Oxygen family Questions in English

(D) Ozone $(O_3)$ exists in different states with distinct colors:
$1$. In the gaseous state,it is a pale blue gas.
$2$. In the liquid state,it is a deep blue liquid.
$3$. In the solid state,it is a violet-black solid.
Therefore,all the given options are correct.

(N/A) The stable form of sulfur at room temperature is rhombic sulfur,which transforms into monoclinic sulfur when heated to $369 \ K$.
Rhombic sulfur ($\alpha$-sulfur):
This allotrope is yellow in color,with a melting point of $385.8 \ K$ and a specific gravity of $2.06$.
It is prepared by evaporating the solution of roll sulfur in $CS_2$.
It is insoluble in water but slightly soluble in benzene,alcohol,and ether.
It is readily soluble in $CS_2$.
Monoclinic sulfur ($\beta$-sulfur):
Its melting point is $393 \ K$ and its specific gravity is $1.98$.
It is soluble in $CS_2$.
This form is prepared by melting rhombic sulfur in a dish and cooling it until a crust is formed. Two holes are made in the crust,and the remaining liquid is poured out.
Upon removing the crust,colorless needle-shaped crystals of $\beta$-sulfur are formed. It is stable above $369 \ K$ and transforms into $\alpha$-sulfur below this temperature.
Conversely,$\alpha$-sulfur is stable below $369 \ K$ and transforms into $\beta$-sulfur above this temperature.

(B) Sulfur exhibits allotropy. The two most common allotropes are rhombic sulfur and monoclinic sulfur.
Rhombic sulfur is also known as $\alpha$-sulfur.
Monoclinic sulfur is also known as $\beta$-sulfur.
Therefore,the correct identification is that rhombic sulfur is $\alpha$-sulfur and monoclinic sulfur is $\beta$-sulfur.

(A) $\alpha$-sulphur,also known as rhombic sulphur,is the most stable allotrope of sulphur at room temperature. It is yellow in color.

(A) The $\alpha$-sulphur (rhombic sulphur) is formed by evaporating the solution of roll sulphur in $CS_2$. Its specific density is $2.06 \ g/cm^3$.
The $\beta$-sulphur (monoclinic sulphur) is prepared by melting rhombic sulphur in a dish and cooling until a crust is formed. Its specific density is $1.98 \ g/cm^3$.
Therefore,the specific densities are $2.06 \ g/cm^3$ and $1.98 \ g/cm^3$ respectively.

(N/A) Preparation:
$1$. Sulphur dioxide is formed together with a little $(6-8 \%)$ sulphur trioxide when sulphur is burnt in air or oxygen: $S_{(s)} + O_{2(g)} \rightarrow SO_{2(g)}$
$2$. In the laboratory,it is readily generated by treating a sulphite with dilute sulphuric acid: $SO_{3(aq)}^{2-} + 2H_{(aq)}^{+} \rightarrow H_{2}O_{(l)} + SO_{2(g)}$
$3$. Industrially,$SO_{2}$ is obtained as a by-product of the roasting of sulphide ores: $4FeS_{2(s)} + 11O_{2(g)} \rightarrow 2Fe_{2}O_{3(s)} + 8SO_{2(g)}$
Properties:
$(i)$ Physical properties: Sulphur dioxide is a colourless gas with a pungent smell and is highly soluble in water. It boils at $263 \ K$ and liquefies at room temperature under a pressure of $2 \ atm$. The shape of $SO_{2}$ is angular. It is a resonance hybrid of two canonical forms.
$(ii)$ Chemical properties:
- $SO_{2}$ when passed through water forms a solution of sulphurous acid: $SO_{2(g)} + H_{2}O_{(l)} \rightleftharpoons H_{2}SO_{3(aq)}$
- It reacts with $NaOH$ solution to form sodium sulphite,which reacts with more $SO_{2}$ to form sodium hydrogen sulphite: $2NaOH + SO_{2} \rightarrow Na_{2}SO_{3} + H_{2}O$; $Na_{2}SO_{3} + H_{2}O + SO_{2} \rightarrow 2NaHSO_{3}$
- $SO_{2}$ reacts with chlorine in the presence of charcoal (catalyst) to give sulphuryl chloride: $SO_{2(g)} + Cl_{2(g)} \rightarrow SO_{2}Cl_{2(l)}$
- In the presence of vanadium$(V)$ oxide,it oxidizes to $SO_{3}$: $SO_{2(g)} + \frac{1}{2}O_{2(g)} \xrightarrow{V_{2}O_{5}} SO_{3(g)}$
- Moist sulphur dioxide acts as a reducing agent: $2Fe^{3+} + SO_{2} + 2H_{2}O \rightarrow 2Fe^{2+} + SO_{4}^{2-} + 4H^{+}$; $5SO_{2} + 2MnO_{4}^{-} + 2H_{2}O \rightarrow 5SO_{4}^{2-} + 4H^{+} + 2Mn^{2+}$
Uses:
- In the refining of petroleum and sugar.
- In the bleaching of wool and silk.
- As an anti-chlor,disinfectant,and preservative.
- Industrial chemicals such as $H_{2}SO_{4}$,$NaHSO_{3}$,and $Ca(HSO_{3})_{2}$ are prepared from $SO_{2}$.
- Liquid $SO_{2}$ is used as a solvent to dissolve a number of inorganic and organic chemicals.

