Oxygen family Questions in English

(B) Ozone $(O_3)$ reacts with potassium iodide $(KI)$ solution in the presence of water to liberate iodine $(I_2)$:
$2KI + H_2O + O_3 \rightarrow 2KOH + I_2 + O_2$
The liberated iodine $(I_2)$ reacts with starch to form a blue-colored complex,which is a standard test for the presence of ozone.

(D) The "Springs reaction" is a method for the preparation of sodium thiosulphate $(Na_2S_2O_3)$.
In this reaction,sodium sulphide $(Na_2S)$ and sodium sulphite $(Na_2SO_3)$ react with iodine $(I_2)$ to produce sodium thiosulphate and sodium iodide $(NaI)$.
The balanced chemical equation is: $Na_2S + Na_2SO_3 + I_2 \to Na_2S_2O_3 + 2NaI$.
Therefore,the correct option is $(D)$.

(D) The structure of rhombic sulphur is molecular in nature.
It consists of $S_8$ molecules,which adopt a puckered $8-$ membered ring structure (crown shape).
The $S-S$ bond distance is $2.12 \ \mathring{A}$.

(D) In the lead chamber process for the manufacture of $H_2SO_4$,the oxidation of $SO_2$ to $SO_3$ is catalyzed by oxides of nitrogen ($NO$ and $NO_2$).
The reaction is: $2SO_2 + O_2 \xrightarrow{NO} 2SO_3$.

(C) The contact process involves the catalytic oxidation of sulphur dioxide to sulphur trioxide.
The catalyst used in this process is $V_2O_5$ (Vanadium pentoxide).
The chemical reaction is: $2SO_{2(g)} + O_{2(g)} \xrightarrow{V_2O_5} 2SO_{3(g)}$.

(B) The test for ozone $O_3$ is performed using mercury $(Hg)$.
When ozone reacts with mercury,it forms mercurous oxide $(Hg_2O)$.
The reaction is: $6Hg + O_3 \to 3Hg_2O$.
During this reaction,mercury loses its meniscus and starts sticking to the glass surface,which is a characteristic test for ozone.

(A) In the contact process for the industrial manufacture of $H_2SO_4$,vanadium pentoxide $(V_2O_5)$ is used as a catalyst for the oxidation of $SO_2$ to $SO_3$.

(C) When red hot iron reacts with sulfur dioxide,it undergoes a redox reaction to form iron$(II)$ oxide and iron$(II)$ sulfide.
The balanced chemical equation is:
$3Fe + SO_2 \to 2FeO + FeS$

(C) In xerox machines,the photoconductor drum is typically coated with a thin layer of $Selenium$ $(Se)$. $Selenium$ is a semiconductor that exhibits photoconductivity,meaning its electrical conductivity increases when exposed to light. This property is essential for the electrostatic imaging process used in xerocopying.

(D) $1$. Ozone $(O_3)$ is formed in the upper atmosphere by the action of $UV$ radiation on dioxygen $(O_2)$.
$2$. Ozone is thermodynamically unstable with respect to oxygen,making it more reactive.
$3$. Ozone $(O_3)$ is diamagnetic (all electrons are paired),whereas dioxygen $(O_2)$ is paramagnetic (contains two unpaired electrons in antibonding molecular orbitals).
$4$. Ozone layer absorbs harmful $UV$ radiation,not gamma radiation. Therefore,the statement regarding gamma radiation is incorrect.

(A) Sulfur exhibits allotropy. At room temperature,$Rhombic$ $sulfur$ (also known as $\alpha-sulfur$) is the most stable crystalline form. Monoclinic sulfur $(\beta-sulfur)$ is stable only above $369 \ K$.

