Halogen family Questions in English

(D) The strongest oxoacid of chlorine is perchloric acid,which has the chemical formula $HClO_4$.
In the structure of $HClO_4$,the chlorine atom is central and is bonded to four oxygen atoms.
Specifically,one oxygen atom is part of an $OH$ group,and the other three oxygen atoms are double-bonded to the chlorine atom $(Cl=O)$.
Therefore,the total number of oxygen atoms bonded to chlorine is $4$.

(B) Fluorine $(F)$ is the most electronegative element in the periodic table.
It has a small atomic size and lacks $d$-orbitals in its valence shell.
Due to its high electronegativity,it can only gain one electron to complete its octet and cannot exhibit positive oxidation states or expand its octet.
Therefore,it always exhibits an oxidation state of $-1$ in all its compounds.

(D) The thermal stability of interhalogen compounds depends on the difference in electronegativity between the two halogen atoms. $A$ larger difference in electronegativity leads to a more stable bond.
Thermal stability decreases in the order: $ClF > ICl > IBr > BrCl > BrF$.
Therefore,the compound with the highest thermal stability is $ClF$.

(B) The thermal stability of interhalogen compounds depends upon the difference in electronegativity between the combining atoms. $A$ larger electronegativity difference leads to a stronger bond and higher thermal stability.
The electronegativity values are: $F = 4.0, Cl = 3.2, Br = 3.0, I = 2.7$.
The electronegativity differences are:
$ClF: |4.0 - 3.2| = 0.8$
$BrCl: |3.2 - 3.0| = 0.2$
$IBr: |3.0 - 2.7| = 0.3$
$ICl: |3.2 - 2.7| = 0.5$
Comparing these differences,the order of stability is $ClF > ICl > IBr > BrCl$.

(D) The physical states of the given interhalogen compounds at $25^{\circ} C$ are as follows:
$ICl$ is a ruby-red or brown-red solid.
$IBr$ is a black solid.
$IF_3$ is a yellow powder (solid).
$IF_5$ is a colourless liquid (Note: While $IF_5$ is often described as a liquid at room temperature,among the given options,it is the one that is not a solid,as it has a melting point of $9.4^{\circ} C$ and exists as a liquid at $25^{\circ} C$).

(C) The bond dissociation enthalpy of $H-X$ bond decreases in the order: $HF > HCl > HBr > HI$.
Thermal stability of hydrogen halides is directly proportional to the bond dissociation enthalpy.
Therefore,the thermal stability decreases in the order: $HF > HCl > HBr > HI$.
Thus,$HI$ has the lowest thermal stability among the given options.

(C) Based on the physical states of interhalogen compounds at room temperature $(25^{\circ} C)$:
$1$. $ClF$ is a colourless gas.
$2$. $BrF$ is a pale brown gas.
$3$. $ClF_3$ is a colourless gas.
$4$. $IF_3$ is a yellow powder (solid).
Therefore,$IF_3$ is the compound that is not in the gaseous phase at $25^{\circ} C$.

(D) The acid strength of oxoacids of chlorine increases with the increase in the oxidation state of the central chlorine atom.
The increasing order of acid strength is: $HOCl < HClO_2 < HClO_3 < HClO_4$.
In $HClO_4$,the chlorine atom is in the $+7$ oxidation state,which makes it the most stable conjugate base due to the maximum dispersal of negative charge.
Therefore,perchloric acid $(HClO_4)$ is the strongest acid among the given oxoacids of chlorine.

(C) The acidic strength of hydrohalic acids $(HX)$ depends on the bond dissociation enthalpy of the $H-X$ bond.
As the size of the halogen atom increases from $F$ to $I$,the bond length increases and the bond dissociation enthalpy decreases.
Therefore,the acidic strength increases in the order: $HF < HCl < HBr < HI$.
Hence,the weakest halogen acid is $HF$.

(B) Thermal stability of hydrogen halides decreases down the group as the bond dissociation energy decreases.
As the size of the halogen atom increases from $F$ to $I$,the bond length of $H-X$ increases,resulting in a decrease in bond strength.
Therefore,the order of thermal stability is $HF > HCl > HBr > HI$.
Thus,$HI$ is the thermally least stable hydrogen halide.

