Oxygen family Questions in English

(C) The elements of Group $16$ (chalcogens) generally show $-2$ oxidation state due to their electronic configuration $ns^2 np^4$.
$Oxygen$ $(O)$ shows $-2$ (in most oxides),$-1$ (in peroxides),$+1$ and $+2$ (in $OF_2$) oxidation states.
$Selenium$ $(Se)$ and $Tellurium$ $(Te)$ show $-2, +2, +4$ and $+6$ oxidation states.
$Polonium$ $(Po)$ is a metal and primarily shows $+2$ and $+4$ oxidation states; it does not exhibit the $-2$ oxidation state.
Therefore,the correct answer is $Po$.

(D) Statement $I$ is correct. The boiling point of group $16$ hydrides follows the order $H_2O > H_2Te > H_2Se > H_2S$.
Statement $II$ is also correct. While $H_2O$ has the lowest molecular mass among these hydrides,it exhibits the highest boiling point due to the presence of extensive intermolecular $H$-bonding.
The trend from $H_2Te$ to $H_2S$ is primarily determined by the decrease in molecular mass,which reduces the magnitude of van der Waals forces.

(B) $2 KClO_3 \xrightarrow{MnO_2} 2 KCl + 3 O_2$ $(W)$
$P_4 + 5 O_2 \longrightarrow P_4O_{10}$ $(X)$
$P_4O_{10} + 4 HNO_3 \longrightarrow 4 HPO_3$ $(Z)$ $+ 2 N_2O_5$ $(Y)$
Therefore,$W = O_2$,$X = P_4O_{10}$,$Y = N_2O_5$,and $Z = HPO_3$.

(C) peroxide linkage is defined as an oxygen-oxygen single bond $(-O-O-)$.
$1$. $H_2S_2O_7$ (Oleum/Pyrosulfuric acid) has an $S-O-S$ linkage.
$2$. $H_2S_2O_8$ (Peroxodisulfuric acid/Marshall's acid) has an $S-O-O-S$ linkage,which is a peroxide linkage.
$3$. $H_2S_2O_5$ (Pyrosulfurous acid) has an $S-S$ linkage.
$4$. $H_2SO_5$ (Peroxomonosulfuric acid/Caro's acid) has an $S-O-O-H$ linkage,which contains a peroxide linkage.
Therefore,both $H_2S_2O_8$ and $H_2SO_5$ contain peroxide linkages.

(D) $SO_2$ contains sulfur in the $+4$ oxidation state. \\ Since sulfur can exist in oxidation states ranging from $-2$ to $+6$,$SO_2$ can act as an oxidizing agent (by being reduced to $S$ or $H_2S$) as well as a reducing agent (by being oxidized to $SO_3$ or $SO_4^{2-}$). \\ Therefore,the statement that $SO_2$ cannot act as a reducing agent is incorrect.

(B) Oxygen exists as a diatomic molecule,$O_2$ (Atomicity $= 2$),which is held by weak van der Waals forces.
Sulphur exists as a polyatomic molecule,$S_8$ (Atomicity $= 8$),which has a much larger molecular mass and stronger van der Waals forces.
Therefore,the melting and boiling points of sulphur are significantly higher than those of oxygen due to the difference in their atomicity.

(A) $TeO_2$ acts as an oxidizing agent because $Te$ in $+4$ oxidation state can be reduced to lower oxidation states.
$TeH_2$ is acidic in nature because the bond dissociation energy of the $Te-H$ bond decreases down the group,making it easy to release $H^+$ ions.

(D) The acidic strength of hydrides of group $16$ elements increases down the group due to the decrease in bond dissociation enthalpy. Thus,the order is $H_2Se < H_2Te$. Therefore,Statement $I$ is false.
The bond dissociation enthalpy decreases down the group as the bond length increases. Thus,$H_2Se$ has a higher bond dissociation enthalpy $(276 \ kJ/mol)$ compared to $H_2Te$ $(238 \ kJ/mol)$. Therefore,Statement $II$ is true.

