
$Cr^{2+}$	$-0.41$
$Mn^{2+}$	$+1.57$

The liberation of hydrogen gas from dilute acids by metal ions depends on the reduction potential of the $M^{3+}/M^{2+}$ couple.
If the $E^o$ value is negative,the $M^{2+}$ ion can be oxidized to $M^{3+}$ while reducing $H^+$ to $H_2$ gas.
Among the given ions,$Mn^{2+}$ has a highly positive $E^o$ value $(+1.57 \ V)$,which means it is very stable and does not easily lose an electron to reduce $H^+$ ions.
Therefore,$Mn^{2+}$ does not liberate hydrogen gas.

(C) $(a) \ NO^{+} = 7+8-1 = 14 \ e^-$. $O_2 = 16 \ e^-$. Thus,they are not isoelectronic. This statement is false.
$(b)$ Boron forms only covalent compounds due to its extremely high ionization energy. This statement is true.
$(c)$ In aqueous solution,$Ti^{3+}$ is more stable than $Ti^{+}$. This statement is false.
$(d) \ LiAlH_4$ is a well-known versatile reducing agent in organic synthesis. This statement is true.
Note: The question asks for the statement that is not true. Both $(a)$ and $(c)$ are technically false,but $(c)$ is a more standard textbook error regarding stability in aqueous media.

(D) . $BeO$ is amphoteric in nature. It reacts with both acids and bases. Oxides of other $s-$ block metal elements are basic in nature and react only with acids.
$B$. The oxides $NO, N_2O, CO$ are neutral,while most other non-metal oxides are acidic.
$C$. All $d-$ block elements are transition metals.
$D$. This statement is incorrect because $d-$ block metal oxides exhibit a range of properties: some are basic,some are acidic,and others are amphoteric (e.g.,$CrO_3$ is acidic,$Cr_2O_3$ is amphoteric,and $CrO$ is basic).

(A) When ferrous sulphate heptahydrate $(FeSO_4 \cdot 7H_2O)$ is heated,it first loses water of crystallization to form anhydrous ferrous sulphate $(FeSO_4)$.
Upon further heating,anhydrous ferrous sulphate decomposes to form ferric oxide $(Fe_2O_3)$,sulphur dioxide $(SO_2)$,and sulphur trioxide $(SO_3)$.
The chemical reaction is as follows:
$2FeSO_4(s) \xrightarrow{\Delta} Fe_2O_3(s) + SO_2(g) + SO_3(g)$

(C) Amphoteric oxides are those that react with both acids and bases.
$Al_2O_3$ (Aluminum oxide),$Cr_2O_3$ (Chromium$(III)$ oxide),and $ZnO$ (Zinc oxide) are well-known amphoteric oxides.
$Fe_2O_3$ (Ferric oxide) is basic in nature and does not react with bases,hence it is not amphoteric.

(C) Iron becomes passive when treated with concentrated nitric acid $(conc. HNO_3)$.
This is due to the formation of a thin,protective,and non-reactive oxide layer $(Fe_2O_3)$ on the surface of the metal,which prevents further reaction with the acid.

(A) The passivity of iron is caused by the formation of a thin,non-porous,and protective layer of oxide on its surface when it is treated with concentrated nitric acid $(HNO_3)$.
This protective layer is composed of ferric oxide $(Fe_2O_3)$.
This layer prevents further reaction of the metal with the acid,making the iron passive.

(B) When molten silver absorbs oxygen,it dissolves in the liquid metal. Upon cooling,the solubility of oxygen decreases,and the gas is released. This rapid evolution of oxygen gas causes the surface of the solidifying silver to break,scattering small particles of silver. This phenomenon is known as the $spitting$ of silver.

(C) The stability of copper$(II)$ halides is limited. While $CuF_2$,$CuCl_2$,and $CuBr_2$ are stable,$CuI_2$ and $Cu(CN)_2$ are unstable at room temperature.
$CuI_2$ spontaneously decomposes into $CuI$ and $I_2$.
$Cu(CN)_2$ spontaneously decomposes into $CuCN$ and $(CN)_2$.
Therefore,$X^-$ can be $CN^-$ or $I^-$.

(D) Metals like $Ni$,$Cr$,and $Co$ become passive when treated with concentrated $HNO_3$ due to the formation of a thin,protective,and non-porous oxide layer on their surface.

