Oxygen family Questions in English

(C) The $5^{th}$ period starts after the $4^{th}$ noble gas,which is Krypton ($Kr$,$Z=36$).
Group $16$ elements in the $5^{th}$ period follow the configuration of the noble gas $Kr$ $([Kr] 4d^{10} 5s^2 5p^4)$.
The atomic number is calculated as $36 (Kr) + 2 (5s^2) + 10 (4d^{10}) + 4 (5p^4) = 52$.
Thus,the element is Tellurium $(Te)$ with atomic number $52$.

(D) For peroxodisulphuric acid $(H_2S_2O_8)$: The structure contains $4$ $S=O$ bonds and $2$ $S-OH$ bonds.
For pyrosulphuric acid $(H_2S_2O_7)$: The structure contains $4$ $S=O$ bonds and $2$ $S-OH$ bonds.
Therefore,the number of $S=O$ and $S-OH$ bonds in $H_2S_2O_8$ are $4$ and $2$ respectively,and in $H_2S_2O_7$ are $4$ and $2$ respectively.

(C) The number of $\pi$-bonds in the structures are as follows:
$1.$ $H_2SO_3$ (Sulfurous acid): It has one $S=O$ double bond,so it contains $1$ $\pi$-bond.
$2.$ $H_2SO_4$ (Sulfuric acid): It has two $S=O$ double bonds,so it contains $2$ $\pi$-bonds.
$3.$ $H_2S_2O_7$ (Pyrosulfuric acid): It has four $S=O$ double bonds (two on each sulfur atom),so it contains $4$ $\pi$-bonds.
Therefore,the correct decreasing sequence of the number of $\pi$-bonds is $H_2S_2O_7 (4) > H_2SO_4 (2) > H_2SO_3 (1)$.

(A) The $S_6$ molecule has a chair-form hexagon ring with the same bond length as that in $S_8$,but with somewhat smaller bond angles. Therefore,the bond angles in $S_8$ (crown shape,$\approx 107^{\circ}$) and $S_6$ (chair shape,$\approx 102^{\circ}$) are different. Thus,statement $A$ is incorrect.

(B) When $KMnO_4$ is added to concentrated $H_2SO_4$,it reacts to form manganese heptoxide $(Mn_2O_7)$.
$2KMnO_4 + 2H_2SO_4 \rightarrow Mn_2O_7 + H_2O + 2KHSO_4$
$Mn_2O_7$ is a green,oily,and highly explosive liquid.

(C) To determine which oxoacid does not contain a $S-S$ bond,we examine their structures:
$1$. $H_2S_2O_3$ (Thiosulphuric acid): Contains a $S-S$ bond $(HO-SO_2-SH)$.
$2$. $H_2S_2O_4$ (Dithionous acid): Contains a $S-S$ bond $(HO-SO-SO-OH)$.
$3$. $H_2S_2O_7$ (Pyrosulphuric acid or Oleum): Its structure is $HO-SO_2-O-SO_2-OH$. It contains a $S-O-S$ linkage,not a $S-S$ bond.
$4$. $H_2S_4O_6$ (Tetrathionic acid): Contains a chain of sulphur atoms ($S-S-S-S$ linkage).
Therefore,$H_2S_2O_7$ is the correct answer.

(C) The thermal decomposition of ferric sulphate $(Fe_2(SO_4)_3)$ yields ferric oxide and sulphur trioxide: $Fe_2(SO_4)_3 \xrightarrow{\Delta} Fe_2O_3 + 3SO_3$.
The thermal decomposition of ferrous sulphate heptahydrate $(FeSO_4 \cdot 7H_2O)$ first yields anhydrous $FeSO_4$,which then decomposes to ferric oxide,sulphur dioxide,and sulphur trioxide: $2FeSO_4 \cdot 7H_2O \xrightarrow{\Delta} Fe_2O_3 + SO_2 + SO_3 + 14H_2O$.
Since both reactions produce $SO_3$,the correct option is $C$.

(B) The structure of peroxodisulphate ion $(S_2O_8^{2-})$ contains a peroxide linkage $(-O-O-)$ between the two sulphur atoms.
In this ion,the structure is $O_3S-O-O-SO_3^{2-}$.
Other options like $S_2O_3^{2-}$ (thiosulphate),$S_2O_5^{2-}$ (pyrosulphite),and $S_2O_7^{2-}$ (pyrosulphate) do not contain a peroxide linkage.

