Oxygen family Questions in English

(A) To determine the number of oxo groups $(=O)$,we look at the structures of the given oxoacids:
$1$. $H_2SO_4$ (Sulfuric acid): Structure is $(HO)_2SO_2$,which contains $2$ oxo groups.
$2$. $H_2SO_3$ (Sulfurous acid): Structure is $(HO)_2SO$,which contains $1$ oxo group.
$3$. $H_3PO_3$ (Phosphorous acid): Structure is $(HO)_2P(H)=O$,which contains $1$ oxo group.
$4$. $H_3PO_4$ (Phosphoric acid): Structure is $(HO)_3P=O$,which contains $1$ oxo group.
Thus,$H_2SO_4$ has the highest number of oxo groups.

(D) The acidity of hydrides of group $16$ elements increases down the group.
This is because the bond dissociation enthalpy of the $H-E$ bond decreases as the size of the central atom $(E)$ increases.
Since the size of $S$ is larger than $O$,the $H-S$ bond is significantly weaker than the $H-O$ bond.
Consequently,$H_2S$ can release $H^+$ ions more easily than $H_2O$,making it acidic.

(C) Sulfuric acid $(H_2SO_4)$ is a strong dehydrating agent.
It has a very strong affinity for water because it reacts with water to form stable hydrates such as $H_2SO_4 \cdot H_2O$,$H_2SO_4 \cdot 2H_2O$,etc.
This process is highly exothermic,which is why it is used to remove water from many substances.

(A) When $SO_2$ gas is passed through bromine water,it acts as a reducing agent and reduces bromine $(Br_2)$ to bromide ions $(Br^-)$,causing the reddish-brown color of bromine water to disappear (decolorization).
The chemical reaction is: $SO_2 + Br_2 + 2H_2O \rightarrow H_2SO_4 + 2HBr$.
Therefore,the correct gas is $SO_2$.

(C) The structure of $H_2S_2O_8$ (peroxodisulfuric acid) contains a peroxide linkage,which is an $O-O$ bond.
The structure is $HO-SO_2-O-O-SO_2-OH$.
Other options like $H_2S_2O_3$,$H_2S_2O_6$,and $H_2S_4O_6$ do not contain an $O-O$ bond.

(A) When $SO_2$ is passed through an aqueous solution of sodium hydroxide $(NaOH)$,it reacts to form sodium sulfite $(Na_2SO_3)$ and water. If $SO_2$ is in excess,it further reacts with $Na_2SO_3$ to form sodium bisulfite $(NaHSO_3)$.
The balanced chemical equation for the reaction with limited $SO_2$ is: $2NaOH(aq) + SO_2(g) \rightarrow Na_2SO_3(aq) + H_2O(l)$.
However,in the context of standard chemistry problems regarding the reaction of $SO_2$ with $NaOH$ to form an acidic salt,the formation of $NaHSO_3$ is the characteristic reaction when $SO_2$ is in excess or when considering the stoichiometry of the bisulfite formation: $NaOH(aq) + SO_2(g) \rightarrow NaHSO_3(aq)$.

(B) $Ozone$ $(O_3)$ is a powerful oxidising agent because it is thermodynamically unstable and readily decomposes to give nascent oxygen $(O_3 \to O_2 + [O])$.
The magnetic property of $O_3$ is diamagnetic,whereas $O_2$ is paramagnetic due to the presence of two unpaired electrons in its antibonding molecular orbitals.
While both statements are factually correct,the magnetic property of $O_3$ is not the reason for its high oxidising power. Therefore,the Reason is not the correct explanation of the Assertion.

(C) Sulphur exhibits several allotropic forms such as rhombic sulphur $(S_8)$,monoclinic sulphur,and others. Therefore,the statement that $S_8$ is the only allotropic form of sulphur is incorrect.

(A) little more than $47\%$ of the earth's crust consists of oxygen. The most common rock constituents of the earth's crust are nearly all oxides.

(A) The chemical formula for peroxodisulphuric acid (also known as Marshall's acid) is $H_2S_2O_8$.
In its structure,two $SO_3$ groups are linked by a peroxide linkage,which is an $O-O$ single bond.
The structure is $HO-SO_2-O-O-SO_2-OH$.
Therefore,it contains an $O-O$ bond.