(D) $1$. Heat of atomisation: $Zn, Cd, Hg$ have low enthalpies of atomisation due to their $d^{10}$ configuration. The order is $Zn > Cd > Hg$ because metallic bonding strength decreases down the group. This is correct.
$2$. Acidic strength: Acidic character of oxides increases with the oxidation state of the metal. $Mn$ in $Mn_2O_7$ is $+7$,in $MnO_2$ is $+4$,and in $MnO$ is $+2$. Thus,$Mn_2O_7 > MnO_2 > MnO$ is correct.
$3$. Oxidising nature: Oxidising power decreases down the group in the same oxidation state for $Cr, Mo, W$. Thus,$CrO_3 > MoO_3 > WO_3$ is correct.
$4$. Ionic character: According to Fajan's rule,smaller cations have higher polarising power,leading to more covalent character. $Zn^{2+}$ is the smallest,so $ZnCl_2$ is the most covalent and $HgCl_2$ is the most ionic. The correct order of ionic character is $HgCl_2 > CdCl_2 > ZnCl_2$. Wait,actually,$HgCl_2$ is covalent due to high polarising power of $Hg^{2+}$. The order of ionic character is $ZnCl_2 > CdCl_2 > HgCl_2$. Therefore,the given order $HgCl_2 > CdCl_2 > ZnCl_2$ is incorrect.

(D) An amalgam is an alloy of mercury $(Hg)$ with other metals.
Iron $(Fe)$ does not form an amalgam with mercury because it is insoluble in mercury.
In contrast,metals like gold $(Au)$,silver $(Ag)$,zinc $(Zn)$,copper $(Cu)$,and tin $(Sn)$ readily form amalgams.

(C) Let us analyze the reactions:
$A$: $Fe_2(SO_4)_3$ on heating decomposes to $Fe_2O_3$ and $SO_3$,not $Fe$.
$B$: $FeSO_4$ on heating decomposes to $Fe_2O_3$,$SO_2$,and $SO_3$,not $Fe$.
$C$: $FeCl_3$ with $H_2$ gives $FeCl_2$ and $HCl$. $FeCl_2$ with $Zn$ gives $Fe$ and $ZnCl_2$. This is a valid sequence.
$D$: The reduction of $Fe_3O_4$ with $CO$ at $600 ^\circ C$ typically yields $FeO$,and further reduction at $700 ^\circ C$ yields $Fe$. Both $C$ and $D$ represent valid chemical transformations.

(A) $Cu^{2+}(aq.)$ acts as an oxidizing agent.
It oxidizes $I^{-}(aq.)$,$CN^{-}(aq.)$,and $SCN^{-}(aq.)$ to their respective products $I_2$,$(CN)_2$,and $(SCN)_2$ while getting reduced to $CuX$.
However,$Cl^{-}(aq.)$ is a much weaker reducing agent compared to the others and cannot be oxidized by $Cu^{2+}(aq.)$ at room temperature $(R.T.)$.
Therefore,$X$ cannot be $Cl^{-}(aq.)$.

(A) Transition metal hydroxides like $Ni(OH)_2$,$Cd(OH)_2$,and $Cu(OH)_2$ form soluble ammine complexes with excess $NH_3$ solution.
For example,$Cu(OH)_2 + 4NH_3 \rightarrow [Cu(NH_3)_4]^{2+} + 2OH^-$.
However,$Fe(OH)_2$ does not form a stable soluble ammine complex in aqueous $NH_3$ solution and remains as a precipitate.

(B) $PbO_2$ is a lead$(IV)$ oxide,which is an acidic oxide. It does not react with cold dilute $HNO_3$ because it is not basic enough to undergo an acid-base neutralization reaction with dilute nitric acid.

(B) The reaction sequence is as follows:
$1$. $P$ is a coloured solution containing $Fe^{2+}$ ions (pale green).
$2$. Upon exposure to air,$Fe^{2+}$ is oxidized to $Fe^{3+}$,forming solution $Q$ (yellowish).
$3$. When $KOH$ is added to $Q$,$Fe^{3+}$ forms a reddish-brown precipitate of $Fe(OH)_3$ $(R)$.
$4$. $Fe(OH)_3$ is insoluble in both excess $NaOH$ and excess $NH_3$ solution.
Therefore,$P$ contains $Fe^{2+}(aq.)$.