(B) When $SO_2$ is moist,it reacts with water to form sulfurous acid $(H_2SO_3)$.
$SO_2 + H_2O \rightarrow H_2SO_3$
In this state,it acts as a strong reducing agent because it can easily be oxidized to sulfuric acid $(H_2SO_4)$ by losing electrons or by reacting with oxygen.

(N/A) Sulfur forms many oxoacids such as $H_{2}SO_{3}$,$H_{2}S_{2}O_{3}$,$H_{2}S_{2}O_{4}$,$H_{2}S_{2}O_{5}$,$H_{2}S_{x}O_{6}$ ($x=2$ to $5$),$H_{2}SO_{4}$,$H_{2}S_{2}O_{7}$,$H_{2}SO_{5}$,and $H_{2}S_{2}O_{8}$.
Some of these acids are unstable and cannot be isolated. They are often identified in aqueous solution or in the form of their salts. The structures of some important oxoacids are shown in the image.

(N/A) The contact process involves the following key steps:
$(i)$ Burning of sulphur or sulphide ores in the presence of air to generate $SO_{2}$.
$(ii)$ Purification of $SO_{2}$ gas by removing dust and impurities like arsenic compounds.
$(iii)$ Catalytic oxidation of $SO_{2}$ to $SO_{3}$ using oxygen in the presence of a $V_{2}O_{5}$ catalyst:
$2 SO_{2(g)} + O_{2(g)} \rightarrow 2 SO_{3(g)}$,$\Delta_{r}H^{\ominus} = -196.6 \ kJ/mol$.
This reaction is exothermic,reversible,and involves a decrease in volume. Thus,low temperature and high pressure are favourable for maximum yield. In practice,the process is carried out at $2 \ bar$ pressure and $720 \ K$ temperature.
$(iv)$ Absorption of $SO_{3}$ in concentrated $H_{2}SO_{4}$ to produce oleum $(H_{2}S_{2}O_{7})$:
$SO_{3} + H_{2}SO_{4} \rightarrow H_{2}S_{2}O_{7}$ (Oleum).
$(v)$ Dilution of oleum with water to obtain sulphuric acid of the desired concentration ($96-98 \%$ pure).

(N/A) The contact process for the industrial manufacturing of sulphuric acid involves the following major steps:
$(i)$ Burning of sulphur or sulphide ores in the presence of air to generate $SO_{2}$ gas.
$(ii)$ Purification of $SO_{2}$ gas by removing dust and other impurities such as arsenic compounds.
$(iii)$ Catalytic oxidation of $SO_{2}$ with $O_{2}$ to form $SO_{3}$ in the presence of a vanadium pentoxide $(V_{2}O_{5})$ catalyst:
$2SO_{2(g)} + O_{2(g)} \rightleftharpoons 2SO_{3(g)}, \Delta_{r}H^{\ominus} = -196.6 \ kJ/mol$
This reaction is exothermic,reversible,and involves a decrease in volume. Therefore,low temperature and high pressure are favourable for maximum yield. In practice,the process is carried out at $2 \ bar$ pressure and $720 \ K$ temperature.
$(iv)$ Absorption of $SO_{3}$ in concentrated $H_{2}SO_{4}$ to produce oleum $(H_{2}S_{2}O_{7})$:
$SO_{3} + H_{2}SO_{4} \rightarrow H_{2}S_{2}O_{7}$
$(v)$ Dilution of oleum with water to obtain sulphuric acid of the desired concentration.
The sulphuric acid obtained by the contact process is $96-98 \%$ pure.