(D) Among the hydrides of group $16$ elements,the boiling point of $H_2O$ is the highest due to strong intermolecular hydrogen bonding.
For the remaining hydrides ($H_2S$,$H_2Se$,$H_2Te$),the boiling point increases with the increase in molecular mass and size,which leads to stronger van der Waals forces.
Thus,the order is $H_2O > H_2Te > H_2Se > H_2S$.

(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 size of $S$ is larger than $O$,the $H-S$ bond is weaker than the $H-O$ bond,making it easier for $H_2S$ to release $H^+$ ions compared to $H_2O$.

(A) Peroxoacids of sulphur are those that contain at least one peroxy linkage $(-O-O-)$.
$H_2SO_5$ (peroxomonosulphuric acid or Caro's acid) contains one peroxy linkage.
$H_2S_2O_8$ (peroxodisulphuric acid or Marshall's acid) also contains one peroxy linkage.
Therefore,both $H_2SO_5$ and $H_2S_2O_8$ are peroxoacids of sulphur.

(D) The Assertion is incorrect because the dinegative anion of sulphur $(S^{2-})$ is actually very common (e.g.,in metal sulphides like $ZnS$,$FeS$),whereas the formation of $O^{2-}$ is energetically demanding due to high electron-electron repulsion in the small oxygen atom.
The Reason is also incorrect because the covalency of oxygen is generally two,but this is not the reason for the stability or commonality of its anions.
Therefore,both the Assertion and the Reason are incorrect.

(C) The reaction between $H_2S$ and $SO_2$ in the presence of $Fe_2O_3$ catalyst is given by: $2H_2S + SO_2 \to 2H_2O + 3S \downarrow$.
In this reaction,$H_2S$ acts as a reducing agent and is oxidized to elemental sulphur $(S)$,while $SO_2$ acts as an oxidizing agent and is reduced.
Therefore,the Assertion is correct,but the Reason is incorrect because $SO_2$ acts as an oxidizing agent in this reaction,not a reducing agent.

(C) The assertion is true because ozone $(O_3)$ is indeed an allotrope of oxygen.
The reason is false. While liquid oxygen is pale blue,it is paramagnetic in its triplet state due to the presence of two unpaired electrons in its antibonding molecular orbitals. In its singlet state,all electrons are paired,making it diamagnetic,not paramagnetic.

(A) Oxygen has a small atomic size and small bond length,which allows it to form stable $p\pi-p\pi$ multiple bonds,resulting in the formation of diatomic $O_2$ molecules.
In contrast,sulphur has a larger atomic size and longer bond length,making effective $p\pi-p\pi$ overlap difficult. Consequently,sulphur prefers to form single bonds with other sulphur atoms,leading to the formation of puckered $S_8$ ring structures.
Thus,both the Assertion and the Reason are correct,and the Reason correctly explains why oxygen exists as $O_2$ while sulphur exists as $S_8$.

(C) When ozone $(O_3)$ is passed over a silver surface,it oxidizes silver to silver oxide $(Ag_2O)$,which is black in color.
The reaction is as follows:
$2Ag + O_3 \to Ag_2O + O_2$
Thus,the blackening of the silver surface is due to the formation of $Ag_2O$.

(C) The thermal stability of hydrides of group $16$ elements depends on the bond dissociation enthalpy of the $E-H$ bond.
As the size of the central atom $E$ increases down the group $(O < S < Se < Te < Po)$,the bond length increases and the bond dissociation enthalpy decreases.
Therefore,the thermal stability decreases down the group.
The correct order is: $H_2 O > H_2 S > H_2 Se > H_2 Te > H_2 Po$.

(A) Oleum is also known as pyrosulphuric acid.
Its chemical formula is $H_2S_2O_7$.