(C) The structures of the given oxoacids are as follows:
$1$. $H_2S_2O_8$ (Peroxodisulfuric acid): Contains an $S-O-O-S$ linkage.
$2$. $H_2S_2O_7$ (Pyrosulfuric acid): Contains an $S-O-S$ linkage.
$3$. $H_2S_2O_3$ (Thiosulfuric acid): Contains an $S-S$ bond.
$4$. $H_2S_2O_6$ (Dithionic acid): Contains an $S-S$ bond.
Note: Both $H_2S_2O_3$ and $H_2S_2O_6$ contain $S-S$ bonds. However,in standard competitive chemistry questions of this type,$H_2S_2O_3$ is the most common example cited for the presence of an $S-S$ bond. Given the options,$H_2S_2O_3$ is the primary answer.

(A) To determine the number of oxo groups,we look at the structures of the given acids:
$1$. $H_2SO_4$ (Sulfuric acid): The central $S$ atom is bonded to two $OH$ groups and two terminal $O$ atoms via double bonds. Thus,it has $2$ oxo groups.
$2$. $H_2SO_3$ (Sulfurous acid): The central $S$ atom is bonded to two $OH$ groups and one terminal $O$ atom via a double bond. Thus,it has $1$ oxo group.
$3$. $H_3PO_3$ (Phosphorous acid): The central $P$ atom is bonded to two $OH$ groups,one $H$ atom,and one terminal $O$ atom via a double bond. Thus,it has $1$ oxo group.
$4$. $H_3PO_4$ (Phosphoric acid): The central $P$ atom is bonded to three $OH$ groups and one terminal $O$ atom via a double bond. Thus,it has $1$ oxo group.
Comparing the number of oxo groups,$H_2SO_4$ has the maximum number $(2)$.

(C) The hydrolysis of peroxodisulphuric acid $(H_2S_2O_8)$ proceeds in two steps:
$1$. $H_2S_2O_8 + H_2O \rightarrow H_2SO_4 + H_2SO_5$
$2$. $H_2SO_5 + H_2O \rightarrow H_2SO_4 + H_2O_2$
However,for one mole of $H_2S_2O_8$,the initial hydrolysis reaction yields one mole of sulphuric acid $(H_2SO_4)$ and one mole of peroxomonosulphuric acid $(H_2SO_5)$.

(A) When $SO_2$ is passed through an aqueous solution of sodium hydroxide $(NaOH)$,it reacts to form sodium bisulphite $(NaHSO_3)$.
The balanced chemical equation is: $NaOH(aq) + SO_2(g) \to NaHSO_3(aq)$.

(B) In the contact process,the oxidation of sulfur dioxide $(SO_2)$ to sulfur trioxide $(SO_3)$ is a reversible exothermic reaction:
$2SO_2(g) + O_2(g) \rightleftharpoons 2SO_3(g)$.
This reaction is carried out in the presence of vanadium pentoxide $(V_2O_5)$ as a catalyst to increase the rate of reaction.

(D) The oxidation of $SO_2$ to $SO_3$ can occur through various oxidizing agents:
$1$. $2SO_2 + O_2 \xrightarrow{\text{catalyst}} 2SO_3$
$2$. $SO_2 + O_3 \rightarrow SO_3 + O_2$
$3$. $SO_2 + H_2O_2 \rightarrow SO_3 + H_2O$
Since all three reagents can facilitate this conversion,the correct option is $D$.

(C) Caro's acid is $H_2SO_5$.
Its chemical structure is $HO-SO_2-O-OH$,which contains a peroxy linkage $(-O-O-)$.

(A) The tetrathionate ion $(S_4O_6^{2-})$ has a structure with an $S-S-S-S$ chain,thus it contains $S-S$ bonds.
The thiosulfate ion $(S_2O_3^{2-})$ has a structure where one sulfur atom is bonded to another sulfur atom via a double bond ($S=S$ bond),thus it contains an $S-S$ bond.
The disulfate ion $(S_2O_7^{2-})$ has an $S-O-S$ linkage and no $S-S$ bond.
The peroxodisulfate ion $(S_2O_8^{2-})$ has an $S-O-O-S$ linkage and no $S-S$ bond.
Therefore,both $S_4O_6^{2-}$ and $S_2O_3^{2-}$ contain $S-S$ bonds.

