Oxygen family Questions in English

(C) $H_2S_2O_8$ (Peroxodisulfuric acid or Marshall's acid) contains a peroxide linkage ($O - O$ bond).
Its structure is $HO - SO_2 - O - O - SO_2 - OH$.

(B) The acidity of hydrides of group $16$ elements increases down the group because the bond dissociation enthalpy of the $X-H$ bond decreases from top to bottom.
Since the $O-H$ bond is stronger than the $S-H$ bond,$H_2O$ is less acidic than $H_2S$.
The order of acidity is $H_2O < H_2S < H_2Se < H_2Te$.

(B) The preparation of ozone from oxygen is an endothermic process.
$3O_2(g) \rightleftharpoons 2O_3(g)$; $\Delta H = +142 \ kJ/mol$.
Since the enthalpy change is positive,energy is absorbed during this reaction.

(D) . The oxide ion $(O^{2-})$ is the most common dinegative anion because oxygen has a high electronegativity and a small size,allowing it to easily gain two electrons to complete its octet. Unlike the other elements in the group,oxygen lacks vacant $d$-orbitals,but its high effective nuclear charge makes the formation of $O^{2-}$ energetically favorable in many ionic compounds.

(A) Chalcogens belong to Group $16$ of the periodic table,also known as the oxygen family.
These elements have $6$ valence electrons in their outermost shell.
The general valence shell electronic configuration for Group $16$ elements is $ns^2 np^4$.

(B) Sodium thiosulphate $(Na_2S_2O_3)$ is prepared by boiling a solution of sodium sulphite $(Na_2SO_3)$ with powdered sulphur $(S)$ in an alkaline medium.
The chemical reaction is: $Na_2SO_3 + S \xrightarrow{\Delta} Na_2S_2O_3$.

(C) The correct answer is $(C)$ Sulphur.
Sulphur exhibits several allotropic forms,the most common being:
$(1)$ Rhombic sulphur ($\alpha$-sulphur)
$(2)$ Monoclinic sulphur ($\beta$-sulphur)
$(3)$ Plastic sulphur

(B) The tendency towards catenation depends on the strength of the element-element bond.
As the atomic size increases down the group from $S$ to $Te$,the bond dissociation energy decreases.
Although $O$ is smaller than $S$,the $O-O$ bond energy is lower than the $S-S$ bond energy due to significant inter-electronic repulsions between the lone pairs of the small oxygen atoms.
Therefore,$S$ exhibits the highest tendency towards catenation among the group $16$ elements.

(B) Ozone $(O_3)$ is an allotrope of oxygen. Oxygen belongs to Group $16$ of the periodic table,which is also known as the $VI$ group (or chalcogens). Therefore,ozone belongs to the $VI$ group.

(A) The abundance of elements in the earth's crust is as follows:
$1$. Oxygen $(O)$: $\approx 46.6 \, \%$
$2$. Silicon $(Si)$: $\approx 27.7 \, \%$
$3$. Aluminum $(Al)$: $\approx 8.1 \, \%$
$4$. Iron $(Fe)$: $\approx 5.0 \, \%$
Since oxygen has the highest percentage by mass in the earth's crust,it is the most abundant element.

(C) The correct answer is $(C)$.
$SO_2$ is produced by the reaction of sulfites with dilute acids.
The chemical reaction is:
$Na_2SO_3(s) + H_2SO_4(aq) \to Na_2SO_4(aq) + H_2O(l) + SO_2(g)$

(C) Fractional distillation of liquid air is based on the difference in the boiling points $(b.p.)$ of its components.
Oxygen has a higher boiling point $(90 \ K)$ compared to Nitrogen $(77 \ K)$.
Therefore,Nitrogen boils off first,leaving behind liquid Oxygen.
Hence,Option $C$ is the correct answer.

(A) When oxygen gas is passed through an aqueous solution of sodium sulfite $(Na_2SO_3)$,it undergoes oxidation to form sodium sulfate $(Na_2SO_4)$.
The balanced chemical equation for this reaction is:
$2Na_2SO_3 + O_2 \to 2Na_2SO_4$

(A) Allotropes are different structural modifications of an element where the atoms of the same element are bonded together in different ways.
For some elements,allotropes have different molecular formulae which can persist in different phases.
For example,two allotropes of oxygen,dioxygen $(O_2)$ and ozone $(O_3)$,can both exist in solid,liquid,and gaseous states.