(N/A) The upper stratosphere consists of a considerable amount of ozone $(O_{3})$,which protects us from the harmful ultraviolet $(UV)$ radiations coming from the sun.
These radiations can cause skin cancer in humans. Therefore,it is important to maintain the ozone shield.
In the stratosphere,when $UV$ radiations act on dioxygen $(O_{2})$ molecules,the $UV$ radiations split molecular oxygen into free oxygen $(O)$ atoms. These oxygen atoms then combine with molecular oxygen to form ozone.
$O_{2(g)} \stackrel{UV}{\longrightarrow} O_{(g)} + O_{(g)}$
$O_{(g)} + O_{2(g)} \stackrel{}{\longrightarrow} O_{3(g)}$
Ozone is thermodynamically unstable and decomposes back into molecular oxygen. Thus,a dynamic equilibrium exists between the production and decomposition of ozone molecules.

(N/A) The electronic configuration of Group $15$ elements is $ns^2 np^3$,which represents a half-filled $p$-orbital configuration.
This configuration is extra stable due to exchange energy and symmetry.
In contrast,Group $16$ elements have an electronic configuration of $ns^2 np^4$.
Removing an electron from the stable half-filled $p$-orbital of Group $15$ requires more energy than removing an electron from the $np^4$ configuration of Group $16$ elements.
Therefore,Group $16$ elements have lower first ionisation enthalpy values compared to Group $15$ elements.

(N/A) The acidic character of hydrides of group $16$ elements depends on the bond dissociation enthalpy of the $E-H$ bond.
As we move down the group from $S$ to $Te$,the size of the central atom increases,which leads to an increase in the bond length.
Consequently,the $E-H$ bond dissociation enthalpy decreases down the group.
Since it becomes easier to break the $H-Te$ bond compared to the $H-S$ bond,$H_2Te$ is more acidic than $H_2S$.

(C) In the vapour state,sulphur partly exists as the $S_2$ molecule.
Similar to the $O_2$ molecule,the $S_2$ molecule contains two unpaired electrons in its antibonding $\pi^*$ molecular orbitals.
Due to the presence of these unpaired electrons,$S_2$ exhibits paramagnetic behaviour.

(N/A) $(i)$ When concentrated $H_2SO_4$ is added to calcium fluoride $(CaF_2)$,it undergoes a double displacement reaction to form calcium sulfate and hydrogen fluoride gas:
$CaF_2 + H_2SO_4 \rightarrow CaSO_4 + 2HF$
$(ii)$ When $SO_3$ is passed through water,it reacts to form sulfuric acid $(H_2SO_4)$:
$SO_3 + H_2O \rightarrow H_2SO_4$

(N/A) Sulphur mainly exists in combined form in the earth's crust,primarily as sulphates $[CaSO_4 \cdot 2H_2O$ (gypsum),$MgSO_4 \cdot 7H_2O$ (Epsom salt),$BaSO_4$ (baryte)$]$ and sulphides $[PbS$ (galena),$ZnS$ (zinc blende),$CuFeS_2$ (copper pyrites)$]$.

(A) The thermal stability of hydrides decreases on moving down the group.
This is due to a decrease in the bond dissociation enthalpy ($E-H$ bond) of hydrides as the size of the central atom increases down the group.
Therefore,the correct order of thermal stability is:
$H_2O > H_2S > H_2Se > H_2Te > H_2Po$

(N/A) Ozone is not a very stable compound under normal conditions and decomposes readily on heating to give a molecule of oxygen and nascent oxygen. Nascent oxygen,being a free radical,is very reactive.
$\mathop {{O_3}}\limits_{\text{Ozone}}$ $\xrightarrow{\Delta }\mathop {{O_2}}\limits_{\text{Oxygen}} + \mathop {[O]}\limits_{\text{Nascent oxygen}}$
Therefore,ozone acts as a powerful oxidising agent.

(N/A) Quantitatively,ozone $(O_3)$ can be estimated by reacting it with an excess of potassium iodide $(KI)$ solution buffered with a borate buffer $(pH \ 9.2)$.
$1$. Ozone reacts with iodide ions to liberate iodine $(I_2)$:
$2I^- + H_2O + O_3 \to 2OH^- + I_2 + O_2$
$2$. The liberated iodine is then titrated against a standard solution of sodium thiosulphate $(Na_2S_2O_3)$ using starch as an indicator:
$I_2 + 2Na_2S_2O_3 \to Na_2S_4O_6 + 2NaI$
By measuring the volume of sodium thiosulphate used,the amount of ozone can be calculated.