(C) $OF_2$ is a fluoride of oxygen because the electronegativity of fluorine $(3.98)$ is greater than that of oxygen $(3.44)$.
Therefore,$OF_2$ is named oxygen difluoride,not an oxide of fluorine.
$O_3$ is bent,$ONF$ ($8+7+9 = 24$ electrons) is isoelectronic with $O_2N^-$ ($16+7+1 = 24$ electrons),and $Cl_2O_7$ is the anhydride of $HClO_4$ $(2HClO_4 \rightarrow Cl_2O_7 + H_2O)$.

(B) $SO_2$ is used as a food preservative because it acts as an antioxidant and antimicrobial agent.
$NO_2$ is not used as a food preservative.
Both $NO_2$ and $SO_2$ are soluble in water,form acid rain,and can act as reducing agents.

(A) The acidic strength of hydrides increases as the size of the central atom increases,which weakens the $M-H$ bond.
Since the atomic size increases from $S$ to $Se$ to $Te$,the bond dissociation enthalpy decreases,leading to an increase in acidic strength.
The order is $H_2S < H_2Se < H_2Te$.
Acidic nature $\propto \frac{1}{\text{Bond dissociation enthalpy}}$.

(B) The thermal decomposition of ferric sulphate is a standard laboratory method for the preparation of sulphur trioxide:
$Fe_2(SO_4)_3 \xrightarrow{\Delta} Fe_2O_3 + 3SO_3$
Analysis of other options:
$(A)$ $CaSO_4 + C \xrightarrow{\Delta} CaO + SO_2 + CO$ (Produces $SO_2$)
$(B)$ $Fe_2(SO_4)_3 \xrightarrow{\Delta} Fe_2O_3 + 3SO_3$ (Produces $SO_3$)
$(C)$ $S + 2H_2SO_4 \xrightarrow{\Delta} 3SO_2 + 2H_2O$ (Produces $SO_2$)
$(D)$ $H_2SO_4 + PCl_5 \xrightarrow{\Delta} SO_3HCl + POCl_3 + HCl$ (Produces chlorosulphonic acid)
Thus,$SO_3$ is obtained by heating $Fe_2(SO_4)_3$.

(B) The acidic strength of hydrides of group $16$ elements increases down the group as the bond dissociation energy decreases: $H_2O < H_2S < H_2Se < H_2Te$.
Since $pK_a = -\log(K_a)$,a higher acidic strength corresponds to a lower $pK_a$ value.
Therefore,the correct order for $pK_a$ values should be $H_2O > H_2S > H_2Se > H_2Te$.
Option $B$ shows an increasing order of $pK_a$ values,which is incorrect.

(A) Sulfur dioxide $(SO_2)$ acts as a strong reducing agent.
When $SO_2$ gas is passed through an acidified solution of potassium permanganate $(KMnO_4)$,it reduces the purple-colored $Mn^{7+}$ ion to the colorless $Mn^{2+}$ ion.
The balanced chemical equation is:
$2KMnO_4 + 5SO_2 + 2H_2O \rightarrow K_2SO_4 + 2MnSO_4 + 2H_2SO_4$
Since $MnSO_4$ is a colorless solution,the purple color of $KMnO_4$ disappears.

(A) $ONCl$ has $8 + 7 + 17 = 32$ electrons,while $ONO^{-}$ $(NO_2^-)$ has $7 + 8 + 8 + 1 = 24$ electrons. Therefore,they are not isoelectronic.
$O_3$ has a bent structure due to the presence of a lone pair on the central oxygen atom.
Ozone is indeed violet-black in the solid state.
Ozone is diamagnetic as all electrons are paired.
Thus,the statement in option $A$ is incorrect.

(D) The oxidation state of sulphur ranges from $-2$ to $+6$ in its various compounds. For example,in $H_2S$,the oxidation state of sulphur is $-2$. Therefore,the statement that the oxidation state of sulphur is never less than $+4$ is incorrect.

