Oxygen family Questions in English

(A) The acidic character of group-$16$ hydrides increases down the group.
This is because the bond dissociation enthalpy of the $M-H$ bond decreases as the size of the central atom increases down the group.
As the $M-H$ bond becomes weaker,it becomes easier to release a proton $(H^+)$.
Therefore,the correct order is $H_2O < H_2S < H_2Se < H_2Te$.
Hence,the correct option is $(A)$.

(A) The reactions are:
$1) P_4 + 8SOCl_2 \longrightarrow 4PCl_3 + 4SO_2 + 2S_2Cl_2$
$2) P_4 + 10SO_2Cl_2 \longrightarrow 4PCl_5 + 10SO_2$
In both reactions,the common product '$x$' is sulfur dioxide $(SO_2)$.
In $SO_2$,the central atom is sulfur $(S)$.
The electronic configuration of sulfur is $[Ne] 3s^2 3p^4$.
In $SO_2$,sulfur forms two double bonds with two oxygen atoms.
Sulfur has $6$ valence electrons. It uses $4$ electrons for bonding with oxygen atoms (two double bonds).
Therefore,$6 - 4 = 2$ electrons remain,which form $1$ lone pair on the sulfur atom.
Thus,the number of lone pair of electrons on the central atom of $SO_2$ is $1$.

(A) White phosphorus $(P_4)$ reacts with sulphuryl chloride $(SO_2Cl_2)$ to form phosphorus pentachloride $(PCl_5)$ and sulphur dioxide ($SO_2$,which is $X$).
$P_4 + 10 SO_2Cl_2 \longrightarrow 4 PCl_5 + 10 SO_2$ $(X)$
Chlorine $(Cl_2)$ reacts with sulphur dioxide ($SO_2$,$X$) in the presence of wood charcoal to form sulphuryl chloride ($SO_2Cl_2$,which is $Y$).
$SO_2$ $(X)$ $+ Cl_2 \xrightarrow{\text{Wood charcoal}} SO_2Cl_2$ $(Y)$
Therefore,$X$ is $SO_2$ and $Y$ is $SO_2Cl_2$.

(C) $H_2S_2O_7$ (pyrosulphuric acid) has an $S-O-S$ linkage in its structure. The structure consists of two $SO_3$ units joined by an oxygen atom,represented as $HO-SO_2-O-SO_2-OH$.

(C) Reducing power decreases from $SO_2$ to $TeO_2$ on moving down the group.
$(B)$ The order of boiling points of hydrides of $16^{\text{th}}$ group elements is $H_2S < H_2Se < H_2Te < H_2O$. Due to $H$-bonding,$H_2O$ has a higher boiling point.
$(C)$ Both rhombic and monoclinic sulphur contain $S_8$ molecules. Therefore,statement $(C)$ is incorrect.
$(D)$ The bond angle of ozone is $117^{\circ}$.
Thus,option $(C)$ is the incorrect statement.

(C) Statement $(i)$ is correct: Oxygen shows $-2$ oxidation state in most compounds,$-1$ in peroxides (e.g.,$H_2O_2$),$+1$ in $O_2F_2$,and $+2$ in $OF_2$.
Statement $(ii)$ is incorrect: The thermal stability of group $16$ hydrides decreases down the group due to the increase in bond length and decrease in bond dissociation enthalpy. The correct order is $H_2O > H_2S > H_2Se > H_2Te$.
Statement $(iii)$ is correct: The reducing nature increases down the group as the thermal stability of the hydrides decreases. The order is $H_2S < H_2Se < H_2Te$.

(B) In molecular oxygen $(O_2)$, the bond order is $2$, resulting in a shorter bond length $(121 \text{ pm})$.
In ozone $(O_3)$, due to resonance, the $O-O$ bonds have partial double bond character with a bond order of $1.5$, resulting in a longer bond length $(128 \text{ pm})$.
Therefore, the statement that the $O-O$ bond length in $O_3$ is identical to that of molecular oxygen is incorrect.

(C) The formula for peroxydisulphuric acid (Marshall's acid) is $H_2S_2O_8$.
The formula for pyrosulphuric acid (Oleum) is $H_2S_2O_7$.
The number of oxygen atoms in $H_2S_2O_8$ is $8$.
The number of oxygen atoms in $H_2S_2O_7$ is $7$.
The sum of oxygen atoms $= 8 + 7 = 15$.