(B) $SO_{2}$ is a colourless and pungent smelling gas.
It can be detected with the help of acidified potassium permanganate solution. When $SO_{2}$ is passed through an acidified potassium permanganate solution,it decolourises the solution as it reduces $MnO_{4}^{-}$ ions to $Mn^{2+}$ ions.
The chemical reaction is:
$5SO_{2} + 2MnO_{4}^{-} + 2H_{2}O \longrightarrow 5SO_{4}^{2-} + 4H^{+} + 2Mn^{2+}$

(N/A) Sulphuric acid $(H_2SO_4)$ is a vital industrial chemical with numerous applications. Three important areas of its use are:
$(i)$ Fertilizer industry: It is used in the production of fertilizers such as ammonium sulphate and calcium superphosphate.
$(ii)$ Chemical manufacturing: It is used in the manufacture of pigments,paints,and synthetic detergents.
$(iii)$ Energy storage: It is used as an electrolyte in lead-acid storage batteries.

(N/A) The manufacture of sulphuric acid by the Contact process involves three main steps:
$1.$ Burning of sulphur or sulphide ores to form $SO_2$.
$2.$ Catalytic oxidation of $SO_2$ to $SO_3$ using $O_2$ in the presence of $V_2O_5$ catalyst: $2SO_2(g) + O_2(g) \rightleftharpoons 2SO_3(g)$,$\Delta H = -196.6 \ kJ/mol$.
$3.$ Absorption of $SO_3$ in $H_2SO_4$ to produce oleum $(H_2S_2O_7)$,which is then diluted with water.
The key step for maximizing yield is the second step. According to Le Chatelier's principle:
- Since the reaction is exothermic,a low temperature (approximately $720 \ K$) favors the forward reaction.
- Since there is a decrease in the number of moles of gas ($3$ moles of reactants to $2$ moles of product),a high pressure (approximately $2 \ bar$) favors the forward reaction.

(N/A) The elements of group $16$ are collectively called chalcogens.
$(i)$ Electronic configuration: Elements of group $16$ have six valence electrons each. The general electronic configuration of these elements is $ns^{2} np^{4}$,where $n$ varies from $2$ to $6$.
$(ii)$ Oxidation state: As these elements have six valence electrons $(ns^{2} np^{4})$,they typically exhibit an oxidation state of $-2$. Oxygen,due to its high electronegativity,shows $-2$ as its most common state,but also exhibits $-1$ (in $H_{2}O_{2}$),$0$ (in $O_{2}$),and $+2$ (in $OF_{2}$). The stability of the $-2$ oxidation state decreases down the group due to decreasing electronegativity. Heavier elements show $+2, +4$,and $+6$ oxidation states due to the availability of $d$-orbitals.
$(iii)$ Formation of hydrides: These elements form hydrides of the general formula $H_{2}E$,where $E = O, S, Se, Te, Po$. Oxygen and sulfur also form hydrides of the type $H_{2}E_{2}$. These hydrides are generally volatile.

(N/A) Oxygen has a smaller atomic size compared to sulphur. Due to its small size,it can effectively form $p\pi-p\pi$ multiple bonds to exist as a discrete $O_2$ molecule. The intermolecular forces in $O_2$ are weak van der Waals forces,which result in it being a gas at room temperature.
On the other hand,sulphur has a larger atomic size and does not form stable $S=S$ double bonds. Instead,it forms single $S-S$ bonds and exists as a puckered $S_8$ ring structure. These large $S_8$ molecules are held together by stronger van der Waals forces,making sulphur a solid.