(B) The acidic nature of oxides of non-metals decreases down the group as the metallic character increases.
In group $16$,the order of elements is $S > Se > Te$.
Therefore,the acidic strength of their dioxides follows the order: $SO_2 > SeO_2 > TeO_2$.

(D) $1$. $MnO_2$ acts as a catalyst for the decomposition of $KClO_3$ to $O_2$ is a correct statement.
$2$. The conversion of $SO_2$ to $SO_3$ in the contact process is an exothermic reaction,which is a correct statement.
$3$. White phosphorus is indeed poisonous and has a characteristic garlic-like odor,which is a correct statement.
$4$. $\gamma-SO_3$ exists as a cyclic trimer $(SO_3)_3$ in the solid state,not a chain structure. Therefore,this statement is incorrect.

(C) The hydrolysis of $SO_2Cl_2$ (sulfuryl chloride) is given by the reaction: $SO_2Cl_2 + 2H_2O \rightarrow H_2SO_4 + 2HCl$.
The product $H_2SO_4$ (sulfuric acid) is an $oxy$-$acid$ with the structure $(HO)_2SO_2$.
It has two $OH$ groups attached to the central sulfur atom,which can release two $H^+$ ions in aqueous solution,thus its basicity is $2$.

(B) The pale yellow solid molecule $(A)$ with a crown shape is sulfur $(S_8)$.
When $S_8$ is heated with concentrated $H_2SO_4$,it undergoes oxidation to produce sulfur dioxide gas $(SO_2)$:
$S_8 + 16H_2SO_4 \rightarrow 8SO_2 + 8SO_2 + 16H_2O$ (simplified as $S + 2H_2SO_4 \rightarrow 3SO_2 + 2H_2O$).
$SO_2$ is a colorless gas with a suffocating smell.
When $SO_2$ is passed through starch iodate paper,it reduces iodate $(IO_3^-)$ to iodine $(I_2)$,which reacts with starch to form a blue-colored complex.
Therefore,the gas $(B)$ is $SO_2$.

(B) Ozone $(O_3)$ reacts with mercury $(Hg)$ to form mercurous oxide $(Hg_2O)$,which dissolves in mercury.
This causes the mercury to lose its meniscus and stick to the glass walls,a phenomenon known as 'tailing of mercury'.
$2Hg + O_3 \rightarrow Hg_2O + O_2$

(A) The thermal decomposition of ferrous sulphate is given by: $2 FeSO_4 \rightarrow SO_2 + SO_3 + Fe_2O_3$.
Here,gas $(B)$ is $SO_2$,gas $(C)$ is $SO_3$,and oxide $(D)$ is $Fe_2O_3$.
Gas $(B)$ $(SO_2)$ is a reducing agent and turns acidified $K_2Cr_2O_7$ paper green due to the formation of $Cr_2(SO_4)_3$.
Gas $(C)$ $(SO_3)$ forms a trimer $(SO_3)_3$ which has a cyclic structure with no $S-S$ bonds.
Compound $(D)$ $(Fe_2O_3)$ reacts with $HCl$ to form $FeCl_3$ $(E)$,which exists as a dimer $(Fe_2Cl_6)$ and acts as a Lewis acid.
Thus,the correct sequence is $FeSO_4, SO_2, SO_3, Fe_2O_3, FeCl_3$.

(B) $SO_3$ exists in three allotropic forms:
$a$) $\alpha-SO_3$: It forms long,transparent,ice-like crystals. The melting point of this form is $17^{\circ} C$.
$b$) $\beta-SO_3$: It forms needle-like,silky white crystals. It melts at $32.5^{\circ} C$. Above $50^{\circ} C$,it changes to the $\alpha$-form.
$c$) $\gamma-SO_3$: It is similar to the $\beta$-form and is obtained by completely drying $\beta-SO_3$. It melts at $62.2^{\circ} C$ under $2$ atmospheric pressure.