(C) The correct option is $C$.
Ozone $(O_3)$ is prepared by passing silent electric discharge through pure,dry oxygen.
The reaction is endothermic and reversible:
$3O_2 \rightleftharpoons 2O_3$ $(\Delta H = +142 \ kJ/mol)$.

(D) Ozone protects the earth's inhabitants by absorbing $UV$ radiations,not $\gamma$ radiations.
Ozone $(O_3)$ is diamagnetic,whereas dioxygen $(O_2)$ is paramagnetic due to the presence of two unpaired electrons in its antibonding molecular orbitals.
Ozone is thermodynamically unstable and decomposes to give atomic oxygen,making it more reactive than stable dioxygen.
Ozone is formed in the upper atmosphere by the action of $UV$ radiation on dioxygen.
Therefore,the statement in option $D$ is incorrect.

(D) Ozone $(O_3)$ decomposes readily to give nascent oxygen $([O])$.
Due to the release of nascent oxygen,it acts as a powerful oxidizing agent and a bleaching agent.
It oxidises lead sulphide $(PbS)$ to lead sulphate $(PbSO_4)$,potassium iodide $(KI)$ to iodine $(I_2)$,and mercury $(Hg)$ to mercurous oxide $(Hg_2O)$.
Therefore,the statement that it cannot act as a bleaching agent is incorrect.

(B) The reaction of ozone $(O_3)$ with potassium iodide $(KI)$ solution is an oxidation-reduction reaction.
Ozone acts as a strong oxidizing agent and liberates iodine $(I_2)$ from the potassium iodide solution.
The chemical reaction is as follows:
$O_3 \to O_2 + [O]$
$2KI + H_2O + [O] \to 2KOH + I_2$
Combining these,the overall reaction is:
$2KI + H_2O + O_3 \to 2KOH + I_2 + O_2$
Thus,$I_2$ is produced.

(B) Ozone $(O_3)$ is a strong oxidizing agent. It reacts with tetramethyl base paper (often referred to as tetramethyl base or tetramethyl benzidine paper) to produce a violet color,which is a standard test for the detection of ozone.

(A) When copper $(Cu)$ is heated with concentrated sulphuric acid $(H_2SO_4)$,it acts as an oxidizing agent and oxidizes copper to copper$(II)$ sulphate,while itself being reduced to sulphur dioxide $(SO_2)$.
The balanced chemical equation is:
$Cu(s) + 2H_2SO_4(conc.) \to CuSO_4(aq) + 2H_2O(l) + SO_2(g)$

(B) Sulphur dioxide $(SO_2)$ is an acidic oxide.
It reacts with a strong base like $KOH$ to form potassium sulphite and water.
The reaction is: $2KOH + SO_2 \rightarrow K_2SO_3 + H_2O$.
Therefore,$KOH$ solution is used to absorb sulphur dioxide.

(D) When $SO_2$ gas is passed through an acidified $K_2Cr_2O_7$ solution,it acts as a reducing agent and reduces $Cr(VI)$ to $Cr(III)$.
The balanced chemical equation is:
$K_2Cr_2O_7 + H_2SO_4 + 3SO_2 \to K_2SO_4 + Cr_2(SO_4)_3 + H_2O$
In this reaction,the orange-colored $K_2Cr_2O_7$ solution turns green due to the formation of chromium$(III)$ sulfate,$Cr_2(SO_4)_3$.

(C) When $SO_2$ is passed through a solution of cupric chloride $(CuCl_2)$,it acts as a reducing agent.
The reaction is: $2CuCl_2 + SO_2 + 2H_2O \rightarrow Cu_2Cl_2 + H_2SO_4 + 2HCl$.
In this reaction,$Cu^{2+}$ is reduced to $Cu^+$,resulting in the formation of a white precipitate of cuprous chloride $(Cu_2Cl_2)$.
As the blue $Cu^{2+}$ ions are converted to $Cu^+$,the solution loses its blue colour and becomes colourless.

(A) $SO_2$ acts as a bleaching agent due to its reducing property.
In the presence of moisture,$SO_2$ reacts to produce nascent hydrogen: $2H_2O + SO_2 \to H_2SO_4 + 2[H]$.
This nascent hydrogen reduces the coloured matter to a colourless substance: $\text{Coloured matter} + 2[H] \to \text{Colourless matter}$.