(N/A) Sulphuric acid is manufactured by the contact process. It involves the following steps:
Step $(i):$
Sulphur or sulphide ores are burnt in air to form $SO_2$.
Step $(ii):$
By a reaction with oxygen,$SO_2$ is converted into $SO_3$ in the presence of $V_2O_5$ as a catalyst.
$2SO_{2(g)} + O_{2(g)} \xrightarrow{V_2O_5} 2SO_{3(g)}$
Step $(iii):$
$SO_3$ produced is absorbed in $H_2SO_4$ to give $H_2S_2O_7$ (oleum).
$SO_3 + H_2SO_4 \longrightarrow H_2S_2O_7$
This oleum is then diluted with water to obtain $H_2SO_4$ of the desired concentration.
In practice,the plant is operated at $2 \ bar$ pressure and $720 \ K$ temperature. The sulphuric acid thus obtained is $96-98 \%$ pure.

(C) Marshall's acid is also known as peroxydisulfuric acid. Its molecular formula is $H_2S_2O_8$.

(N/A) Group $16$ of the periodic table consists of oxygen $(O)$,sulfur $(S)$,selenium $(Se)$,tellurium $(Te)$,polonium $(Po)$,and livermorium $(Lv)$.
These elements are collectively known as chalcogens. The term is derived from the Greek word for brass.
Oxygen is the most abundant element on Earth,accounting for approximately $46.6 \%$ of the Earth's crust by mass. Dry air contains $20.946 \%$ oxygen by volume.
Sulfur is found in the Earth's crust in $0.03-0.1 \%$ abundance. It occurs in combined forms as sulfates like gypsum $(CaSO_4 \cdot 2H_2O)$,epsom salt $(MgSO_4 \cdot 7H_2O)$,barite $(BaSO_4)$,and as sulfides like galena $(PbS)$,zinc blende $(ZnS)$,and copper pyrites $(CuFeS_2)$.
Trace amounts of sulfur are found in volcanic gases as hydrogen sulfide $(H_2S)$. Organic materials like eggs,proteins,garlic,onion,mustard,hair,and wool also contain sulfur.
Selenium and tellurium are found as metal selenides and tellurides in sulfide ores.
Polonium is a decay product of thorium and uranium minerals.
Livermorium is a synthetic radioactive element with symbol $Lv$,atomic number $116$,atomic mass $292$,and electronic configuration $[Rn] 5f^{14} 6d^{10} 7s^2 7p^4$. It has a very short half-life.

(N/A) Electronic Configuration: The elements of group $16$ have six electrons in their outermost shell,and their general electronic configuration is $ns^2 np^4$.
Atomic and Ionic Radii: As we move down the group,the number of shells increases,leading to an increase in atomic and ionic radii. However,the oxygen atom is exceptionally small in size.

(N/A) Ionization Enthalpy: Ionization enthalpy decreases down the group due to an increase in atomic size. The ionization enthalpy values of Group-$16$ elements are lower than those of the corresponding elements of Group-$15$ in the same period.
This is because Group-$15$ elements have extra stable,half-filled $p$-orbital electronic configurations.
Electron Gain Enthalpy: Due to the compact nature of oxygen,its electron gain enthalpy is less negative than that of sulfur. However,from sulfur to polonium,the values become less negative.
Electronegativity: Among all elements,oxygen has the second highest electronegativity after fluorine. Electronegativity decreases with an increase in atomic number down the group. This indicates that metallic character increases from oxygen to polonium.

(N/A) Among the elements of Group $16$,oxygen and sulfur are non-metals,selenium and tellurium are metalloids,while polonium is a metal.
Polonium is radioactive and has a short half-life $(13.8 \ days)$. All these elements exhibit allotropy.
As we move down the group,the melting and boiling points increase with an increase in atomic number.
The large difference between the melting and boiling points of oxygen and sulfur can be explained based on their atomicity.
Oxygen exists as a diatomic molecule $(O_2)$,whereas sulfur exists as a polyatomic molecule $(S_8)$.

(N/A) Group $16$ elements exhibit a variety of oxidation states.
The stability of the $(-2)$ oxidation state decreases down the group. Polonium hardly shows the $(-2)$ oxidation state.
Due to the high electronegativity of oxygen,it shows only the $(-2)$ oxidation state in its compounds,except in $OF_2$,where it shows $(+2)$.
Other elements of the group show $+2, +4, +6$ oxidation states,but $+4$ and $+6$ are more common.
Sulfur,selenium,and tellurium usually show the $+4$ oxidation state in their compounds with oxygen and the $+6$ oxidation state in their compounds with fluorine.
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 $+4$ and $+6$ oxidation states is primarily covalent.