(N/A) $(i)$ Physical properties: Sulphuric acid is a colourless,dense,oily liquid with a specific gravity of $1.84$ at $298 \ K$. The freezing point is $283 \ K$ and the boiling point is $611 \ K$. It dissolves in water with the evolution of a large quantity of heat. Hence,while diluting an acid,the acid must be added to water and not water to an acid with constant stirring.
$(ii)$ Chemical properties: The chemical reactions of sulphuric acid are a result of the following characteristics:
$(a)$ Low volatility
$(b)$ Strong acidic character
$(c)$ Strong affinity for water
$(d)$ Ability to act as an oxidising agent
In aqueous medium,sulphuric acid ionises in two steps:
$H_2SO_{4(aq)} + H_2O_{(l)} \rightarrow H_3O^{+}_{(aq)} + HSO^{-}_{4(aq)} ; Ka_1 = \text{very large } (Ka_1 > 10)$
$HSO^{-}_{4(aq)} + H_2O_{(l)} \rightarrow H_3O^{+}_{(aq)} + SO^{2-}_{4(aq)} ; Ka_2 = 1.2 \times 10^{-2}$
The larger value of $Ka_1$ $(Ka_1 > 10)$ means that $H_2SO_4$ is largely dissociated into $H^{+}$ and $HSO^{-}_4$. Greater the value of the dissociation constant $(K_a)$,the stronger is the acid. The acid forms two series of salts: Normal salts (e.g.,sodium sulphate,copper sulphate) and acid sulphates (e.g.,sodium hydrogen sulphate).
Sulphuric acid,because of its low volatility,can be used to manufacture more volatile acids from their corresponding salts:
$2MX + H_2SO_4 \rightarrow 2HX + M_2SO_4$ (where $X = F, Cl, NO_3$ and $M = \text{Metal}$).
Concentrated sulphuric acid is a strong dehydrating agent. Many wet gases can be dried by passing them through sulphuric acid,provided the gases do not react with the acid. Sulphuric acid removes water from organic compounds; it is evident by its charring action on carbohydrates:
$C_{12}H_{22}O_{11} \xrightarrow{H_2SO_4} 11H_2O + 12C$
Hot concentrated sulphuric acid is a moderately strong oxidising agent. It oxidises both metals and non-metals,being reduced to $SO_2$:
$Cu + 2H_2SO_4 \rightarrow CuSO_4 + SO_2 + 2H_2O$
$S + 2H_2SO_4 \rightarrow 3SO_2 + 2H_2O$
$C + 2H_2SO_4 \rightarrow CO_2 + 2SO_2 + 2H_2O$
Sulphuric acid is an important industrial chemical,often called the "King of chemicals".

(A) The $H_2SO_4$ (sulfuric acid) is manufactured by the $Contact \ process$.
In this process,sulfur dioxide $(SO_2)$ is oxidized to sulfur trioxide $(SO_3)$ in the presence of a catalyst (vanadium pentoxide,$V_2O_5$),which is then absorbed in concentrated $H_2SO_4$ to form oleum $(H_2S_2O_7)$.
Finally,oleum is diluted with water to obtain $H_2SO_4$ of the desired concentration.

(A) The Contact Process for the manufacture of sulfuric acid consists of $3$ main stages:
$1$. Burning of sulfur or sulfide ores in air to generate sulfur dioxide $(SO_2)$.
$2$. Conversion of $SO_2$ to sulfur trioxide $(SO_3)$ by the reaction with oxygen in the presence of a catalyst $(V_2O_5)$.
$3$. Absorption of $SO_3$ in $H_2SO_4$ to give oleum $(H_2S_2O_7)$,which is then diluted with water to produce sulfuric acid $(H_2SO_4)$.
Thus,the final product obtained is sulfuric acid $(H_2SO_4)$.