(C) The absorption of sulphur trioxide $(SO_3)$ in concentrated sulphuric acid $(H_2SO_4)$ results in the formation of oleum,also known as pyrosulphuric acid.
The chemical reaction is: $SO_3 + H_2SO_4 \rightarrow H_2S_2O_7$.

(A) In the contact process,$SO_2$ is primarily obtained by burning sulfur or sulfide ores in excess of air.
$1$. Burning of sulfur: $S(s) + O_2(g) \rightarrow SO_2(g)$
$2$. Roasting of iron pyrites: $4FeS_2(s) + 11O_2(g) \rightarrow 2Fe_2O_3(s) + 8SO_2(g)$
Thus,$S$ and $FeS_2$ are the common source materials.

(A) $SO_2$ acts as a reducing agent because the $+6$ oxidation state of $S$ is more stable than its $+4$ oxidation state.
$TeO_2$ acts as an oxidizing agent because its $+4$ oxidation state is more stable due to the inert pair effect.
$SO_3$ and $TeO_3$ do not show reducing character because $S$ and $Te$ are in their highest oxidation state of $+6$.

(D) $(i)$ $SO_2$ (sulphur dioxide) reacts with potassium iodate to form sulphuric acid,potassium sulphate,and iodine gas: $5SO_2 + 2KIO_3 + 4H_2O \longrightarrow 4H_2SO_4 + K_2SO_4 + I_2$.
$(ii)$ $SO_2$ decolourises acidified $KMnO_4$ solution by reducing $Mn^{7+}$ to $Mn^{2+}$: $5SO_2 + 2KMnO_4 + 2H_2O \longrightarrow K_2SO_4 + 2H_2SO_4 + 2MnSO_4$.
$(iii)$ $SO_2$ reacts with acidified potassium dichromate $(K_2Cr_2O_7)$ to form green-coloured chromium$(III)$ sulphate: $3SO_2 + K_2Cr_2O_7 + H_2SO_4 \longrightarrow K_2SO_4 + Cr_2(SO_4)_3 + H_2O$.
$(iv)$ $SO_2$ acts as a reducing agent and reduces ferric salts $(Fe^{3+})$ to ferrous salts $(Fe^{2+})$,changing the colour from yellow to light green: $Fe_2(SO_4)_3 + SO_2 + 2H_2O \longrightarrow 2FeSO_4 + 2H_2SO_4$. Therefore,the statement in option $D$ is incorrect as $SO_2$ reduces ferric sulphate to ferrous sulphate,not the other way around.

(A) The reaction of sulphur $(S)$ with sodium sulphite $(Na_2SO_3)$ produces sodium thiosulphate $(Na_2S_2O_3)$,which is compound $(A)$.
$S + Na_2SO_3 \rightarrow Na_2S_2O_3 (A)$
When $Na_2S_2O_3$ reacts with excess $AgNO_3$,it forms silver thiosulphate $(Ag_2S_2O_3)$,which is compound $(B)$.
$Na_2S_2O_3 + 2AgNO_3 \rightarrow Ag_2S_2O_3 (B) + 2NaNO_3$
Finally,$Ag_2S_2O_3$ decomposes in water to form a black precipitate of silver sulphide $(Ag_2S)$,which is compound $(C)$.
$Ag_2S_2O_3 + H_2O \rightarrow Ag_2S (C) + H_2SO_4$
Thus,the correct sequence is $(A) = Na_2S_2O_3$,$(B) = Ag_2S_2O_3$,and $(C) = Ag_2S$.

(D) $(a) \ [S_2 O_3]^{2-}$ (thiosulfate) acts as a strong reducing agent $(III)$.
$(b) \ H_2 O_2$ undergoes disproportionation in the presence of $Mn^{2+}$ ions $(IV)$.
$(c) \ H_2 S$ is a toxic gas $(II)$.
$(d) \ S_2$ is paramagnetic $(I)$ due to the presence of two unpaired electrons in its molecular orbitals,similar to $O_2$.
Therefore,the correct matching is $(a)$ $\rightarrow (III), (b)$ $\rightarrow (IV), (c)$ $\rightarrow (II), (d)$ $\rightarrow (I)$.

(D) Statement $(i)$ is correct: Sulphuric acid is manufactured by the contact process.
Statement $(ii)$ is correct: $SO_3$ dissolves in $H_2SO_4$ to form oleum or pyrosulphuric acid $(H_2S_2O_7)$.
Statement $(iii)$ is correct: $H_2SO_4$ is widely used in the manufacture of fertilisers like ammonium sulphate and super phosphate.
Statement $(iv)$ is incorrect: In the reaction $S + 2H_2SO_4 (\text{conc.}) \longrightarrow 3SO_2 + 2H_2O$,sulphur $(S)$ is oxidised from $0$ to $+4$ oxidation state in $SO_2$,while $H_2SO_4$ acts as an oxidising agent and is reduced to $SO_2$. Thus,$S$ is oxidised,not $H_2SO_4$.