(N/A) The anomalous behavior of oxygen compared to other members of the $p$-block in the second period is primarily due to its small size and high electronegativity.
$A$ significant consequence of its small size and high electronegativity is the presence of strong hydrogen bonding in $H_2O$,which is not observed in $H_2S$.
Due to the absence of $d$-orbitals in oxygen,its covalency is limited to $4$,and in practice,it is rarely more than $2$. In contrast,other elements of the group can expand their valence shells,allowing for a covalency greater than $4$.

(N/A) $(i)$ Reactivity with hydrogen: All Group $16$ elements form hydrides of the type $H_2E$ (where $E = O, S, Se, Te, Po$).
Their acidic character increases from $H_2O$ to $H_2Te$. This increase in acidic character is due to the decrease in bond dissociation enthalpy of the $H-E$ bond down the group.
Thermal stability of hydrides decreases from $H_2O$ to $H_2Po$ due to the decrease in $H-E$ bond dissociation enthalpy down the group.
All hydrides except water possess reducing property,and this character increases from $H_2S$ to $H_2Te$.
$(ii)$ Reactivity with oxygen: All these elements form oxides of the types $EO_2$ and $EO_3$ (where $E = S, Se, Te, Po$).
Ozone $(O_3)$ and sulfur dioxide $(SO_2)$ are gases,while selenium dioxide $(SeO_2)$ is a solid. The reducing property of dioxides decreases from $SO_2$ to $TeO_2$; $SO_2$ is a reducing agent,while $TeO_2$ is an oxidizing agent. Sulfur,selenium,and tellurium also form $EO_3$ type oxides $(SO_3, SeO_3, TeO_3)$. Both types of oxides are acidic in nature.
$(iii)$ Reactivity with halogens: Group $16$ elements form a large number of halides of the types $EX_6, EX_4,$ and $EX_2$ (where $E$ is an element of this group and $X$ is a halogen).
The stability of halides decreases in the order: $F^- > Cl^- > Br^- > I^-$.
Among hexahalides,only hexafluorides are stable. All hexafluorides are gaseous and have octahedral geometry.
Sulfur hexafluoride,$SF_6$,is exceptionally stable for steric reasons.
Among tetrafluorides,$SF_4$ is a gas,$SeF_4$ is a liquid,and $TeF_4$ is a solid. These fluorides have $sp^3d$ hybridization and thus possess a trigonal bipyramidal geometry,in which one equatorial position is occupied by a lone pair of electrons. This geometry is also called the see-saw geometry.
All elements except oxygen form dichlorides and dibromides. These dihalides are formed by $sp^3$ hybridization and thus have a tetrahedral structure.
Well-known monohalides are dimeric in nature. Examples include $S_2F_2, S_2Cl_2, S_2Br_2, Se_2Cl_2,$ and $Se_2Br_2$. These dimeric halides undergo disproportionation reactions,e.g.,$2Se_2Cl_2 \rightarrow SeCl_4 + 3Se$.

(N/A) Group-$16$ elements form hydrides of type $H_{2}E$ $(E=O, S, Se, Te, Po)$. Acidic strength increases down the group $(H_{2}Te > H_{2}Se > H_{2}S > H_{2}O)$ due to decreasing $H-E$ bond dissociation enthalpy. Thermal stability decreases down the group $(H_{2}O > H_{2}S > H_{2}Se > H_{2}Te)$. Reducing property increases down the group $(H_{2}Te > H_{2}Se > H_{2}S > H_{2}O)$.
Oxides are of type $EO_{2}$ and $EO_{3}$. $SO_{2}$ is a gas,$SeO_{2}$ is solid. Stability and acidic strength of $EO_{2}$ and $EO_{3}$ decrease down the group ($SO_{2} > SeO_{2} > TeO_{2}$ and $SO_{3} > SeO_{3} > TeO_{3}$).
Halides are of type $EX_{6}, EX_{4}, EX_{2}$. Hexafluorides are stable and octahedral. Tetrafluorides $(EX_{4})$ have $sp^{3}d$ hybridization (see-saw geometry). Dihalides $(EX_{2})$ have $sp^{3}$ hybridization (tetrahedral). Monohalides are dimeric and undergo disproportionation.