(B) The acidity of hydrides of group $16$ elements depends on the bond dissociation enthalpy of the $H-E$ bond.
As we move down the group,the size of the central atom increases,which leads to a decrease in the $H-E$ bond dissociation enthalpy.
Consequently,the ease of releasing $H^+$ ions increases,meaning the acidity increases down the group.
The correct order of acidity $(K_a)$ is: $H_2Te > H_2Se > H_2S > H_2O$.

(C) The acidity of hydrides of group $16$ elements depends on the bond dissociation enthalpy.
As we move down the group,the size of the central atom increases,which leads to a decrease in the bond dissociation enthalpy.
Therefore,the $H-S$ bond is weaker than the $O-H$ bond,making it easier for $H_2S$ to release $H^+$ ions compared to $H_2O$.

(D) Peroxy linkage means that in a compound,there should be a bond between two oxygen atoms $(-O-O-)$.
Among the given oxoacids of sulfur,only $H_2SO_5$ (peroxomonosulphuric acid) and $H_2S_2O_8$ (peroxodisulphuric acid) contain a peroxy linkage.

(C) Group $16$ elements are known as chalcogens.
The elements in group $16$ are Oxygen $(Z=8)$,Sulphur $(Z=16)$,Selenium $(Z=34)$,Tellurium $(Z=52)$,and Polonium $(Z=84)$.
Therefore,the set $8, 16, 34, 52$ corresponds to elements of group $16$.

(A) The structures of the given ions are as follows:
$1$. $S_2O_8^{2-}$ (Peroxodisulphate ion): Contains a peroxide linkage $(S-O-O-S)$. It does not have an $S-S$ bond.
$2$. $S_2O_6^{2-}$ (Dithionate ion): Contains an $S-S$ bond $(O_3S-SO_3^{2-})$.
$3$. $S_2O_5^{2-}$ (Pyrosulphite ion): Contains an $S-S$ bond $(O_2S-SO_3^{2-})$.
$4$. $S_2O_3^{2-}$ (Thiosulphate ion): Contains an $S-S$ bond $(O_3S-S^{2-})$.
Therefore,$S_2O_8^{2-}$ is the ion that does not have an $S-S$ linkage.
Hence,option $A$ is correct.

(B) The thermal stability of hydrides of the oxygen family $(H_2O, H_2S, H_2Se, H_2Te)$ decreases regularly on moving down the group due to the increase in bond length and decrease in bond dissociation enthalpy.
Specifically,the order of thermal stability is: $H_2O > H_2S > H_2Se > H_2Te$.
Other properties like melting point,boiling point,and critical temperature do not change regularly due to the anomalous behavior of $H_2O$ caused by hydrogen bonding.

(D) Sodium thiosulphate $(Na_2S_2O_3)$ can be prepared by the following methods:
$1$. Boiling a solution of sodium sulphite $(Na_2SO_3)$ with elemental sulphur: $Na_2SO_3 + S \rightarrow Na_2S_2O_3$.
$2$. Neutralisation of thiosulphuric acid $(H_2S_2O_3)$ with sodium hydroxide $(NaOH)$: $H_2S_2O_3 + 2NaOH \rightarrow Na_2S_2O_3 + 2H_2O$.
Since both methods $(B)$ and $(C)$ produce sodium thiosulphate,the correct option is $(D)$.

(D) $(1)$ In the gas phase,$SO_2$ has $sp^2$ hybridization with one lone pair on the sulphur atom,resulting in a $V$-shaped (bent) geometry. This statement is correct.
$(2)$ In the gas phase,$SO_3$ has $sp^2$ hybridization with no lone pairs on the sulphur atom,resulting in a trigonal planar geometry. This statement is correct.
$(3)$ $\gamma-SO_3$ exists as a cyclic trimer,$(SO_3)_3$,where sulphur atoms are linked through oxygen bridges. This statement is correct.
Therefore,all three statements are correct.

(C) The correct order of Arrhenius acid character is $H_2S < H_2Se < H_2Te$.
As we move down the group,the bond dissociation energy of $H-E$ bonds decreases,making it easier to release $H^+$ ions.
Therefore,$H_2Te$ is the strongest acid among the given hydrides,and the order provided in option $C$ is incorrect.

(C) Ozone $(O_3)$ is a diamagnetic molecule because all its electrons are paired. Therefore,it is not paramagnetic.