(A) In Group $16$ (Oxygen family),the stability of the $+6$ oxidation state decreases down the group due to the inert pair effect,which makes the $ns^2$ electrons less available for bonding.
Conversely,the stability of the $+4$ oxidation state increases down the group due to the same inert pair effect.
Therefore,statement $(i)$ is correct and $(ii)$ is incorrect.
Regarding bonding,elements in the $+4$ and $+6$ oxidation states typically form covalent bonds with more electronegative elements like oxygen or fluorine.
Thus,statement $(iii)$ is also correct.
The correct statements are $(i)$ and $(iii)$.

(A) Oxygen exists as a diatomic molecule $(O_2)$ due to its small size and ability to form $p\pi-p\pi$ multiple bonds. Sulfur exists as a polyatomic puckered ring structure $(S_8)$ due to its larger size and inability to form stable $p\pi-p\pi$ multiple bonds.
The large difference in the molecular size and the strength of van der Waals forces between $O_2$ (small,weak forces) and $S_8$ (large,stronger forces) leads to a significant difference in their melting and boiling points.
Therefore,both the Assertion and the Reason are true,and the Reason correctly explains the Assertion.

(B) The oxygen family,also known as group $16$ or the chalcogens,consists of the elements oxygen $(O)$,sulfur $(S)$,selenium $(Se)$,tellurium $(Te)$,polonium $(Po)$,and livermorium $(Lv)$.
Among the given options,$Se$ (selenium) belongs to this group.

(C) The elements of group $16$ are known as chalcogens.
These elements are oxygen $(O)$,sulfur $(S)$,selenium $(Se)$,tellurium $(Te)$,and polonium $(Po)$.
They are called chalcogens because they are primarily found in ores.

(C) Group $16$ elements are known as chalcogens and include oxygen $(O)$,sulfur $(S)$,selenium $(Se)$,tellurium $(Te)$,polonium $(Po)$,and livermorium $(Lv)$.
Among the given options,$Po$ (Polonium) belongs to group $16$.

(D) . Besides $EO_2$ type,$S$,$Se$,and $Te$ also form $EO_3$ type oxides.
Group $16$ elements consist of non-metals ($O$,$S$),metalloids ($Se$,$Te$),and a metal $(Po)$.
All elements of this group require $2$ electrons to attain a stable noble gas configuration.
The reducing character of dioxides decreases from $SO_2$ to $TeO_2$ because the stability of the $+4$ oxidation state increases down the group due to the inert pair effect.

(D) The correct answer is $(d)$.
$Se$ $(selenium)$ does not belong to group $15$; it belongs to group $16$ of the periodic table.
$Se$ is located in the $4^{th}$ period and $16^{th}$ group with an atomic number of $34$.
The electronic configuration of $Se$ is $[Ar] 3d^{10} 4s^2 4p^4$.

(A) Ionization enthalpy is the energy required to remove the most loosely bound electron from an isolated gaseous atom.
As we move down a group in the periodic table,the atomic size increases due to the addition of new shells.
This increase in atomic size results in a decrease in the effective nuclear attraction on the valence electrons,making it easier to remove an electron.
Therefore,the ionization enthalpy decreases down the group.
For Group $16$ elements $(O, S, Se, Te, Po)$,the decreasing order of ionization enthalpy is $S > Se > Te > Po$.

(D) In group $16$,the density of elements increases as we move down the group.
The order of density for group $16$ elements is $O < S < Se < Te < Po$.
Therefore,among the given options,$Te$ (Tellurium) has the highest density.

(B) $As_2O_3$,$P_2O_5$,and $SO_2$ are covalent compounds that exist as molecular solids or gases at room temperature,making them volatile.
$ZnO$ is an ionic solid with a high melting point and is non-volatile.

(C) The lead chamber process involves the oxidation of $SO_2$ to $SO_3$ using air in the presence of nitrogen oxides as a catalyst.
Specifically,$NO$ (nitric oxide) acts as the catalyst in this process.

(C) Oxygen is the most abundant element in the Earth's crust,accounting for approximately $46.6\%$ of its total mass by weight.