(A) The abundance of sulfur in the Earth's crust is approximately $0.03 - 0.1 \%$ by mass. It is the $16^{th}$ most abundant element in the Earth's crust.

(N/A) $(i)$ Laboratory preparation :
$(1)$ By thermal decomposition of salts such as chlorates,nitrates and permanganates.
$2 KClO_{3} \stackrel{\Delta}{\longrightarrow} 2 KCl + 3 O_{2}$
$2 NaNO_{3} \stackrel{\Delta}{\longrightarrow} 2 NaNO_{2} + O_{2}$
$2 KMnO_{4} \stackrel{\Delta}{\longrightarrow} K_{2}MnO_{4} + MnO_{2} + O_{2}$
$(2)$ By the thermal decomposition of the oxides of metals low in the electrochemical series and higher oxides of some metals.
$2 Ag_{2}O_{(s)} \stackrel{\Delta}{\longrightarrow} 4 Ag_{(s)} + O_{2(g)}$
$2 Pb_{3}O_{4(s)} \stackrel{\Delta}{\longrightarrow} 6 PbO_{(s)} + O_{2(g)}$
$2 HgO_{(s)} \stackrel{\Delta}{\longrightarrow} 2 Hg_{(l)} + O_{2(g)}$
$2 PbO_{2(s)} \stackrel{\Delta}{\longrightarrow} 2 PbO_{(s)} + O_{2(g)}$
$(3)$ By decomposition of hydrogen peroxide in presence of finely divided metals and manganese dioxide.
$2 H_{2}O_{2(aq)} \xrightarrow{MnO_{2}} 2 H_{2}O_{(l)} + O_{2(g)}$ ($MnO_{2}$ acts as catalyst)
$(ii)$ Commercial production : Industrially,dioxygen is obtained from air by first removing carbon dioxide and water vapour and the remaining gases are liquefied and fractionally distilled to give dinitrogen and dioxygen.
$(iii)$ On large scale it is prepared by electrolysis of acidulated water which releases hydrogen at cathode and oxygen at the anode.
$(A)$ Physical properties : Dioxygen is a colourless and odourless gas.
The water solubility is $3.08 \ cm^{3}$ per $100 \ cm^{3}$ water at $293 \ K$ which is just sufficient for the vital support of marine and aquatic life.
It has three stable isotopes: $^{16}O, ^{17}O, ^{18}O$.
It liquefies at $90 \ K$ and freezes at $55 \ K$.
Molecular oxygen,$(O_{2})$ is paramagnetic due to presence of unpaired electrons in $\pi^{*}$ orbitals.
$(B)$ Chemical properties :
$(i)$ Reaction with metals : Dioxygen directly reacts with nearly all metals except $Au$ and $Pt$ to form oxides.
For example :
$2 Ca + O_{2} \rightarrow 2 CaO$
$4 Al + 3 O_{2} \rightarrow 2 Al_{2}O_{3}$
$4 Fe + 3 O_{2} \rightarrow 2 Fe_{2}O_{3}$
$(ii)$ Reaction with non-metals:
$C + O_{2} \rightarrow CO_{2}$
$P_{4} + 5 O_{2} \rightarrow P_{4}O_{10}$
$(iii)$ Reaction with compounds :
$2 ZnS + 3 O_{2} \rightarrow 2 ZnO + 2 SO_{2}$
$CH_{4} + 2 O_{2} \rightarrow CO_{2} + 2 H_{2}O$
$2 SO_{2} + O_{2} \xrightarrow{V_{2}O_{5}} 2 SO_{3}$ (Catalytic oxidation)
$4 HCl + O_{2} \xrightarrow{CuCl_{2}} 2 Cl_{2} + 2 H_{2}O$ (Oxidation)
In addition to its importance in normal respiration and combustion processes,oxygen is used in oxyacetylene welding and in the manufacturing of steel.
Oxygen cylinders are used in hospitals,high altitude flying and in mountaineering.
The combustion of fuels,e.g.,hydrazines in liquid oxygen provides tremendous thrust in rockets.