(A) $SO_3$ is an acidic oxide because it reacts with water to form a strong acid,sulfuric acid.
The reaction is: $H_2O + SO_3 \rightarrow H_2SO_4$.

(D) In the contact process for the manufacture of $H_2SO_4$,$SO_3$ gas is absorbed in $98\% \ H_2SO_4$ to form oleum $(H_2S_2O_7)$.
This oleum is then diluted with water to obtain $H_2SO_4$ of the desired concentration.
Therefore,the final acid obtained directly from the absorption tower is oleum,which is $H_2S_2O_7$.

(B) Sulfuric acid,with the chemical formula $H_2SO_4$,is historically known as "oil of vitriol".
This name was given by medieval European chemists because it was prepared by the distillation of "green vitriol" (iron$(II)$ sulfate,$FeSO_4 \cdot 7H_2O$) in an iron container.

(D) The dissolution of sulfur trioxide $(SO_3)$ in concentrated sulfuric acid $(H_2SO_4)$ produces oleum,also known as fuming sulfuric acid.
The chemical reaction is: $H_2SO_4 + SO_3 \to H_2S_2O_7$ (Oleum).

(C) $Oleum$ is also known as fuming $H_2SO_4$. It is prepared by dissolving $SO_3$ in $H_2SO_4$. Its chemical formula is $H_2S_2O_7$.

(D) . In $S_2O_7^{2-}$ (pyrosulfate ion),the two sulfur atoms are linked via an oxygen atom ($S-O-S$ linkage). There is no direct $S-S$ bond.
The structures are as follows:
$1$. $S_2O_4^{2-}$: $O_2S-SO_2^{2-}$ (contains $S-S$ bond)
$2$. $S_2O_5^{2-}$: $O_2S-SO_3^{2-}$ (contains $S-S$ bond)
$3$. $S_2O_3^{2-}$: $S-SO_3^{2-}$ (contains $S-S$ bond)
$4$. $S_2O_7^{2-}$: $O_3S-O-SO_3^{2-}$ (contains $S-O-S$ bond,no $S-S$ bond)

(D) When sulphur is boiled with an aqueous solution of sodium sulphite $(Na_2SO_3)$,it undergoes a reaction to form sodium thiosulphate $(Na_2S_2O_3)$.
The chemical equation for this reaction is:
$Na_2SO_3 + S \to Na_2S_2O_3$

(A) The group $16$ hydrides $(H_2O, H_2S, H_2Se, H_2Te)$ show a trend in boiling points.
$H_2O$ has an exceptionally high boiling point compared to the other hydrides of the group.
This is due to the presence of extensive intermolecular hydrogen bonding in $H_2O$ molecules,which is not possible in other hydrides like $H_2S$ due to the lower electronegativity of the central atom.
Therefore,$H_2O$ molecules are associated,leading to higher energy requirements for vaporization.

(C) The acidic strength of hydrides of group $16$ elements increases down the group as the bond dissociation enthalpy decreases.
As we move from $O$ to $Te$ in the group,the atomic size increases,which leads to a decrease in the bond strength of the $E-H$ bond.
Therefore,the bond between $H$ and $Te$ in ${H_2}Te$ is the weakest,making it the easiest to dissociate and release ${H^{+}}$ ions.
Thus,the correct order of acidic strength is ${H_2}O < {H_2}S < {H_2}Se < {H_2}Te$.

(B) Oxygen differs from the rest of the members of Group-$16$ (chalcogens) because it lacks $d$-orbitals in its valence shell. Due to this,it cannot exhibit higher oxidation states like $+4$ and $+6$,which are characteristic of other members like sulfur,selenium,and tellurium. Therefore,the property in which oxygen differs is the ability to show oxidation states of $+2, +4, +6$.

(B) In the contact process for the industrial preparation of sulphuric acid,$SO_2$ is oxidized to $SO_3$ in the presence of a catalyst,vanadium pentoxide $(V_2O_5)$.
The chemical reaction is: $2SO_2(g) + O_2(g) \xrightarrow{V_2O_5} 2SO_3(g)$.

(C) In the contact process for the manufacture of $H_2SO_4$,the oxidation of $SO_2$ to $SO_3$ is the key step.
This reaction is catalyzed by vanadium pentoxide,$V_2O_5$.