(C) The acidity of hydrides of group $16$ elements increases down the group.
As we move down the group from $O$ to $Te$,the atomic size of the central atom increases.
This leads to a decrease in the bond dissociation enthalpy of the $E-H$ bond (where $E = O, S, Se, Te$).
Consequently,the ease of releasing $H^+$ ions increases,making the hydrides more acidic.
Therefore,the correct order of acidity is $H_2O < H_2S < H_2Se < H_2Te$ or $H_2Te > H_2Se > H_2S > H_2O$.

(D) The acidic character of the hydrides of group $16$ elements increases down the group as the bond dissociation enthalpy decreases.
Therefore,the order of acidic strength is $H_2O < H_2S < H_2Se < H_2Te$.
Thus,$H_2Te$ is the most acidic molecule among the given options.

(B) Among the given elements,$S$ (Sulfur) exhibits the highest tendency for catenation due to its moderate size and electronic configuration.
This property allows it to form a large number of allotropic forms,such as $S_8$,$S_6$,and other cyclic structures.

(A) Selenium exists in two allotropic forms: red (non-metallic) and grey (metallic).

(B) The formation of ozone from oxygen is represented by the reaction: $3O_2(g) \rightarrow 2O_3(g)$.
This reaction is endothermic,meaning the enthalpy change $(\Delta H)$ is positive $(+142 \ kJ \ mol^{-1})$.
Since the number of moles of gas decreases from $3$ to $2$,the entropy change $(\Delta S)$ is negative.
$O_2$ is more stable than $O_3$ because $O_3$ is an endothermic compound.
$O_2$ is paramagnetic due to the presence of two unpaired electrons in its antibonding molecular orbitals,whereas $O_3$ is diamagnetic.

(A) Ozone $(O_3)$ acts as a strong oxidizing agent because it easily releases nascent oxygen $(O_3 \rightarrow O_2 + [O])$.
It is not a reducing agent.
Ozone has an angular (bent) shape with a bond angle of approximately $117^{\circ}$.
It acts as a good bleaching agent due to its oxidizing property.
It protects the Earth's surface from harmful ultraviolet $(UV)$ radiation by absorbing it in the stratosphere.

(B) $1$. Oxygen has the electronic configuration $[He] 2s^2 2p^4$,which contains $2$ unpaired electrons in the $2p$ orbitals.
$2$. Sulphur has the electronic configuration $[Ne] 3s^2 3p^4$,which also contains $2$ unpaired electrons in the $3p$ orbitals.
$3$. Oxygen is highly electronegative and typically exhibits an oxidation state of $-2$. It does not show $+4$ or $+6$ oxidation states because it lacks $d$-orbitals.
$4$. Sulphur can show $-2, +2, +4,$ and $+6$ oxidation states due to the availability of vacant $d$-orbitals.
$5$. Therefore,the statement that both oxygen and sulphur show $-2, +4,$ and $+6$ oxidation states is false.

(D) Group $16$ elements,also known as chalcogens,include Oxygen $(O)$,Sulfur $(S)$,Selenium $(Se)$,Tellurium $(Te)$,and Polonium $(Po)$.
Astatine $(At)$ is a radioactive element that belongs to Group $17$ (the halogens).

(A) Oxygen: Its hydride is water $(H_2O)$,which is a colourless,odourless liquid.
Nitrogen: Its hydride is ammonia $(NH_3)$,which is colourless but has a strong,pungent odour.
Sulphur: Its hydride is hydrogen sulphide $(H_2S)$,which is colourless but has a distinct,rotten-egg smell.
Selenium: Its hydride is hydrogen selenide $(H_2Se)$,which has an extremely unpleasant odour.
Therefore,only oxygen forms a hydride that is both colourless and odourless.

(B) The thermal stability of the hydrides of group $16$ elements depends on the bond dissociation energy of the $E-H$ bond.
As the size of the central atom increases down the group $(O < S < Se < Te)$,the bond length increases and the bond dissociation energy decreases.
Therefore,the thermal stability decreases in the order: $H_2O > H_2S > H_2Se > H_2Te$.
Thus,$H_2Te$ has the lowest thermal stability.