(D) The reaction of bromine with fluorine produces interhalogen compounds based on the conditions of temperature and the ratio of reactants.
$Br_2 + F_2 (excess) \rightarrow 2BrF$
$Br_2 + 3F_2 \rightarrow 2BrF_3$
$Br_2 + 5F_2 \rightarrow 2BrF_5$
$BrF_7$ is not formed because bromine is not large enough to accommodate seven fluorine atoms around it due to steric hindrance.

(D) When chlorine gas reacts with hot and concentrated sodium hydroxide $(NaOH)$,it undergoes a disproportionation reaction.
The balanced chemical equation for this reaction is:
$3Cl_2 + 6NaOH \rightarrow 5NaCl + NaClO_3 + 3H_2O$
Thus,the products obtained are sodium chloride $(NaCl)$,sodium chlorate $(NaClO_3)$,and water $(H_2O)$.

(C) At room temperature,$F_2$ and $Cl_2$ are gases,$Br_2$ is a liquid,and $I_2$ and $At$ are solids. Therefore,the correct answer is $Bromine$.

(C) $IF_3$ (Iodine trifluoride) is an interhalogen compound.
It is a yellow solid at room temperature,whereas $ClF$,$IF_7$,and $ClF_3$ are colourless gases.

(C) The acidic strength of hydrogen halides depends on the bond dissociation enthalpy of the $H-X$ bond.
As we move down the group from $F$ to $I$,the atomic size increases,which leads to a decrease in the bond dissociation enthalpy.
Consequently,the ease of releasing an $H^{+}$ ion increases down the group.
Therefore,the acidic strength increases in the order: $HF < HCl < HBr < HI$.
Thus,$HF$ has the lowest acidic strength.

(C) The oxidation states of $Cl$ in the given acids are as follows:
$1$. Hypochlorous acid $(HClO)$: $+1$
$2$. Chlorous acid $(HClO_2)$: $+3$
$3$. Chloric acid $(HClO_3)$: $+5$
$4$. Perchloric acid $(HClO_4)$: $+7$
Therefore,the $Cl$ atom in Perchloric acid $(HClO_4)$ has the highest oxidation state of $+7$.

(A) The thermal stability of oxyacids of chlorine increases with an increase in the oxidation state of the chlorine atom.
The oxidation states of chlorine in the given oxyacids are:
$HClO$ $(+1)$,$HClO_{2}$ $(+3)$,$HClO_{3}$ $(+5)$,and $HClO_{4}$ $(+7)$.
Therefore,the order of thermal stability is $HClO < HClO_{2} < HClO_{3} < HClO_{4}$.
Thus,$HClO_{4}$ has the highest thermal stability.

(D) $I$. Haber process: $NH_3$ production.
$II$. Ostwald's process: $HNO_3$ production.
$III$. Deacon process: $Cl_2$ production by oxidation of $HCl$ gas.
$IV$. Contact process: $H_2SO_4$ production.

(B) The reactivity of halogens towards dihydrogen decreases down the group. Fluorine is the most reactive halogen and reacts with dihydrogen even in the dark at very low temperatures. Therefore,$F_2$ combines with $H_2$ at the lowest temperature.

(A) $Cl_{2}$ acts as a powerful bleaching and oxidising agent due to the formation of nascent oxygen in the presence of moisture.
The reaction is: $Cl_{2} + H_{2}O \rightarrow HCl + HOCl$
$HOCl \rightarrow HCl + [O]$
This nascent oxygen $[O]$ is responsible for the bleaching action of $Cl_{2}$.

(D) Fluorine $(F)$ is the most electronegative element and has a small atomic size.
It lacks $d$-orbitals in its valence shell,which prevents it from expanding its octet or forming polyhalide ions like $I_3^-$ or $Br_3^-$.
Therefore,$F$ does not form polyhalide ions.

(B) $Cl$ occurs commonly in both the earth's crust and seawater.
The most common compound of chlorine in seawater is $NaCl$,and a small amount of $KCl$ is also present.
The common minerals of $Cl$ are Rocksalt $(NaCl)$,Sylvine $(KCl)$,and Carnallite $(KCl \cdot MgCl_2 \cdot 6H_2O)$.
Cryolite $(Na_3AlF_6)$ is a mineral of fluorine,not chlorine.