(D) Polymorphism (or allotropy) is the property of an element to exist in more than one crystalline form.
$O$ exists as $O_2$ and $O_3$.
$S$ exists in various allotropic forms like rhombic and monoclinic sulfur.
$Se$ also exists in several allotropic forms (red,grey,and black selenium).
Therefore,all these elements exhibit polymorphism.

(D) All the elements of the oxygen family (Group $16$) exhibit the property of allotropy,meaning they exist in more than one form,which is known as being polymorphic.

(D) The triatomic species of elemental oxygen is known as ozone,which has the chemical formula $O_3$.

(D) When $H_2S$ gas is passed through concentrated nitric acid,it acts as an oxidizing agent and oxidizes $H_2S$ to elemental sulfur $(S)$.
The chemical reaction is as follows:
$H_2S + 2HNO_3 \to 2NO_2 + S + 2H_2O$
The sulfur produced in this reaction is typically in the form of a yellow precipitate,which is not specifically identified as rhombic,prismatic,or amorphous in the context of this general chemical equation.
Therefore,the correct option is $D$.

(D) The $16^{th}$ group elements are known as chalcogens (oxygen family).
$O$,$S$,and $Se$ belong to the $16^{th}$ group.
$Na$ belongs to the $1^{st}$ group and is known as an alkali metal.
Therefore,$Na$ is not a chalcogen.

(D) In the contact process for the manufacture of $H_2SO_4$,the $SO_2$ gas obtained from burning sulfur or roasting sulfide ores contains impurities like dust particles.
These impurities must be removed to prevent the poisoning of the catalyst $(V_2O_5)$.
The Tyndall box is specifically used to detect the presence of these dust particles in the gas stream by observing the scattering of light (Tyndall effect).

(B) Permono sulphuric acid $(H_2SO_5)$ is commonly known as Caro's acid.
Marshall's acid is peroxodisulphuric acid $(H_2S_2O_8)$.

(B) In the presence of moisture,$SO_2$ acts as a reducing agent because it gets oxidized to $SO_4^{2-}$ ions.
$SO_2 + 2H_2O \rightarrow SO_4^{2-} + 4H^+ + 2e^-$.
Since it releases electrons in this process,it acts as a reductant and undergoes the loss of electrons.

(B) $SO_3$ is not absorbed directly in water because it forms a dense fog of $H_2SO_4$ droplets which is difficult to condense.
To avoid this,$SO_3$ is absorbed in $98\% \; H_2SO_4$ to form oleum $(H_2S_2O_7)$,which is then diluted with water to obtain $H_2SO_4$ of the desired concentration.
Therefore,$98\% \; H_2SO_4$ is the most efficient agent for the absorption of $SO_3$.

(D) $SO_2$ acts as a bleaching agent due to its reducing property.
$SO_2 + 2H_2O \to H_2SO_4 + 2[H]$
$\text{Coloured matter} + 2[H] \to \text{Colourless matter}$
This reduction process removes the oxygen from the coloured substance,making it colourless.

(B) Ozone $(O_3)$ acts as a strong oxidizing agent.
When ozone reacts with $H_2O_2$,$Hg$,and $KI$,it releases oxygen gas $(O_2)$.
For example:
$O_3 + H_2O_2 \to H_2O + 2O_2$
$Hg + O_3 \to HgO + O_2$
$2KI + H_2O + O_3 \to 2KOH + I_2 + O_2$
However,when ozone reacts with $SO_2$,it oxidizes $SO_2$ to $SO_3$ without evolving oxygen gas:
$3SO_2 + O_3 \to 3SO_3$

(A) In the contact process,$SO_3$ gas is absorbed in concentrated $H_2SO_4$ $(98\%)$ to form oleum $(H_2S_2O_7)$.
This is because $SO_3$ reacts directly with water to form a dense fog of $H_2SO_4$ droplets,which is difficult to condense.
Therefore,$98\% \ H_2SO_4$ is used as the absorbing agent to produce oleum,which is then diluted with water to get $H_2SO_4$ of the desired concentration.
The reaction is: $H_2SO_4 + SO_3 \to H_2S_2O_7$.

(C) When $SO_2$ reacts with $H_2S$ in an acidic medium,$SO_2$ acts as an oxidizing agent and $H_2S$ acts as a reducing agent.
The chemical reaction is: $2H_2S + SO_2 \to 2H_2O + 3S$.
As a result,sulphur is precipitated.

(D) . The colour of liquid $O_2$ is pale blue.