(A) Transition metals act as good catalysts because they provide a large surface area for reactants to adsorb and possess variable oxidation states,which allow them to form unstable intermediate compounds.
In the contact process for the manufacture of $H_2SO_4$,$V_2O_5$ (vanadium pentoxide) is used as a catalyst for the oxidation of $SO_2$ to $SO_3$.
Since the reason provides a specific example that supports the general statement in the assertion,the reason is the correct explanation of the assertion.

(N/A) $AgF_2$ contains $Ag$ in the $+2$ oxidation state.
Since $+2$ is an unstable oxidation state for $Ag$,it has a strong tendency to gain an electron to return to the more stable $+1$ oxidation state $(Ag^{2+} + e^- \rightarrow Ag^+)$.
Because it readily gains an electron,it acts as a powerful oxidizing agent.

(C) $Pt$ (Platinum) is a noble metal and is chemically inert under standard conditions.
It does not react with oxygen directly.
In contrast,$Zn$,$Ti$,and $Fe$ are reactive metals that form their respective oxides upon heating with oxygen.

(N/A) The oxidising power of these oxoanions depends on the stability of the lower oxidation states to which they are reduced.
As we move from $VO_2^+$ to $MnO_4^-$,the stability of the reduced species increases,which makes the reduction of the parent oxoanion more favorable.
Specifically,the reduction potentials increase in the order $VO_2^+ < Cr_2O_7^{2-} < MnO_4^-$,reflecting the increasing ease with which these species accept electrons.

(N/A) The following reactions occur when $Cr^{2+}$ and $Fe^{2+}$ act as reducing agents:
$Cr^{2+} \longrightarrow Cr^{3+} + e^-$
$Fe^{2+} \longrightarrow Fe^{3+} + e^-$
The standard electrode potential value $E^\circ_{Cr^{3+}/Cr^{2+}}$ is $-0.41 \ V$,while $E^\circ_{Fe^{3+}/Fe^{2+}}$ is $+0.77 \ V$.
$A$ more negative $E^\circ$ value indicates a greater tendency to undergo oxidation.
Since the $E^\circ$ value for $Cr^{3+}/Cr^{2+}$ is negative,$Cr^{2+}$ is easily oxidized to $Cr^{3+}$.
Conversely,the positive $E^\circ$ value for $Fe^{3+}/Fe^{2+}$ indicates that $Fe^{2+}$ is not easily oxidized to $Fe^{3+}$.
Therefore,$Cr^{2+}$ is a stronger reducing agent than $Fe^{2+}$.

(N/A) $(i)$ $Cr^{2+}$ is strongly reducing in nature. It has a $d^{4}$ configuration. While acting as a reducing agent,it gets oxidized to $Cr^{3+}$ (electronic configuration,$d^{3}$). This $d^{3}$ configuration can be written as $t_{2g}^{3}$ configuration,which is a more stable configuration. In the case of $Mn^{3+} (d^{4})$,it acts as an oxidizing agent and gets reduced to $Mn^{2+} (d^{5})$. This has an exactly half-filled $d$-orbital and is highly stable.
$(ii)$ $Co^{2+}$ is stable in aqueous solutions. However,in the presence of strong field complexing reagents,it is oxidized to $Co^{3+}$. Although the $3^{rd}$ ionization energy for $Co$ is high,the higher amount of crystal field stabilization energy $(CFSE)$ released in the presence of strong field ligands overcomes this ionization energy.
$(iii)$ The ions in $d^{1}$ configuration tend to lose one more electron to get into stable $d^{0}$ configuration. Also,the hydration or lattice energy is more than sufficient to remove the only electron present in the $d$-orbital of these ions. Therefore,they act as reducing agents.

(N/A) $(i)$ In the case of a lower oxide of a transition metal,the metal atom has a low oxidation state. This means that some of the valence electrons of the metal atom are not involved in bonding. As a result,it can donate electrons and behave as a base.
On the other hand,in the case of a higher oxide of a transition metal,the metal atom has a high oxidation state. This means that the valence electrons are involved in bonding and so,they are unavailable. There is also a high effective nuclear charge. As a result,it can accept electrons and behave as an acid.
For example,$MnO$ is basic and $Mn_2O_7$ is acidic.
$(ii)$ Oxygen and fluorine act as strong oxidising agents because of their high electronegativities and small sizes. Hence,they bring out the highest oxidation states from the transition metals. In other words,a transition metal exhibits higher oxidation states in oxides and fluorides. For example,in $OsF_6$ and $V_2O_5$,the oxidation states of $Os$ and $V$ are $+6$ and $+5$ respectively.
$(iii)$ Oxygen is a strong oxidising agent due to its high electronegativity and small size.
So,oxo-anions of a metal have the highest oxidation state. For example,in $MnO_4^-$,the oxidation state of $Mn$ is $+7$.