(D) $O_3$ and $SO_2$ both have a bent (angular) shape,so this statement is incorrect.
$(B)$ The molecular formula of pyrosulphuric acid (oleum) is $H_2S_2O_7$,not $H_2S_2O_8$. Thus,this statement is incorrect.
$(C)$ In the presence of moisture,$SO_2$ acts as a reducing agent,not an oxidising agent,as it gets oxidised to $H_2SO_4$ or $SO_3$. The reaction is: $SO_2 + 2H_2O \longrightarrow SO_4^{2-} + 4H^+ + 2e^-$.
$(D)$ $V_2O_5$ acts as a catalyst in the contact process for the oxidation of $SO_2$ to $SO_3$. The reaction is: $2SO_2 + O_2 \xrightarrow{V_2O_5} 2SO_3$. This statement is correct.

(A) $1$. Thermal stability: As the size of the central atom increases down the group,the $M-H$ bond length increases and bond dissociation enthalpy decreases. Thus,thermal stability decreases: $H_2O > H_2S > H_2Se > H_2Te$. Statement $a$ is correct.
$2$. Acidic nature: Acidic strength depends on the ease of release of $H^+$ ions,which increases as the $M-H$ bond strength decreases down the group. Thus,acidic nature increases: $H_2O < H_2S < H_2Se < H_2Te$. Statement $b$ is correct.
$3$. Reducing nature: Reducing character depends on the ability to provide hydrogen,which increases as the $M-H$ bond strength decreases. Thus,reducing nature increases: $H_2S < H_2Se < H_2Te$. Statement $c$ is correct.
Therefore,all statements $a, b,$ and $c$ are correct.

(D) $1$. Rhombic and monoclinic sulphur are allotropes of sulphur and both are soluble in $CS_2$. Thus,option $A$ is correct.
$2$. Both rhombic and monoclinic sulphur consist of puckered $S_8$ ring structures. Thus,option $B$ is correct.
$3$. $SO_2$ is highly soluble in water,forming sulphurous acid $(H_2SO_3)$. Thus,option $C$ is correct.
$4$. In the vapour state,$S_2$ is paramagnetic,similar to $O_2$,because it contains two unpaired electrons in its antibonding $\pi^*$ molecular orbitals. Thus,option $D$ is incorrect.

(B) The statement that the oxidation state of sulphur is never less than $+4$ is incorrect. Sulphur exhibits a wide range of oxidation states in its compounds,ranging from $-2$ (as in $H_2S$) to $+6$ (as in $H_2SO_4$).
At high temperatures (around $1000 \ K$),$S_2$ is the dominant species and it is paramagnetic,similar to $O_2$.
Rhombic sulphur is the most stable allotrope at room temperature and is readily soluble in $CS_2$.

(A) Only $H_2S_2O_6$ (dithionic acid) contains an $S-S$ bond. The structure is as follows:
$HO-S(=O)_2-S(=O)_2-OH$

(D) $1$. Thermal stability decreases down the group as the bond dissociation energy decreases: $H_2O > H_2S > H_2Se > H_2Te > H_2Po$. This is correct.
$2$. Reducing property increases down the group as the bond dissociation energy decreases: $H_2S < H_2Se < H_2Te < H_2Po$. This is correct.
$3$. Boiling point: $H_2O$ has the highest boiling point due to hydrogen bonding. For the rest,boiling point increases with increasing molecular mass: $H_2S < H_2Se < H_2Te < H_2O$. This is correct.
$4$. Melting point: The trend for melting point is $H_2S < H_2Se < H_2O < H_2Te$. $H_2O$ does not have the highest melting point compared to $H_2Te$ because of the crystal structure differences. Thus,the statement in option $D$ is incorrect.

(D) The $S_2O_7^{2-}$ (pyrosulphate or disulphate) ion possesses an $S-O-S$ bridge bond.
In its structure,two $SO_4$ tetrahedra share a common oxygen atom,resulting in an $S-O-S$ linkage.

(D) Group $16$ elements are known as chalcogens because they are ore-forming elements.