(D) The reaction sequence is as follows:
$1$. An inorganic compound (like $Ca(OH)_2$) reacts with $SO_2$ to form $(A)$ $(CaSO_3)$.
$2$. $(A)$ $(CaSO_3)$ reacts with $Na_2CO_3$ to give $(B)$ $(Na_2SO_3)$.
$3$. Alternatively,the standard industrial process for preparing sodium thiosulphate involves: $2NaOH + SO_2 \rightarrow Na_2SO_3 + H_2O$.
$4$. Then,$Na_2SO_3 + S \rightarrow Na_2S_2O_3$ (Compound $(C)$).
$5$. $Na_2S_2O_3$ (Sodium thiosulphate) is widely used in photography as a fixing agent (hypo solution).

(A) $1$. $SO_2$ is a non-combustible gas and does not support combustion,so option $A$ is incorrect.
$2$. The maximum permitted atmospheric concentration of $SO_2$ for humans is indeed $5 \ ppm$.
$3$. In the contact process,$SO_2$ is oxidized to $SO_3$ using $V_2O_5$ as a catalyst at high temperature (around $720 \ K$) and low pressure (around $2 \ bar$),not low temperature and high pressure.
$4$. $SO_2$ has sulfur in the $+4$ oxidation state,which can be oxidized to $+6$ or reduced to $0$,thus it acts as both an oxidizing and reducing agent.
$5$. Both $A$ and $C$ are technically incorrect statements based on standard chemical properties and industrial processes.

(D) $1$. Arsenic,antimony,and bismuth are primarily found as sulphide minerals,so statement $A$ is incorrect.
$2$. Holme's signals use calcium phosphide $(Ca_3P_2)$ and calcium carbide $(CaC_2)$. When thrown into the sea,they produce phosphine $(PH_3)$,which undergoes spontaneous combustion. White phosphorus itself is not used directly in this manner,making statement $B$ incorrect.
$3$. Polonium is a radioactive element and it does exhibit allotropy. Statement $C$ is incorrect.
$4$. Oxygen exists in two allotropic forms: dioxygen $(O_2)$ and ozone $(O_3)$. Thus,statement $D$ is correct.

(B) Tailing of mercury is the reaction: $2Hg + O_3 \longrightarrow Hg_2O + O_2$. This is a characteristic test for ozone.
$(B)$ Carbon suboxide $(C_3O_2)$ is acidic in nature,not amphoteric. Carbon monoxide $(CO)$ is neutral.
$(C)$ $Ni_3B$ contains isolated boron atoms,and $FeB$ contains zig-zag chains of boron atoms with $B-B$ bonds.
$(D)$ $V_3B_2$ contains pairs of boron atoms ($B-B$ bond) and $NaB_{15}$ contains icosahedral arrangements of boron atoms.

(B) The acidity of hydrides of group $16$ elements increases from top to bottom in the periodic table.
This is because the bond dissociation enthalpy of the $X-H$ bond decreases as the size of the central atom increases from $O$ to $Te$.
Since the $S-H$ bond is weaker than the $O-H$ bond,$H_2S$ releases $H^+$ ions more easily than $H_2O$,making it more acidic.
Therefore,the correct reason is that the $O-H$ bond is stronger than the $S-H$ bond.

(C) The decomposition of ozone $(2O_3 \to 3O_2)$ is an exothermic process because the formation of $O_2$ is more stable than $O_3$,releasing energy. Therefore,the statement in option $C$ is $INCORRECT$. The hydrolysis of $XeF_6$ $(XeF_6 + 3H_2O \to XeO_3 + 6HF)$ involves no change in oxidation states,making it a non-redox process.

(B) The reaction between phosphorus pentachloride $(PCl_5)$ and concentrated sulfuric acid $(H_2SO_4)$ proceeds as follows:
$PCl_5 + H_2SO_4 \to SO_2Cl_2 + POCl_3 + 2HCl$
Thus,the product formed is sulfuryl chloride $(SO_2Cl_2)$.