(C) To determine the number of lone pairs on the chlorine atom,we examine the oxidation state and bonding of chlorine in each oxyacid:
$1$. $HOCl$ $(HClO)$: Chlorine is in the $+1$ oxidation state. It forms $1$ single bond with $O$ and has $3$ lone pairs.
$2$. $HOClO$ $(HClO_{2})$: Chlorine is in the $+3$ oxidation state. It forms $1$ single bond with $OH$ and $1$ double bond with $O$,leaving $2$ lone pairs.
$3$. $HOClO_{2}$ $(HClO_{3})$: Chlorine is in the $+5$ oxidation state. It forms $1$ single bond with $OH$ and $2$ double bonds with $O$,leaving $1$ lone pair.
$4$. $HOClO_{3}$ $(HClO_{4})$: Chlorine is in the $+7$ oxidation state. It forms $1$ single bond with $OH$ and $3$ double bonds with $O$. Since chlorine has $7$ valence electrons and all are involved in bonding,there are $0$ lone pairs on the chlorine atom.
Therefore,$HOClO_{3}$ is the correct answer.

(A) The oxidizing power of oxoacids of chlorine depends on the oxidation state of the central chlorine atom.
Lower oxidation state of the central atom corresponds to higher oxidizing power.
The oxidation states of $Cl$ in $HClO$,$HClO_{2}$,and $HClO_{3}$ are $+1$,$+3$,and $+5$ respectively.
Therefore,the decreasing order of oxidizing power is $HClO > HClO_{2} > HClO_{3}$.

(C) The correct answer is $I$ and $F$.
Interhalogen compounds of the type $XX_{7}^{\prime}$ are formed when a large halogen atom $(X)$ reacts with a smaller,highly electronegative halogen atom $(X^{\prime})$.
Iodine heptafluoride $(IF_{7})$ is the only known interhalogen compound of this type.
It is formed by the reaction of iodine $(I)$ with fluorine $(F)$.
According to $VSEPR$ theory,$IF_{7}$ has a pentagonal bipyramidal structure.

(B) $Cl_2$ (Chlorine) is a common bleaching agent used in the paper industry to whiten wood pulp.
It acts as an oxidizing agent to remove lignin,which gives wood its brown color.
While chlorine dioxide $(ClO_2)$ is often preferred in modern $ECF$ (Elemental Chlorine Free) bleaching processes to reduce the formation of organochlorides,$Cl_2$ is historically and industrially recognized as the primary bleaching agent for this purpose.

(D) When halide ions combine with halogen molecules or interhalogen compounds,polyhalide ions are formed.
Fluorine $(F)$ does not form polyhalide ions because it lacks $d-$orbitals and cannot expand its octet or show higher oxidation states to accommodate the extra halogen atoms.

(D) The general electronic configuration of Group-$17$ elements is $ns^2 np^5$.
Except for fluorine,all other elements in this group can exhibit higher oxidation states by utilizing their vacant $d$-orbitals.
They can promote electrons from the $p$ and $s$ orbitals to the $d$-orbitals,allowing them to show oxidation states up to $+7$ (e.g.,in $HClO_4$ or $IF_7$).

(B) The order $HI < HBr < HCl < HF$ represents the increasing trend for thermal stability,ionic character,and dipole moment.
However,the reducing power of hydrogen halides increases as the bond dissociation enthalpy decreases,which follows the order $HF < HCl < HBr < HI$.
Therefore,the property that does not correspond to the given order is reducing power.

(B) Glass contains silica $(SiO_{2})$,which reacts with hydrofluoric acid $(HF)$.
$SiO_{2} + 4HF \longrightarrow SiF_{4} + 2H_{2}O$
$SiF_{4} + 2HF \longrightarrow H_{2}SiF_{6}$ (fluorosilicic acid)
$HF$ is commonly used for the etching of glass.

(D) The ionic character of metal halides follows the order $MF > MCl > MBr > MI$,where $M$ is a monovalent metal.
According to Fajan's rule,the smaller the size of the anion,the greater is the ionic character of the bond.
Since the fluoride ion $(F^-)$ is the smallest among the halide ions,it forms metal halides with the highest ionic character.