(C) Among the given elements,$Sulfur$ exhibits the maximum tendency for catenation.
Due to this property,it forms the maximum number of polyanions (e.g.,$S_n^{2-}$).

(C) The bond dissociation energy depends on the bond strength. Among the group $16$ elements,the $S - S$ bond is the strongest. The $O - O$ bond is weaker than the $S - S$ bond due to the small size of the oxygen atom,which leads to significant inter-electronic repulsion between the lone pairs of electrons on the bonded atoms. Thus,the order of bond dissociation energy is $S - S > Se - Se > Te - Te > O - O$.

(B) Oxygen $(O)$ is the best oxidizing agent among the given elements.
This is due to its high electronegativity and small atomic size,which allows it to gain electrons more readily than the other elements in the group.

(A) Polonium $(Po)$ is the only element in group $16$ that is a metal.
It exists in a metallic lattice in its monoatomic state.
It also exhibits two crystalline forms,known as $\alpha-Po$ and $\beta-Po$.

(C) $O_2$ is paramagnetic due to the presence of two unpaired electrons in its antibonding molecular orbitals. $O_3$ is diamagnetic as all its electrons are paired. Therefore,the statement that both are paramagnetic is incorrect.

(C) Selenium is used as a photoconductor in xerox machines due to its light-sensitive properties.

(A) When $H_2S$ is burned in a limited supply of oxygen,it forms water and sulfur:
$2H_2S + O_2 \to 2H_2O + 2S$
When burned in excess oxygen,it forms sulfur dioxide:
$2H_2S + 3O_2 \to 2H_2O + 2SO_2$
Given the options,the reaction producing water and sulfur is the standard representation for the combustion of $H_2S$ in limited oxygen.

(D) The statement that ozone protects organisms by absorbing gamma radiation is incorrect. $O_3$ protects the Earth's surface from harmful ultraviolet $(UV)$ radiation,not gamma radiation.

(C) The structure of dithionic acid $(H_2S_2O_6)$ is $HO-SO_2-SO_2-OH$.
In this structure,there is a direct $S - S$ bond between the two sulfur atoms.
$H_2S_2O_7$ (pyrosulfuric acid) has an $S - O - S$ linkage.
$H_2S_2O_8$ (peroxydisulfuric acid) has an $S - O - O - S$ linkage.
Therefore,the correct compound is $H_2S_2O_6$.

(B) The thermal decomposition of potassium chlorate $(KClO_3)$ occurs as follows:
$2KClO_3 \xrightarrow{\Delta} 2KCl + 3O_2$
Thus,heating $KClO_3$ yields potassium chloride $(KCl)$ and oxygen gas $(O_2)$.

(C) Oleum is obtained by dissolving $SO_3$ in $H_2SO_4$.
Its chemical formula is $H_2S_2O_7$.

(D) Ozone $(O_3)$ acts as a powerful oxidizing agent due to the liberation of nascent oxygen $(O_3 \rightarrow O_2 + [O])$.
Because of this strong oxidizing property,it acts as an effective bleaching agent for substances like ivory,delicate fabrics,and oils.
Therefore,the statement that it cannot act as a bleaching agent is incorrect.

(A) Rhombic sulfur $(S_8)$ is the most stable crystalline form of sulfur at room temperature.

(B) Among the given elements,$Te$ (Tellurium) is a metalloid and exhibits significantly higher electrical resistance compared to metals like $Pb$,$Fe$,and $Po$ at room temperature.

(B) Sulfur dioxide acts as a reducing agent when it reacts with bromine water. The balanced chemical equation for the reaction is: $SO_2 + Br_2 + 2H_2O \to H_2SO_4 + 2HBr$. Thus,the products formed are sulfuric acid $(H_2SO_4)$ and hydrobromic acid $(HBr)$.

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

(C) When ozone $(O_3)$ reacts with mercury $(Hg)$,it oxidizes mercury to mercurous oxide $(Hg_2O)$ and releases oxygen $(O_2)$.
The balanced chemical equation is:
$O_3 + 2Hg \to Hg_2O + O_2$

(C) The structure of thiosulfuric acid $(H_2S_2O_3)$ contains a sulfur-sulfur double bond $(S=S)$.
In $H_2S_2O_3$,one oxygen atom of sulfuric acid $(H_2SO_4)$ is replaced by a sulfur atom,resulting in the $S=S$ linkage.

(A) In the solid state,$SO_3$ exists as a cyclic trimer,which is represented as $(SO_3)_3$ or $S_3O_9$.