(A) The melting and boiling points of elements of group $16$ increase with an increase in atomic number due to an increase in the magnitude of van der Waals forces.
Therefore,the boiling point of sulfur is higher than that of oxygen.
Thus,the statement that the boiling point of sulfur is lower than oxygen is $FALSE$.

(C) The thermal stability of hydrides of group $16$ elements decreases down the group as the size of the central atom increases.
As the size of the central atom increases,the bond length between the central atom and hydrogen increases,which leads to a decrease in bond dissociation enthalpy.
The order of thermal stability is $H_2O > H_2S > H_2Se > H_2Te$.
Therefore,the correct order is $H_2Te < H_2Se < H_2S < H_2O$.

(C) Oleum is also known as fuming sulfuric acid.
It is formed by dissolving sulfur trioxide $(SO_3)$ in concentrated sulfuric acid $(H_2SO_4)$.
The chemical formula for Oleum is $H_2S_2O_7$ (also written as $H_2SO_4 \cdot SO_3$).

(D) In industry,sulphur dioxide is produced as a byproduct by roasting sulphide ores like zinc sulphide and iron pyrites in air.
$2 ZnS_{(s)} + 3 O_{2(g)} \xrightarrow{\Delta} 2 ZnO_{(s)} + 2 SO_{2(g)}$
$4 FeS_{2(s)} + 11 O_{2(g)} \xrightarrow{\Delta} 2 Fe_2O_{3(s)} + 8 SO_{2(g)}$

(D) In the contact process for the manufacture of sulphuric acid,$SO_3$ gas is absorbed in concentrated $H_2SO_4$ to form oleum $(H_2S_2O_7)$.
The chemical reaction is:
$SO_3 + H_2SO_4 \rightarrow H_2S_2O_7$ (oleum)

(C) $SO_2$ is used as a food preservative,as an antichlor,and in the manufacture of $H_2SO_4$ (via the contact process).
However,$SO_2$ does not form oleum with concentrated $H_2SO_4$.
Oleum $(H_2S_2O_7)$ is formed by the reaction of sulfur trioxide $(SO_3)$ with concentrated $H_2SO_4$.

(A) To identify the oxyacid with an $S-S$ linkage,let us examine the structures:
$1$. $H_{2}SO_{3}$ (Sulphurous acid): Structure is $(HO)_{2}S=O$. No $S-S$ bond.
$2$. $H_{2}S_{2}O_{4}$ (Dithionous acid): Structure is $(HO)S(=O)-S(=O)(OH)$. This contains an $S-S$ linkage.
$3$. $H_{2}S_{2}O_{5}$ (Disulphurous acid): Structure is $(HO)S(=O)-O-S(=O)_{2}(OH)$. This contains an $S-O-S$ linkage.
$4$. $H_{2}S_{2}O_{2}$ (Thiosulphurous acid): Structure is $H-S(=O)-S-H$. This contains an $S-S$ linkage,but $H_{2}S_{2}O_{4}$ is the standard oxyacid commonly cited for this property in textbooks.
Therefore,$H_{2}S_{2}O_{4}$ is the correct answer.

(D) The structure of thiosulphuric acid,$H_2S_2O_3$,is $HO-S(=O)(=S)-OH$.
It contains one $S=S$ bond and one $S=O$ bond.
Therefore,the correct option is $D$.

(B) When concentrated $H_2SO_4$ reacts with $PCl_5$,it undergoes a chlorination reaction to produce sulphuryl chloride $(SO_2Cl_2)$ along with phosphorus oxychloride $(POCl_3)$ and hydrogen chloride $(HCl)$.
The balanced chemical equation is:
$H_2SO_4 + 2 PCl_5 \longrightarrow SO_2Cl_2 + 2 POCl_3 + 2 HCl$

(C) The structure of $H_{2}S_{2}O_{6}$ (dithionic acid) is $HO-SO_{2}-SO_{2}-OH$.
In this structure,each sulphur atom is bonded to two oxygen atoms via double bonds $(S=O)$.
Therefore,there are a total of four $S=O$ bonds in $H_{2}S_{2}O_{6}$.
The structure is as follows:
$O=S(OH)(=O)-S(OH)(=O)=O$.