(B) The physical properties of concentrated sulfuric acid $(H_2SO_4)$ are as follows:
$1$. Melting point: $283 \ K$
$2$. Boiling point: $611 \ K$
$3$. Specific gravity: $1.84 \ g/cm^3$ at $298 \ K$
Therefore,the correct values correspond to option $B$.

(A) In the contact process,sulfur dioxide $(SO_2)$ is oxidized to sulfur trioxide $(SO_3)$ in the presence of vanadium pentoxide $(V_2O_5)$ as a catalyst.
The reaction is: $2SO_2(g) + O_2(g) \xrightarrow{V_2O_5} 2SO_3(g)$.

(B) $1.$ In the contact process for the manufacture of sulfuric acid,$SO_2$ is oxidized to $SO_3$ using vanadium pentoxide $(V_2O_5)$ as a catalyst. Thus,statement $1$ is True $(T)$.
$2.$ Ziegler catalyst (e.g.,$TiCl_4 + Al(C_2H_5)_3$) is used for the polymerization of ethene,not for the oxidation of $SO_2$. Thus,statement $2$ is False $(F)$.

(A) The ability of oxygen to stabilize higher oxidation states is due to its ability to form multiple bonds with metals ($p\pi-d\pi$ bonding).
Fluorine,being a monovalent atom,can only form single bonds,which limits its ability to stabilize higher oxidation states compared to oxygen,which can form double bonds with metal atoms.

(A) Sulfur dioxide $(SO_2)$ is a colorless gas with a pungent and suffocating odor,similar to that of a burnt matchstick. It is highly irritating to the respiratory system.

(N/A) If $SO_3$ is absorbed directly in water,a large amount of heat is released,which leads to the formation of an acid fog.
This acid fog is extremely difficult to condense,making the process inefficient and hazardous.

(N/A) The conversion of ozone $(O_3)$ to oxygen $(O_2)$ is an exothermic process,which means it is accompanied by the liberation of heat $(\Delta H < 0)$.
Additionally,the decomposition of $2$ moles of $O_3$ into $3$ moles of $O_2$ leads to an increase in entropy $(\Delta S > 0)$.
According to the Gibbs free energy equation,$\Delta G = \Delta H - T\Delta S$,both a negative $\Delta H$ and a positive $\Delta S$ result in a large negative value for the Gibbs free energy change $(\Delta G)$.
Since a negative $\Delta G$ indicates a spontaneous process,ozone is thermodynamically less stable than oxygen.

(A) The solid $'A'$ is sulphur $(S_8)$ and the gas $'B'$ is sulphur dioxide $(SO_2)$.
$1$. Combustion of sulphur in air:
$S_8(s) + 8O_2(g) \rightarrow 8SO_2(g)$
$2$. Reaction with lime water:
$SO_2(g) + Ca(OH)_2(aq) \rightarrow CaSO_3(s) + H_2O(l)$
(Calcium sulphite makes the solution milky)
$3$. Decolourisation of acidified $KMnO_4$:
$5SO_2 + 2MnO_4^{-} + 2H_2O \rightarrow 5SO_4^{2-} + 4H^{+} + 2Mn^{2+}$
$4$. Reduction of $Fe^{3+}$ to $Fe^{2+}$:
$2Fe^{3+} + SO_2 + 2H_2O \rightarrow 2Fe^{2+} + SO_4^{2-} + 4H^{+}$

(N/A) $(i)$ Oxygen: Oxygen is the most abundant of all the elements on earth. It forms about $46.6 \%$ by mass of the earth's crust. Dry air contains $20.946 \%$ oxygen by volume.
$(ii)$ Sulphur: The abundance of sulphur in the earth's crust is only $0.03-0.1 \%$. Combined sulphur exists primarily as sulphates such as gypsum $(CaSO_4 \cdot 2H_2O)$,epsom salt $(MgSO_4 \cdot 7H_2O)$,baryte $(BaSO_4)$ and sulphides such as galena $(PbS)$,zinc blende $(ZnS)$,and copper pyrites $(CuFeS_2)$. Traces of sulphur occur as hydrogen sulphide in volcanoes and in organic materials such as eggs,proteins,garlic,onion,mustard,hair,and wool.
$(iii)$ Selenium and tellurium: Selenium and tellurium are found as metal selenides and tellurides in sulphide ores.
$(iv)$ Polonium and livermorium: Polonium occurs in nature as a decay product of thorium and uranium minerals. Livermorium is a synthetic radioactive element. It has been produced only in a very small amount and has a very short half-life (only a small fraction of one second),which limits the study of its properties.