(A) The chemical formula for thiosulfuric acid is $H_2S_2O_3$.
In this structure,one oxygen atom of sulfuric acid $(H_2SO_4)$ is replaced by a sulfur atom.
The central sulfur atom is bonded to two hydroxyl groups $(-OH)$,one double-bonded oxygen $(=O)$,and one double-bonded sulfur $(=S)$.
Thus,the structure is $HO-S(=O)(=S)-OH$.

(A) The tetrathionate ion is represented by the formula $S_4O_6^{2-}$.
Its structure consists of a chain of four sulfur atoms,where the two terminal sulfur atoms are each bonded to three oxygen atoms (one double-bonded and two single-bonded with negative charges),and the two central sulfur atoms are bonded to each other.
The correct structure is shown in option $A$.

(A) The elements of group $16$,excluding polonium,are called chalcogens (ore-forming elements) because a large number of metal ores are found as oxides and sulfides of these elements.

(A) $SF_6$ is a sterically hindered molecule where the sulfur atom is surrounded by six fluorine atoms,making it kinetically inert towards hydrolysis by water. Thus,$SF_6$ does not react with water.

(A) The dithionate ion is represented by the formula ${S_2O_6^{2-}}$.
In this structure,two sulfur atoms are directly bonded to each other ($S-S$ bond).
Each sulfur atom is further bonded to three oxygen atoms.
The correct structure is shown in option $A$.

(B) Oleum is pyrosulphuric acid,which has the chemical formula $H_2S_2O_7$.
Its structure consists of two $SO_3$ units linked by an oxygen atom,with two hydroxyl groups attached to the sulfur atoms.
The structure is:
$HO-S(=O)_2-O-S(=O)_2-OH$.

(D) $P_4O_{10}$ is a powerful dehydrating agent.
When concentrated $H_2SO_4$ is heated with $P_4O_{10}$,it undergoes dehydration to form sulfur trioxide $(SO_3)$.
The reaction is: $H_2SO_4 + P_4O_{10} \rightarrow SO_3 + H_3PO_4$ (or simply $H_2SO_4 \xrightarrow{P_4O_{10}} SO_3 + H_2O$).
$SO_3$ is also known as the anhydride of $H_2SO_4$.

(A) An oxo group is defined as the $M=O$ bond.

$A$. $H_2SO_4$: Contains $2$ oxo groups ($O=S=O$ structure).	$B$. $H_2SO_3$: Contains $1$ oxo group ($O=S$ structure).
$C$. $H_3PO_3$: Contains $1$ oxo group ($O=P$ structure).	$D$. $H_3PO_4$: Contains $1$ oxo group ($O=P$ structure).

Comparing the structures,$H_2SO_4$ has the highest number of $M=O$ bonds,which is $2$.

(C) $SO_2$ acts as a bleaching agent due to its reducing property. It removes oxygen from colored substances to make them colorless.

(B) In the contact process for the manufacture of sulfuric acid,the gas mixture is passed through a Tyndall box to detect the presence of dust particles. If dust particles are present,they scatter light,making the beam visible,which indicates that the purification process is incomplete.

(B) Mercury$(II)$ oxide $(HgO)$ decomposes upon moderate heating to release oxygen gas. The reaction is: $2HgO \xrightarrow{450\,^{\circ}C} 2Hg + O_2$.

(C) The hydrolysis of peroxodisulfuric acid $(H_2S_2O_8)$ proceeds as follows:
$H_2S_2O_8 + H_2O \to H_2SO_4 + H_2SO_5$
In this reaction,one mole of peroxodisulfuric acid reacts with one mole of water to produce one mole of sulfuric acid $(H_2SO_4)$ and one mole of peroxomonosulfuric acid $(H_2SO_5)$.

(D) The oxidizing nature of $H_2SO_4$ is demonstrated by its ability to oxidize other substances while being reduced itself.
In the reaction $2HI + H_2SO_4 \to I_2 + SO_2 + 2H_2O$,the oxidation state of $I$ increases from $-1$ in $HI$ to $0$ in $I_2$,and the oxidation state of $S$ decreases from $+6$ in $H_2SO_4$ to $+4$ in $SO_2$.
Thus,$H_2SO_4$ acts as an oxidizing agent.
Options $A$,$B$,and $C$ represent acid-base or substitution reactions where the oxidation states of the elements do not change.