(C) In group $16$,the electron gain enthalpy becomes less negative as we move down the group due to an increase in atomic size.
However,oxygen $(O)$ has an exceptionally low electron gain enthalpy due to its small size and strong inter-electronic repulsions.
Therefore,sulfur $(S)$ has the highest (most negative) electron gain enthalpy,and polonium $(Po)$ or tellurium $(Te)$ depending on the context,but generally,oxygen is the lowest in the group.
Comparing the options,sulfur $(S)$ has the highest and oxygen $(O)$ has the lowest electron gain enthalpy in group $16$.
Thus,option $(C)$ is correct.

(C) In group $16$,the metallic character increases as we move down the group from $O$ to $Po$.
Metallic oxides are generally basic in nature,while non-metallic oxides are acidic.
Since $Po$ is a metal,its oxide $PoO$ is the most basic among the given options.
As we move down the group,the electronegativity of the element decreases,which leads to an increase in the basic character of the corresponding oxides.

(C) As the size of the central atom increases (order of size $O < S < Se < Te$),the $H-A$ (where $A$ is the central atom) bond length increases.
Consequently,the $H-A$ bond dissociation energy decreases.
Therefore,$H_2Te$ releases a proton $(H^+)$ most readily,making it the most acidic among the given compounds.
The acidic strength order is $H_2O < H_2S < H_2Se < H_2Te$.

(C) Among the allotropes of sulphur,rhombic (yellow) sulphur is more stable than the monoclinic sulphur and is stable at room temperature.
Above $369 \ K$,rhombic sulphur transforms into monoclinic.
$N_2$ is a colourless gas.
$O_3$ is less stable than $O_2$ because its decomposition into $O_2$ is exothermic.
$Cl_2$ is a greenish-yellow gas.

(D) On moving down the group from $O$ to $Te$,the molecular mass increases,which leads to an increase in the magnitude of van der Waals forces of attraction,causing the boiling point to increase.
However,$H_2O$ exhibits an anomalously high boiling point due to the presence of strong intermolecular hydrogen bonding.
The boiling point order for group $16$ hydrides is: $H_2S < H_2Se < H_2Te < H_2O$.
Therefore,$H_2S$ has the lowest boiling point $(X)$ and $H_2O$ has the highest boiling point $(Y)$.

(C) Ozone $(O_3)$ is prepared from oxygen $(O_2)$ by passing silent electric discharge through pure,dry oxygen. This process is endothermic and requires energy to break the strong $O=O$ bond in the oxygen molecule. The reaction is as follows:
$3O_2 \xrightarrow{\text{Silent electric discharge}} 2O_3$ $(\Delta H = +142 \ kJ/mol)$
Since the reaction is endothermic,a silent electric discharge is used to prevent the decomposition of ozone back into oxygen due to excessive heat. Thus,option $(c)$ is correct.

(A) The formation of ozone from oxygen is an endothermic reaction,not an exothermic reaction.
The correct reaction is: $3 O_2 \underset{\text{silent electric discharge}}{\rightleftharpoons} 2 O_3 ; \Delta H = +284.5 \ kJ$.
Therefore,the statement $3 O_2 \underset{\text{silent electric discharge}}{\rightleftharpoons} 2 O_3 ; \Delta H = -284.5 \ kJ$ is incorrect.

(C) $1$. $TeO_2$ acts as an oxidizing agent because $Te$ in $+4$ oxidation state is less stable than in $+2$ oxidation state due to the inert pair effect.
$2$. $SeO_3$ is an acidic oxide as it reacts with water to form selenic acid $(H_2SeO_4)$.
$3$. $SeO_2$ is a white crystalline solid at room temperature,not a gas.
$4$. $SO_2$ acts as a reducing agent because sulfur can be oxidized from $+4$ to $+6$ oxidation state.
Therefore,the statement "$SeO_2$ is a gas" is incorrect.

(C) The $S-O-S$ bond is present in disulphuric acid (also known as pyrosulphuric acid or oleum),which has the chemical formula $H_2S_2O_7$.
In its structure,two $SO_3$ units are linked through an oxygen atom,forming an $S-O-S$ bridge.

(C) Both rhombic and monoclinic sulphur consist of $S_8$ molecules.
These $S_8$ molecules have a puckered ring structure,often described as a crown shape,not a planar structure.
Therefore,the Assertion $(A)$ is true,but the Reason $(R)$ is false.