(D) Marshall's acid is also known as peroxydisulfuric acid.
Its chemical structure consists of two sulfur atoms linked by a peroxide bridge $(-O-O-)$.
The molecular formula of Marshall's acid is $H_{2}S_{2}O_{8}$.
Comparing this with the given options,option $D$ is correct.

(A) The most stable allotrope of sulphur is $Rhombic$ or $Octahedral$ $Sulphur$ $(S_8)$.
It is also known as $\alpha-sulphur$ and is the only form of sulphur that is stable at room temperature.

(A) The chalcogen family consists of elements in Group $16$ of the periodic table,which are Oxygen $(O)$,Sulfur $(S)$,Selenium $(Se)$,Tellurium $(Te)$,and Polonium $(Po)$.
Astatine $(At)$ belongs to Group $17$,which is the halogen family.
Therefore,$At$ does not belong to the chalcogen family.

(B) Ozone acts as a reducing agent in reactions where it reduces peroxides to oxides,while itself being reduced to oxygen.
In the reaction $BaO_{2(s)} + O_{3(g)} \rightarrow BaO_{(s)} + 2 O_{2(g)}$,ozone reduces barium peroxide $(BaO_2)$ to barium oxide $(BaO)$.

(A) Due to the high $H-O$ bond dissociation enthalpy,$H_2O$ does not act as a reducing agent.
In contrast,$H_2S$,$H_2Se$,and $H_2Te$ act as reducing agents.
The reducing character increases in the order: $H_2S < H_2Se < H_2Te$.
This increase in reducing power is due to the decrease in $H-E$ bond strength as the size of the central atom $E$ increases down the group.

(A) The general electronic configuration of group $16$ elements (chalcogens) is $ns^2 np^4$.
Among the given options,$Se$ (Selenium) belongs to group $16$ with the valence shell configuration $4s^2 4p^4$.
$Br$ belongs to group $17$,while $Xe$ and $Kr$ belong to group $18$.

(B) The reaction between lead dioxide $(PbO_2)$ and concentrated nitric acid $(HNO_3)$ is a redox reaction.
$PbO_2$ acts as an oxidizing agent and oxidizes water to oxygen gas.
The balanced chemical equation is:
$2PbO_2 + 4HNO_3 \rightarrow 2Pb(NO_3)_2 + 2H_2O + O_2$
Thus,the gas evolved is oxygen $(O_2)$.

(C) When excess $PCl_{5}$ reacts with concentrated $H_{2}SO_{4}$,the hydroxyl groups of the acid are replaced by chlorine atoms to form sulphuryl chloride $(SO_{2}Cl_{2})$.
The chemical reaction is as follows:
$SO_{2}(OH)_{2} + 2PCl_{5} \rightarrow SO_{2}Cl_{2} + 2POCl_{3} + 2HCl$

(C) The reaction between excess $PCl_{5}$ and concentrated $H_{2}SO_{4}$ is as follows:
$SO_{2}(OH)_{2} + 2PCl_{5} \rightarrow SO_{2}Cl_{2} + 2POCl_{3} + 2HCl$
In this reaction,$PCl_{5}$ acts as a chlorinating agent and replaces the hydroxyl groups of $H_{2}SO_{4}$ with chlorine atoms to form sulphuryl chloride $(SO_{2}Cl_{2})$.

(A) $P_{2}O_{5}$ is a powerful dehydrating agent. When concentrated $H_{2}SO_{4}$ is heated with $P_{2}O_{5}$,it undergoes dehydration to form sulphur trioxide $(SO_{3})$.
The chemical reaction is as follows:
$2H_{2}SO_{4} + P_{2}O_{5} \rightarrow 2SO_{3} + 2H_{3}PO_{4}$ (or $2H_{2}SO_{4} + 2P_{2}O_{5} \rightarrow 2SO_{3} + 4HPO_{3}$).