(A) Dioxygen $(O_2)$ is a colorless and odorless gas.
It liquefies at $90 \ K$ and freezes at $55 \ K$.

(A) Oxygen has $3$ stable isotopes in nature. These are $O^{16}$,$O^{17}$,and $O^{18}$.

(N/A) binary compound of oxygen with another element is called an oxide. As mentioned earlier,oxygen reacts with most elements of the periodic table to form oxides. In many cases,an element forms two or more oxides. These oxides show a wide variety in their nature and properties.
Oxides can be simple (e.g.,$MgO, Al_{2}O_{3}$) or mixed (e.g.,$Pb_{3}O_{4}, Fe_{3}O_{4}$). Simple oxides can be classified on the basis of their acidic,basic,or amphoteric character.
An oxide that combines with water to give an acid is termed an acidic oxide (e.g.,$SO_{2}, Cl_{2}O_{7}, CO_{2}, N_{2}O_{5}$). For example,$SO_{2}$ combines with water to form $H_{2}SO_{3}$ acid.
$SO_{2} + H_{2}O \rightarrow H_{2}SO_{3}$
As a general rule,only non-metal oxides are acidic,but oxides of some metals in high oxidation states also have acidic character (e.g.,$Mn_{2}O_{7}, CrO_{3}, V_{2}O_{5}$).
Oxides that give a base with water are known as basic oxides (e.g.,$Na_{2}O, CaO, BaO$).
For example,$CaO$ reacts with water to form $Ca(OH)_{2}$ base.
$CaO + H_{2}O \rightarrow Ca(OH)_{2}$
In general,metallic oxides are basic.
Some metallic oxides exhibit dual behavior. They show characteristics of both acidic and basic oxides. These are called amphoteric oxides. They react with both acids and bases.
For example,$Al_{2}O_{3}$ reacts with both acids and bases.
$Al_{2}O_{3(s)} + 6HCl_{(aq)} + 9H_{2}O_{(l)} \rightarrow 2[Al(H_{2}O)_{6}]^{3+}_{(aq)} + 6Cl^{-}_{(aq)}$
$Al_{2}O_{3(s)} + 6NaOH_{(aq)} + 3H_{2}O_{(l)} \rightarrow 2Na_{3}[Al(OH)_{6}]_{(aq)}$
Some oxides which are neither acidic nor basic are known as neutral oxides. $CO, NO,$ and $N_{2}O$ are examples of neutral oxides.

(N/A) Preparation: When a slow dry stream of oxygen is passed through a silent electrical discharge, a conversion of oxygen to ozone $(10 \%)$ occurs. The product is known as ozonised oxygen.
$3 O_{2(g)} \rightarrow 2 O_{3(g)}$; $\Delta H^{\circ} = +142 \ kJ \ mol^{-1}$
Since the formation of ozone from oxygen is an endothermic process, it is necessary to use a silent electrical discharge in its preparation to prevent its decomposition.
Physical Properties: Pure ozone is a pale blue gas, dark blue liquid, and violet-black solid. It has a characteristic odour. Ozone has two resonating structures. The two oxygen-oxygen bond lengths are identical $(128 \ pm)$ and the bond angle is $117^{\circ}$. Its molecular shape is angular (bent).
Chemical Properties:
$1$. Oxidising agent: $PbS_{(s)} + 4 O_{3(g)} \rightarrow PbSO_{4(s)} + 4 O_{2(g)}$
$2$. Reaction with iodide ions: $2 I^-_{(aq)} + H_2O_{(l)} + O_{3(g)} \rightarrow 2 OH^-_{(aq)} + I_{2(s)} + O_{2(g)}$
$3$. Ozone layer depletion: $NO_{(g)} + O_{3(g)} \rightarrow NO_{2(g)} + O_{2(g)}$
Uses: It is used as a germicide, disinfectant, and for sterilising water. It is also used for bleaching oils, ivory, flour, and starch. It acts as an oxidising agent in the manufacture of potassium permanganate.