(N/A) The stability of $M^{2+}$ ions in an aqueous medium depends on three factors:
$(i)$ Enthalpy of atomisation
$(ii)$ Summation of first and second ionization enthalpies
$(iii)$ Hydration enthalpy
An element in the $M^{2+}$ state in an aqueous medium is more stable if the electrode potential $(M^{2+}/M)$ value is more negative. Across the period,the tendency to form $M^{2+}$ ions generally decreases.
Except for copper,all elements of the first transition series show negative values of electrode potentials. The exceptional behavior of copper is due to its high enthalpy of atomisation and very high summation of first and second ionization enthalpies,which is not compensated by its hydration enthalpy $(Cu^{2+})$.
Because of its positive electrode potential,copper does not liberate hydrogen gas from dilute acids and reacts only with oxidizing acids such as nitric acid and hot concentrated sulphuric acid.
The electrode potentials of $Mn$,$Ni$,and $Zn$ are more negative than expected. The electrode potential values of $Mn$ and $Zn$ are lowered because of their stable electronic configurations (half-filled $d^5$ for $Mn^{2+}$ and fully-filled $d^{10}$ for $Zn^{2+}$),while $Ni$ has an exceptionally more negative electrode potential due to its high hydration enthalpy.

(N/A) The $E_{M^{3+}/M^{2+}}^{o}$ values are determined by the stability of the $M^{3+}$ and $M^{2+}$ oxidation states.
$1$. $Sc^{3+}$ has a noble gas configuration,making it very stable,which results in a very low (highly negative) $E_{M^{3+}/M^{2+}}^{o}$ value.
$2$. The high value for $Mn^{3+}/Mn^{2+}$ $(+1.57 \ V)$ is due to the extra stability of the half-filled $d^5$ configuration in $Mn^{2+}$.
$3$. The low value for $Fe^{3+}/Fe^{2+}$ $(+0.77 \ V)$ is due to the extra stability of the $Fe^{3+}$ $(d^5)$ ion.
$4$. The high value for $Co^{3+}/Co^{2+}$ $(+1.97 \ V)$ is due to the high Crystal Field Stabilization Energy $(CFSE)$ of the $Co^{2+}$ $(d^7)$ system in an aqueous medium.
$5$. Elements with high positive $E_{M^{3+}/M^{2+}}^{o}$ values (like $Mn^{3+}$ and $Co^{3+}$) act as strong oxidizing agents. Conversely,elements with low or negative values (like $Ti^{2+}, V^{2+}, Cr^{2+}$) act as strong reducing agents and can liberate $H_2$ from acids.
Example: $2Cr^{2+}_{(aq)} + 2H^+_{(aq)} \rightarrow 2Cr^{3+}_{(aq)} + H_{2(g)} \uparrow$

Element	Electrode potential ($E_{M^{3+}/M^{2+}}^{o}$ in $V$)
$Ti$	$-0.37$
$V$	$-0.26$
$Cr$	$-0.41$
$Mn$	$+1.57$
$Fe$	$+0.77$
$Co$	$+1.97$

(N/A) Transition metals exhibit a wide range of chemical reactivity. Many of them are sufficiently electropositive to dissolve in mineral acids,although some are noble (e.g.,$Ag$,$Au$,$Pt$).
For the $3d$ series,the reactivity is related to the standard electrode potential $(E^{\circ})$ values. Metals with more negative $E^{\circ}$ values are more reactive.
For example,$Cr^{2+}$ is a strong reducing agent and can liberate hydrogen from acids:
$2Cr^{2+}_{(aq)} + 2H^+_{(aq)} \rightarrow 2Cr^{3+}_{(aq)} + H_{2(g)}$
In this reaction,$Cr^{2+}$ is oxidized to $Cr^{3+}$ (oxidation state changes from $+2$ to $+3$),while $H^+$ is reduced to $H_2$ (oxidation state changes from $+1$ to $0$).