The elements of group-$16$ have six electrons in their valence shell. Hence,the general valence shell electronic configuration is $ns^{2} np^{4}$.

Element	Electronic configuration
$_{8}O$	$[He] 2s^{2} 2p^{4}$
$_{16}S$	$[Ne] 3s^{2} 3p^{4}$
$_{34}Se$	$[Ar] 3d^{10} 4s^{2} 4p^{4}$
$_{52}Te$	$[Kr] 4d^{10} 5s^{2} 5p^{4}$
$_{84}Po$	$[Xe] 4f^{14} 5d^{10} 6s^{2} 6p^{4}$
$_{116}Lv$	$[Rn] 5f^{14} 6d^{10} 7s^{2} 7p^{4}$

(N/A) $(i)$ Atomic radii: The atomic radii of group-$16$ elements are smaller than the corresponding group-$15$ elements. This is due to an increase in effective nuclear charge.
In group-$16$,moving down the group,the atomic radii and ionic radii increase due to the addition of new electron shells at each successive element.
$(ii)$ Ionisation enthalpy: The first ionisation enthalpies of group-$16$ elements are lower than the corresponding group-$15$ elements. This is due to the fact that group-$15$ elements have extra stable half-filled $p$-orbitals.
The ionisation enthalpy of group-$16$ elements is quite high due to their small size and high nuclear charge. However,down the group,the ionisation enthalpy decreases with the increase in atomic size.

(N/A) $(i)$ Electron gain enthalpy: Group-$16$ elements have high negative electron gain enthalpy. Oxygen has a less negative electron gain enthalpy than sulphur due to the compact nature (small size) of the oxygen atom,which leads to inter-electronic repulsions. From sulphur onwards,the electron gain enthalpy becomes less negative down the group up to polonium.
Order of electron gain enthalpy: $O < S > Se > Te > Po$
$(ii)$ Electronegativity: Oxygen is the second most electronegative element after fluorine. Down the group,electronegativity decreases with an increase in atomic number. This implies that metallic character increases from oxygen to polonium.

(N/A) $(i)$ Atomicity: Oxygen is a diatomic gas,while other elements are polyatomic solids. For example,sulphur,selenium,and tellurium exist as polyatomic molecules such as $S_{8}$,$Se_{8}$,etc. This is due to the tendency of oxygen to form $p\pi-p\pi$ bonds because of its small size and high bond enthalpy of the $(O=O)$ bond.
$(ii)$ Metallic character: Down the group,the metallic character of the elements increases. Oxygen and sulphur are non-metals,selenium and tellurium are metalloids,whereas polonium is a metal. Polonium is radioactive with a half-life period of $13.8 \ days$.
$(iii)$ Melting and boiling points: The melting and boiling points gradually increase down the group. The large difference between the melting and boiling points of oxygen and sulphur is due to their atomicity; oxygen is diatomic,whereas sulphur is polyatomic.
$(iv)$ Allotropy: All elements of group-$16$ exhibit allotropy.

(N/A) The elements of group-$16$ exhibit a variety of oxidation states. The stability of the $(-2)$ oxidation state decreases down the group. Polonium hardly shows the $(-2)$ oxidation state.
Since the electronegativity of oxygen is very high,it shows only a negative oxidation state of $(-2)$,except in the case of $OF_{2}$,where its oxidation state is $(+2)$.
The elements of group-$16$ other than oxygen also show $(+2), (+4),$ and $(+6)$ oxidation states,but $(+4)$ and $(+6)$ are the most common. Sulfur,selenium,and tellurium usually show the $(+4)$ oxidation state in their compounds with oxygen and the $(+6)$ oxidation state with fluorine (e.g.,$SF_{6}$).
The stability of the $(+6)$ oxidation state decreases down the group,while the stability of the $(+4)$ oxidation state increases due to the inert pair effect. Bonding in the $(+4)$ and $(+6)$ oxidation states is primarily covalent.