(D) The oxoacid $H_2S_2O_8$ is known as peroxodisulphuric acid (Marshall's acid).
It contains a peroxy linkage $(-O-O-)$ in its structure.
The structure is $HO-SO_2-O-O-SO_2-OH$.

(C) Sulphur dioxide reacts with chlorine in the presence of charcoal to give $SO_2Cl_2$,which is known as sulphuryl chloride.
The chemical reaction is as follows:
$SO_2 + Cl_2 \xrightarrow{\text{charcoal}} SO_2Cl_2$

(A) Peroxydisulphuric acid $(H_2S_2O_8)$ is an inorganic compound.
It is commonly known as Marshall's acid,named after its inventor,Hugh Marshall.
It contains a peroxide linkage $(-O-O-)$ between two sulphur atoms.

(A) When sulphur is boiled with $NaOH$ solution,it undergoes a disproportionation reaction to form sodium thiosulphate and sodium sulphide.
$4S + 6NaOH \longrightarrow Na_2S_2O_3 + 2Na_2S + 3H_2O$

(D) The final acid product obtained during the synthesis of $H_2SO_4$ by contact process is $H_2S_2O_7$,which is known as oleum.
In the contact process,$SO_3$ gas is absorbed in concentrated $H_2SO_4$ to form oleum.
$(i) \ 2SO_2 + O_2 \rightleftharpoons 2SO_3$
$(ii) \ H_2SO_4 + SO_3 \longrightarrow H_2S_2O_7$ (Oleum)

(C) The thermodynamically most stable allotrope of sulphur is rhombic sulphur,also known as $\alpha$-sulphur.
It exists as a molecular crystal composed of puckered $S_8$ ring units.

(D) The catalytic oxidation of $SO_2$ with $O_2$ to give $SO_3$ is the key step in the manufacturing of sulphuric acid by the contact process.
$2SO_{2(g)} + O_{2(g)} \xrightarrow{V_2O_5} 2SO_{3(g)};$
$\Delta_{r}H^{\circ} = -196 \ kJ \ mol^{-1}$

(A) The structures of the given oxoacids of sulphur are as follows:
$1$. Pyrosulphurous acid $(H_2S_2O_5)$: Contains two $S$ atoms with oxidation states $+3$ and $+5$.
$2$. Hyposulphurous acid $(H_2S_2O_4)$: Contains two $S$ atoms with oxidation states $+3$ and $+3$.
$3$. Pyrosulphuric acid $(H_2S_2O_7)$: Contains two $S$ atoms with oxidation states $+6$ and $+6$.
$4$. Peroxodisulphuric acid $(H_2S_2O_8)$: Contains two $S$ atoms with oxidation states $+6$ and $+6$.
Thus,pyrosulphurous acid is the only oxoacid among the options where the two sulphur atoms have different oxidation states.

(A) Among the given options,only $H_2S_2O_6$ (dithionic acid) contains an $S-S$ bond.
Its structure is $HO-SO_2-SO_2-OH$,where the two sulfur atoms are directly linked to each other.

(A) Statement $(A)$ is correct because sulphur vapour exists as $S_2$ at high temperatures,which is paramagnetic due to the presence of two unpaired electrons in its antibonding molecular orbitals,similar to $O_2$.
Statement $(B)$ is incorrect because the reaction of dilute $HCl$ with iron produces $FeCl_2$ and $H_2$ gas,not $FeCl_3$.
The reaction is: $Fe(s) + 2HCl(dil.) \longrightarrow FeCl_2(aq) + H_2(g)$.
The $H_2$ gas produced acts as a reducing agent and prevents the oxidation of $Fe^{2+}$ to $Fe^{3+}$.
Therefore,statement $(A)$ is correct and statement $(B)$ is wrong.

(D) The reducing power of oxides of group $16$ elements decreases down the group.
This is because the stability of the $+4$ oxidation state increases down the group due to the inert pair effect.
$SO_2$ acts as a reducing agent because $S$ can easily be oxidized from $+4$ to $+6$ oxidation state.
In contrast,$TeO_2$ is more stable in the $+4$ state and does not easily oxidize to $+6$.
Therefore,$SO_2$ is the strongest reducing agent among the given options.

(A) When sulphuryl chloride $(SO_{2}Cl_{2})$ reacts with white phosphorus $(P_{4})$,it produces phosphorus pentachloride $(PCl_{5})$ and sulphur dioxide $(SO_{2})$.
The balanced chemical equation for this reaction is:
$P_{4} + 10 SO_{2}Cl_{2} \longrightarrow 4 PCl_{5} + 10 SO_{2}$