(C) $1$. Sulphur exists as $S_8$ molecules in its most stable form,which adopts a puckered ring or cage-like structure. This statement is correct.
$2$. At high temperatures,$S_2$ molecules are formed,which are paramagnetic due to the presence of two unpaired electrons in the antibonding $\pi^*$ orbitals,similar to $O_2$. However,the statement in option $A$ is incomplete/ambiguous compared to $C$.
$3$. The $S-S$ single bond is stronger than the $O-O$ single bond because the lone pairs on oxygen atoms experience significant inter-electronic repulsion due to the small size of the oxygen atom,making the $O-O$ bond weak. Thus,$S-S$ has higher bond energy than $O-O$.
$4$. Sulphur has a higher electron affinity than oxygen because the small size of the oxygen atom leads to strong inter-electronic repulsions,which makes the addition of an electron less favorable compared to sulphur.

(C) $SF_{6}$ is an $sp^{3}d^{2}$ hybridized molecule with an octahedral geometry.
It is sterically protected by six fluorine atoms,making it kinetically inert.
Additionally,there are no vacant $d$-orbitals available for the water molecules to attack the sulphur atom.
Therefore,$SF_{6}$ does not undergo hydrolysis at room temperature.
Thus,the correct answer is option $C$.

(D) $O_3$ (ozone) acts as a dry bleaching agent due to its oxidizing property.
Phosgene,tear gas,and mustard gas are highly poisonous substances.
Sulphuric acid $(H_2SO_4)$ is widely used in petroleum refining for the removal of impurities.
Therefore,both statements $(A)$ and $(C)$ are correct.

(D) $(i)$ $SO_2$ acts as an antichlor and a food preservative. (Correct)
$(ii)$ $O_3$ is used as a disinfectant for water. (Correct)
$(iii)$ $Cl_2$ is used as a disinfectant and for bleaching,but it is not typically classified as an antiseptic. (Incorrect)
$(iv)$ $O_2$ is used in oxy-acetylene welding. (Correct)
$(v)$ $PH_3$ is used in Holme's signals (smoke screens). (Correct)
$(vi)$ $HNO_3$ is used for the pickling of stainless steel. (Correct)
Therefore,statements $(i), (ii), (iv), (v),$ and $(vi)$ are correct. The total number of correct statements is $5$.

(B) The reaction between lead$(IV)$ oxide $(PbO_2)$ and concentrated nitric acid $(HNO_3)$ is an oxidation-reduction reaction.
$PbO_2$ acts as an oxidizing agent and oxidizes water to oxygen gas.
The balanced chemical equation is:
$PbO_2 + 2HNO_3 \rightarrow Pb(NO_3)_2 + H_2O + \frac{1}{2} O_2$
Thus,the gas evolved is oxygen $(O_2)$.

(C) When $SO_2$ is passed through an aqueous solution of $H_2S$,it acts as an oxidizing agent and oxidizes $H_2S$ to elemental sulphur.
The chemical reaction is as follows:
$2H_2S(aq) + SO_2(aq) \rightarrow 2H_2O(l) + 3S(s)$
Thus,sulphur is precipitated out.

(C) When concentrated $H_2SO_4$ is added to charcoal (carbon),it acts as an oxidizing agent and oxidizes carbon to carbon dioxide.
The chemical reaction is:
$C(s) + 2H_2SO_4(conc.) \rightarrow CO_2(g) + 2SO_2(g) + 2H_2O(l)$
In this reaction,$H_2SO_4$ is reduced to $SO_2$ and $C$ is oxidized to $CO_2$.

(D) In $S_2O_7^{2-}$ (pyrosulfate ion),the two sulfur atoms are linked via an oxygen atom ($S-O-S$ linkage).
In $S_2O_3^{2-}$ (thiosulfate),$S_2O_5^{2-}$ (disulfite),and $S_2O_6^{2-}$ (dithionate),there is a direct $S-S$ bond.