(N/A) The elements of group-$16$ form oxides of type $EO_{2}$ and $EO_{3}$,where $E = S, Se, Te$ or $Po$.
$SO_{2}$ is a gas,while $SeO_{2}$ is a solid. $SO_{2}$ acts as a reducing agent,whereas $TeO_{2}$ acts as an oxidising agent.
The stability of $EO_{2}$ type oxides decreases down the group: $SO_{2} > SeO_{2} > TeO_{2}$.
The acidic character of $EO_{2}$ type oxides also decreases down the group: $SO_{2} > SeO_{2} > TeO_{2}$.
These elements also form oxides of type $EO_{3}$ (e.g.,$SO_{3}, SeO_{3}, TeO_{3}$). The acidic strength of these oxides decreases down the group: $SO_{3} > SeO_{3} > TeO_{3}$.

(N/A) The elements of group-$16$ form oxides of type $EO_2$ and $EO_3$,where $E = S, Se, Te$ or $Po$.
$SO_2$ is a gas,while $SeO_2$ is a solid.
$SO_2$ acts as a reducing agent,whereas $TeO_2$ acts as an oxidizing agent.
The stability of these oxides decreases down the group.
Stability order: $SO_2 > SeO_2 > TeO_2$.
Acidic strength order: $SO_2 > SeO_2 > TeO_2$.
These elements also form oxides of type $EO_3$ (e.g.,$SO_3, SeO_3, TeO_3$).
The acidic strength of $EO_3$ type oxides also decreases down the group.
Acidic strength order: $SO_3 > SeO_3 > TeO_3$.

(N/A) The elements of group-$16$ form halides of the type $EX_{6}$,$EX_{4}$,and $EX_{2}$,where $E$ is an element of the group and $X$ is a halogen.
The stability of the halides decreases in the order $F^{-} > Cl^{-} > Br^{-} > I^{-}$. Amongst hexahalides,only hexafluorides are stable. All hexafluorides are gaseous in nature and have an octahedral structure. $SF_{6}$ is exceptionally stable.
Amongst tetrafluorides,$SF_{4}$ is a gas,$SeF_{4}$ is a liquid,and $TeF_{4}$ is a solid. These fluorides exhibit $sp^{3}d$ hybridization and have a trigonal bipyramidal structure,where one of the equatorial positions is occupied by a lone pair of electrons. This geometry is also known as a see-saw geometry.
All elements except oxygen form dichlorides and dibromides. These dihalides are formed by $sp^{3}$ hybridization and have a tetrahedral structure. The well-known monohalides are dimeric in nature.
Examples include $S_{2}F_{2}$,$S_{2}Cl_{2}$,$S_{2}Br_{2}$,$Se_{2}Cl_{2}$,and $Se_{2}Br_{2}$. These dimeric halides undergo disproportionation reactions,for example: $2 Se_{2}Cl_{2} \rightarrow 3 Se + SeCl_{4}$.

(N/A) Oxygen shows anomalous behaviour compared to other elements of Group $16$ due to its small size,high electronegativity,and the absence of $d$-orbitals in its valence shell.
$1$. Oxygen is a diatomic gas $(O_2)$ at room temperature,whereas other members of the group are polyatomic solids.
$2$. Oxygen exhibits an oxidation state of $-2$ in most compounds,except in $OF_2$ where it shows $+2$. Other members of the group show $+4$ and $+6$ oxidation states in addition to $-2$.
$3$. $H_2O$ is a liquid at room temperature due to intermolecular hydrogen bonding,while the hydrides of other elements do not exhibit such strong hydrogen bonding.
$4$. The covalency of oxygen is limited to a maximum of $2$ due to the absence of $d$-orbitals in its valence shell,whereas other elements in the group can expand their covalency beyond $4$.

(N/A) The electron gain enthalpy values for these groups are highly negative. Excluding $F$ and $O$,the negative value of electron gain enthalpy decreases gradually as we move down the group.

(D) The $-O-O-$ linkage is known as a peroxy linkage.
Among the given options,$H_{2}S_{2}O_{8}$ (peroxodisulphuric acid) contains a peroxy linkage.
The structure of $H_{2}S_{2}O_{8}$ is $HO-SO_{2}-O-O-SO_{2}-OH$,which clearly shows the presence of the $-O-O-$ bond.
Therefore,the correct option is $D$.