(N/A) The transition elements and their compounds (mainly oxides) are known for their catalytic properties due to their ability to adopt multiple oxidation states and provide a large surface area.
In the first transition series,elements such as $Fe, Ni, Mn, Co$ utilize their $3d$ and $4s$ electrons to form intermediate bonds with reactant molecules on their surface. This increases the concentration of reactants on the surface and weakens the bonds in the reactant molecules,thereby decreasing the activation energy of the reaction.
Because transition elements can change their oxidation states,they act as effective catalysts by providing alternative reaction pathways.
Example: $Fe(III)$ catalyzes the reaction between iodide and persulphate ions.
$2 I^{-} + S_{2}O_{8}^{2-} \rightarrow I_{2} + 2 SO_{4}^{2-}$
The catalytic action of $Fe^{3+}$ can be shown as:
$2 Fe^{3+} + 2 I^{-} \rightarrow I_{2} + 2 Fe^{2+}$
$2 Fe^{2+} + S_{2}O_{8}^{2-} \rightarrow 2 SO_{4}^{2-} + 2 Fe^{3+}$
Other examples include $V_{2}O_{5}$ in the contact process,$Fe$ in Haber's process,$Ni$ in the hydrogenation of fats,and $TiCl_{4}$ in polymerization.
Thus,the catalytic properties of transition elements are due to the presence of vacant $d$-orbitals,the tendency to exist in variable oxidation states,and the ability to form complexes.

(N/A) Transition metals and their compounds are known for their catalytic activity. This is primarily due to their ability to adopt multiple oxidation states and to form complexes.
Examples:
$(i)$ Vanadium$(V)$ oxide $(V_2O_5)$ in the Contact process.
$(ii)$ Finely divided iron $(Fe)$ in the Haber process.
$(iii)$ Finely divided nickel $(Ni)$ in catalytic hydrogenation.
Explanation of catalytic action:
$(i)$ Adsorption: The reactant molecules get adsorbed on the surface of the catalyst. The transition metal atoms use their $3d$ and $4s$ electrons to form bonds with reactant molecules. This increases the concentration of reactants on the surface and weakens the bonds in the reactant molecules,thereby lowering the activation energy.
$(ii)$ Variable oxidation states: Transition metals can change their oxidation states,which helps in providing an alternative pathway with lower activation energy.
Example: $Fe^{3+}$ ions catalyze the reaction between iodide $(I^-)$ and persulphate $(S_2O_8^{2-})$ ions:
$(i)$ $2I^- + S_2O_8^{2-} \xrightarrow{Fe^{3+}} I_2 + 2SO_4^{2-}$
$(ii)$ $2Fe^{2+} + S_2O_8^{2-} \rightarrow 2Fe^{3+} + 2SO_4^{2-}$

(N/A) The only stable manganese halide in $+7$ oxidation state is $MnF_7$ (though it is highly unstable and often discussed as $MnO_3F$ in related contexts,$MnF_7$ is the theoretical answer for halides).
$(b)$ The $E^{\ominus}_{M^{II}/M}$ value is maximum negative for $Mn$ $(-1.18 \ V)$ in the $3d$ series because of the extra stability of the $d^5$ configuration in $Mn^{2+}$.
$(c)$ $Cr^{2+}$ is a stronger reducing agent than $Fe^{2+}$ because $Cr^{2+}$ changes from $d^4$ to $d^3$ ($t_{2g}^3$ configuration),which is more stable due to half-filled $t_{2g}$ orbitals,while $Fe^{2+}$ changes from $d^6$ to $d^5$.

(N/A) In $AgF_2$,the oxidation number of $Ag$ is $+2$,which is highly unstable.
Therefore,it rapidly gains an electron to return to the more stable $+1$ oxidation state.
$Ag^{2+} + e^- \rightarrow Ag^+$.
Because it readily accepts electrons,$AgF_2$ acts as a strong oxidizing agent.

(N/A) The earlier members of the series show chemical behavior similar to that of calcium. However,with an increase in atomic number,they behave more like aluminium.
The $E^{\circ}$ values for the half-reaction $Ln_{(aq)}^{3+} + 3e^{-} \rightarrow Ln_{(s)}$ range from $-2.2 \ V$ to $-2.4 \ V$,except for $Eu$,which has a value of $-2.0 \ V$.
Lanthanoids react with hydrogen when heated gently in the gas. When heated with carbon,they form carbides of the types $Ln_{3}C$,$Ln_{2}C_{3}$,and $LnC_{2}$.
These metals form trihalides $(LnX_{3})$ with halogens and liberate hydrogen gas when treated with dilute acids. They combine with oxygen to form oxides of the type $Ln_{2}O_{3}$. These oxides are basic in nature and react with water to form hydroxides,$Ln(OH)_{3}$. The basicity of these oxides and hydroxides is intermediate between that of aluminium and calcium compounds:
$Al_{2}O_{3} < Ln_{2}O_{3} < CaO$
$Al(OH)_{3} < Ln(OH)_{3} < Ca(OH)_{2}$
As the atomic number increases,the basicity of the hydroxides decreases due to lanthanoid contraction.
Summary of chemical reactions:
- With $O_{2}$: Forms $Ln_{2}O_{3}$
- With acids: Liberates $H_{2}$
- With halogens: Forms $LnX_{3}$
- With $H_{2}O$: Forms $Ln(OH)_{3} + H_{2}$
- With $C$ (at $2773 \ K$): Forms $LnC_{2}$
- With $N$: Forms $LnN$
- With $S$: Forms $Ln_{2}S_{3}$

(N/A) Lanthanoids react with $H_2$ when heated to form hydrides of the formula $LnH_2$ and $LnH_3$ (where $Ln$ represents a lanthanoid element).
Lanthanoids react with $O_2$ when heated to form oxides of the formula $Ln_2O_3$ (where $Ln$ represents a lanthanoid element).

(N/A) Lanthanoids are highly reactive metals.
$(1)$ They react with dilute acids to liberate hydrogen gas: $2Ln + 6HCl \rightarrow 2LnCl_3 + 3H_2$.
$(2)$ They react with halogens $(X_2)$ upon heating to form trihalides: $2Ln + 3X_2 \rightarrow 2LnX_3$.

(N/A) When $Cu^{2+}$ ions are treated with $KI$,the $Cu^{2+}$ ions are reduced to $Cu^{+}$ ions by iodide ions $(I^-)$.
The resulting $Cu^{+}$ ions react with the remaining iodide ions to form a white precipitate of copper$(I)$ iodide $(Cu_2I_2)$.
The balanced chemical equation for this reaction is:
$2Cu^{2+}_{(aq)} + 4I^{-}_{(aq)} \rightarrow Cu_2I_{2(s)} + I_{2(s)}$

(N/A) The reaction between iodide and persulphate ions is given by: $2 I^{-} + S_{2}O_{8}^{2-} \xrightarrow{Fe^{3+}} I_{2} + 2 SO_{4}^{2-}$
Mechanism:
$(i)$ $2 Fe^{3+} + 2 I^{-} \rightarrow 2 Fe^{2+} + I_{2}$
$(ii)$ $2 Fe^{2+} + S_{2}O_{8}^{2-} \rightarrow 2 Fe^{3+} + 2 SO_{4}^{2-}$
$(b)$ Three processes where transition metals act as catalysts are:
$(i)$ Contact process: $V_{2}O_{5}$ is used as a catalyst.
$(ii)$ Haber's process: Finely divided $Fe$ is used as a catalyst.
$(iii)$ Decomposition of $KClO_{3}$: $MnO_{2}$ is used as a catalyst.

(D) $W(VI)$ (tungsten in $+6$ oxidation state) is indeed more stable than $Cr(VI)$ (chromium in $+6$ oxidation state) because stability increases down the group in group $6$ elements.
$(b)$ $KMnO_4$ is a strong oxidizing agent. In the presence of $HCl$,$KMnO_4$ oxidizes $HCl$ to $Cl_2$ gas. Therefore,$HCl$ cannot be used in permanganate titrations as it interferes with the reaction,making the results unsatisfactory.
$(c)$ Lanthanoid oxides (e.g.,$Y_2O_3$ doped with $Eu$) are widely used as phosphors in television screens and fluorescent lamps.
Thus,only statement $(b)$ is incorrect.