(A) Oleum $(H_2S_2O_7)$,also known as fuming sulfuric acid,is prepared by the absorption of $SO_3$ in concentrated $H_2SO_4$.
The reaction is: $H_2SO_4 + SO_3 \rightarrow H_2S_2O_7$.
It contains an $S-O-S$ linkage,not an $O-O$ linkage.
It has two $OH$ groups in its structure.
Therefore,the correct statement is that it is prepared by the absorption of $SO_3$ in concentrated $H_2SO_4$.

(B) Sulphur exists in various allotropic forms like $\alpha$-sulphur (rhombic) and $\beta$-sulphur (monoclinic),which consist of $S_8$ puckered ring structures. These forms are diamagnetic because all electrons are paired in the $S_8$ molecules.
The $S_2$ form of sulphur is analogous to $O_2$ and is paramagnetic in the vapour state at high temperatures due to the presence of two unpaired electrons in its antibonding $\pi^*$ molecular orbitals.
Therefore,only the $S_2$ form shows paramagnetism.
The number of such allotropic forms is $1$.

(C) The transition temperature for sulphur is $369 \ K$. Below $369 \ K$,$\alpha-$sulphur (rhombic) is stable,and above $369 \ K$,$\beta-$sulphur (monoclinic) is stable.
These two forms can change reversibly between themselves with slow heating or slow cooling,making Statement-$I$ true.
At room temperature,$\alpha-$sulphur (rhombic) is the most stable form,not monoclinic sulphur. Therefore,Statement-$II$ is false.

(B) Group $16$ elements are known as chalcogens.
The members of this group are Oxygen $(O)$,Sulfur $(S)$,Selenium $(Se)$,Tellurium $(Te)$,and Polonium $(Po)$.
Therefore,the correct set is $Se, Te$ and $Po$.

(D) $1$. Sulphurous acid $(H_2SO_3)$: The structure contains $1$ $S=O$ bond.
$2$. Peroxodisulphuric acid $(H_2S_2O_8)$: The structure contains $4$ $S=O$ bonds.
$3$. Pyrosulphuric acid $(H_2S_2O_7)$: The structure contains $4$ $S=O$ bonds.
Therefore,the number of $S=O$ bonds are $1, 4$ and $4$ respectively.

(A) The boiling points of group $16$ hydrides are influenced by both molecular mass and intermolecular hydrogen bonding.
For $H_2S$,$H_2Se$,and $H_2Te$,the boiling point increases with increasing molar mass due to stronger van der Waals forces.
However,$H_2O$ exhibits strong intermolecular hydrogen bonding,which results in a significantly higher boiling point than the other hydrides in the group.
The correct order of boiling points is $H_2S < H_2Se < H_2Te < H_2O$.
Statement $I$ is incorrect because it places $H_2O$ at the beginning of the sequence.
Statement $II$ is incorrect because it suggests a simple trend based on molar mass,ignoring the dominant effect of hydrogen bonding in $H_2O$.

(A) The melting points of group $16$ hydrides $(H_{2}O, H_{2}S, H_{2}Se, H_{2}Te)$ depend on the intermolecular forces of attraction.
$H_{2}O$ has the highest melting point due to strong intermolecular hydrogen bonding.
For the remaining hydrides $(H_{2}S, H_{2}Se, H_{2}Te)$,the melting point increases down the group due to an increase in molecular mass and the resulting increase in van der Waals forces.
Therefore,the correct order is $H_{2}S < H_{2}Se < H_{2}Te < H_{2}O$.

(A) To determine the number of oxoacids with a peroxo $(O-O)$ bond,we examine their structures:
$1$. $H_2SO_3$ (Sulphurous acid): Contains only $S=O$ and $S-OH$ bonds. No peroxo bond.
$2$. $H_2SO_4$ (Sulphuric acid): Contains $S=O$ and $S-OH$ bonds. No peroxo bond.
$3$. $H_2S_2O_8$ (Peroxodisulphuric acid): Contains a central $O-O$ peroxo linkage between two sulphur atoms.
$4$. $H_2S_2O_7$ (Pyrosulphuric acid): Contains an $S-O-S$ linkage. No peroxo bond.
Thus,only $H_2S_2O_8$ contains a peroxo bond.
The total count is $1$.