(A) In $CrO_{4}^{2-}$,the oxidation state of $Cr$ is $x + 4(-2) = -2$,so $x = +6$.
In $Cr_{2}O_{7}^{2-}$,the oxidation state of $Cr$ is $2x + 7(-2) = -2$,so $2x = +12$,$x = +6$.
Since the oxidation states of chromium in both ions are $+6$,the statement in option $A$ is incorrect.

(D) The element that can exhibit oxidation states ranging from $+4$ to $+6$ is chromium $(Cr)$.
Chromium has an electronic configuration of $[Ar] 3d^5 4s^1$.
It can show various oxidation states from $+1$ to $+6$ in its compounds,such as in $CrO_2$ $(+4)$ and $CrO_3$ $(+6)$.

(D) The nature of transition metal oxides depends on the oxidation state of the metal.
$V_{2}O_{3}$ contains $V$ in the $+3$ oxidation state,which is basic in nature.
$CrO$ contains $Cr$ in the $+2$ oxidation state,which is also basic in nature.
Therefore,both $V_{2}O_{3}$ and $CrO$ are basic oxides.
Thus,$X=$ basic and $Y=$ basic.

(B) When dilute $NaOH$ is added to a $Cr^{3+}$ salt solution,it forms a green precipitate of hydrated chromium$(III)$ oxide,which is represented as $Cr_{2}O_{3} \cdot nH_{2}O$ (often written as $Cr(OH)_{3}$ in simpler contexts,but the hydrated oxide is the standard description).
Therefore,the correct option is $B$.

(B) When $Cu^{2+}$ salts react with potassium iodide $(KI)$,the $Cu^{2+}$ ions are reduced to $Cu^{+}$ ions by the iodide ions $(I^{-})$.
The chemical reaction is as follows:
$2Cu^{2+} + 4I^{-} \longrightarrow 2CuI(s) + I_{2}$
Here,$CuI$ is formed as a white precipitate,and $I_{2}$ is liberated.
Therefore,the correct product is $CuI$.

(N/A) $I^{-}$ is a strong reducing agent. It reduces $Cu^{2+}$ ions to $Cu^{+}$ ions.
The reaction is as follows:
$2Cu^{2+} (aq) + 4I^{-} (aq) \rightarrow Cu_{2}I_{2} (s) + I_{2} (s)$
The white precipitate formed is $Cu_{2}I_{2}$ (Copper$(I)$ iodide).

(A) The strength of an oxidizing agent is determined by its reduction potential value. $A$ higher positive reduction potential indicates a stronger oxidizing agent.
Comparing the standard reduction potentials $(E^{0})$ for the given ions:
$E^{0}_{Mn^{3+}/Mn^{2+}} = +1.51 \, V$
$E^{0}_{Fe^{3+}/Fe^{2+}} = +0.77 \, V$
$E^{0}_{Ti^{3+}/Ti^{2+}} = -0.37 \, V$
$E^{0}_{Cr^{3+}/Cr^{2+}} = -0.41 \, V$
Since $Mn^{3+}$ has the highest positive reduction potential,it is the strongest oxidizing agent among the given options.

(B) $(i)$ Manganese exhibits $+7$ oxidation state in its oxide,$Mn_2O_7$. This is a correct statement.
$(ii)$ Ruthenium $(Ru)$ and Osmium $(Os)$ exhibit $+8$ oxidation state in their oxides,$RuO_4$ and $OsO_4$. This is a correct statement.
$(iii)$ Scandium $(Sc)$ has an electronic configuration of $[Ar] 3d^1 4s^2$ and shows only $+3$ oxidation state. It does not show $+4$ oxidation state. This is an incorrect statement.
$(iv)$ Chromium $(Cr)$ in its $+6$ oxidation state (e.g.,in $Cr_2O_7^{2-}$) acts as a strong oxidizing agent. This is a correct statement.
Therefore,the correct statements are $(i), (ii)$ and $(iv)$.

(A) The basicity of oxides of a transition metal decreases as the oxidation state of the metal increases.
In $V_2O_3$,the oxidation state of $V$ is $+3$.
In $V_2O_4$,the oxidation state of $V$ is $+4$.
In $V_2O_5$,the oxidation state of $V$ is $+5$.
As the oxidation state increases,the acidic character increases and the basic character decreases.
Therefore,the order of basicity is $V_2O_3 > V_2O_4 > V_2O_5$.