(A) The thermal stability of hydrides depends on the bond dissociation enthalpy.
As we move down the group-$16$ from $O$ to $Te$,the atomic size of the central atom increases.
This leads to an increase in the bond length of the $E-H$ bond (where $E = O, S, Se, Te$).
As the bond length increases,the bond dissociation enthalpy decreases,making the bond weaker.
Therefore,the thermal stability decreases down the group.
The correct order of thermal stability is $H_2O > H_2S > H_2Se > H_2Te$ or $H_2Te < H_2Se < H_2S < H_2O$.

(D) The reaction of acidified potassium dichromate with hydrogen peroxide produces chromium pentoxide $(CrO_5)$.
$CrO_5$ is unstable in water but is stabilized in amyl alcohol,where it exhibits a characteristic deep blue colour.
Therefore,option $(D)$ is correct.

(A) As we move down the group $16$ from $O$ to $Te$,the atomic size of the central element $E$ increases.
Due to the increase in atomic size,the bond length of the $E-H$ bond increases.
As the bond length increases,the bond strength decreases,which leads to a decrease in the bond dissociation energy.
Therefore,the order of bond dissociation energy is $H_2O > H_2S > H_2Se > H_2Te$.

(C) The structures of the given oxoacids of sulphur are as follows:
$A$. Peroxodisulphuric acid $(H_2S_2O_8)$: Contains two $S-OH$ groups,four $S=O$ bonds,and one peroxide linkage $(S-O-O-S)$. This matches $III$.
$B$. Sulphuric acid $(H_2SO_4)$: Contains two $S-OH$ groups and two $S=O$ bonds. This matches $IV$.
$C$. Pyrosulphuric acid $(H_2S_2O_7)$: Contains two $S-OH$ groups,four $S=O$ bonds,and one $S-O-S$ linkage. This matches $I$.
$D$. Sulphurous acid $(H_2SO_3)$: Contains two $S-OH$ groups and one $S=O$ bond. This matches $II$.
Therefore,the correct matching is $A-III, B-IV, C-I, D-II$.

(C) Statement-$I$: Oxygen exhibits oxidation states ranging from $-2$ to $+2$ (e.g.,in $OF_2$,it is $+2$). Thus,Statement-$I$ is incorrect.
Statement-$II$: Due to the inert pair effect,as we move down group $16$,the stability of the $+6$ oxidation state decreases,while the stability of the $+4$ oxidation state increases. Thus,Statement-$II$ is incorrect.
Therefore,both statements are incorrect.

(C) The anomalous behaviour of oxygen compared to other members of group $16$ is primarily due to its small atomic size,high electronegativity,and the absence of $d$-orbitals in its valence shell.

(B) The acidity of hydrides of group $16$ elements increases down the group because the bond dissociation enthalpy decreases as the size of the central atom increases.
Since the bond dissociation enthalpy of $H_2Te$ is lower than that of $H_2S$,the $H-Te$ bond is weaker than the $H-S$ bond.
Therefore,$H_2Te$ releases $H^+$ ions more easily than $H_2S$,making it more acidic.
Thus,both $(A)$ and $(R)$ are true,and $(R)$ is the correct explanation of $(A).$

(D) Correct: Group $16$ elements form $EO_2$ and $EO_3$ type oxides which are acidic.
$(B)$ Correct: $SO_2$ acts as a reducing agent,whereas $TeO_2$ acts as an oxidising agent.
$(C)$ Incorrect: The reducing property increases from $H_2S$ to $H_2Te$ down the group due to the decrease in bond dissociation enthalpy.
$(D)$ Incorrect: The ozone molecule $(O_3)$ has a bent structure with resonance. The total number of lone pairs is $6$ (terminal oxygen atoms have $2$ and $3$ lone pairs respectively,and the central oxygen has $1$ lone pair).
Therefore,statements $A$ and $B$ are correct.

(A) Oxygen has a small atomic size and high electronegativity,which allows it to form stable $p \pi-p \pi$ multiple bonds with itself $(O_2)$ and other small atoms like $C$ and $N$.
Sulphur has a larger atomic size,making $p \pi-p \pi$ overlap ineffective; therefore,it prefers to form single bonds and exists as $S_8$ puckered rings.
Thus,Assertion $(A)$ is correct because $S_8$ is the stable form of sulphur.
Reason $(R)$ is also correct as it explains why oxygen exists as a diatomic molecule $(O_2)$ while sulphur exists as a polyatomic molecule $(S_8)$.