(C) In the reaction of $NaOH$ with $Cl_2$,hydrogen gas $(H_2)$ is not produced. The reaction depends on the concentration of $NaOH$.
For cold and dilute $NaOH$,the reaction is: $2NaOH + Cl_2 \rightarrow NaCl + NaClO + H_2O$.
For hot and concentrated $NaOH$,the reaction is: $6NaOH + 3Cl_2 \rightarrow 5NaCl + NaClO_3 + 3H_2O$.
Option $C$ is incorrect because it suggests the liberation of $H_2$ and $O_2$ gases,which does not occur.

(A) $MnO_{2} + 4 HCl \xrightarrow{\Delta} MnCl_{2} + 2 H_{2}O + Cl_{2(g)} (A)$
$Cl_{2(g)} + 3 F_{2} (\text{excess}) \xrightarrow{573 K} 2 ClF_{3(g)} (B)$
$3 ClF_{3(g)} + U_{(s)} \xrightarrow{323-363 K} UF_{6(g)} (C) + 3 ClF_{(g)} (D)$
Therefore,the gases $A$,$B$,$C$,and $D$ are $Cl_{2}$,$ClF_{3}$,$UF_{6}$,and $ClF$ respectively.

(C) $HF$ exhibits strong intermolecular hydrogen bonding,which results in a significantly higher boiling point compared to other hydrogen halides.
For the remaining hydrogen halides $(HCl, HBr, HI)$,the boiling point is primarily determined by the magnitude of van der Waals' forces,which increase with an increase in molecular mass.
Therefore,the order of boiling points is $HI > HBr > HCl$.
Combining these,the overall decreasing order of boiling points is $HF > HI > HBr > HCl$.

(C) Down the group,electron gain enthalpies become less negative with the increase in the size of the halogen. However,the electron gain enthalpy of $F$ is less negative than $Cl$. This is due to the small size of the $F$-atom. As a result,there are strong inter-electronic repulsions in the relatively small $2p$-orbitals of $F$,and thus incoming electrons experience much less attraction. Thus,the correct order is $Cl > F > Br > I$. Therefore,the option $F > Cl > Br > I$ for electron gain enthalpy is incorrect.

(D) Fluorine is the most electronegative element in the periodic table,which makes most of its reactions highly exothermic.
It forms only one oxoacid,which is $HOF$ (hypofluorous acid).
Due to the small size of the fluorine atom,the lone pairs on the two fluorine atoms in $F_2$ experience strong inter-electronic repulsion,resulting in a very low $F-F$ bond dissociation enthalpy.
Therefore,the statement that fluorine has a high $F-F$ bond dissociation enthalpy is incorrect.

(A) $F_2$ is a very strong oxidizing agent $(E^\circ = +2.87 \ V)$ and oxidizes water to oxygen: $2F_2 + 2H_2O \rightarrow 4HF + O_2$.
In contrast,$Cl_2$ reacts with water to form a mixture of $HCl$ and $HOCl$.
Fluoride $(F^-)$ is a weak reducing agent,not an oxidizing agent.
$F_2$ is the strongest oxidizing agent among the halogens.

(B) $1$. Oxidising power: The order $F_2 > Cl_2 > Br_2 > I_2$ is correct due to the high reduction potential of fluorine.
$2$. Ionic character of metal halide: According to Fajan's rule,the order of ionic character is $MF > MCl > MBr > MI$. The given order $MI > MBr > MCl > MF$ is incorrect.
$3$. Bond dissociation enthalpy: The correct order is $Cl_2 > Br_2 > F_2 > I_2$. The $F-F$ bond is weaker than $Cl-Cl$ due to inter-electronic repulsion between lone pairs in the small $F_2$ molecule. The given order $F_2 > Cl_2 > Br_2 > I_2$ is incorrect.
$4$. Hydrogen-halogen bond strength: The order $HI < HBr < HCl < HF$ is correct due to the decrease in bond length as the size of the halogen decreases.
Thus,statements $II$ and $III$ are incorrect.

(D) Fused silver nitrate $(AgNO_3)$,shaped into sticks,was traditionally called "lunar caustic".
It is used as a cauterizing agent in medicine.

(C) The van Arkel method for the refining of $Ti$ metal involves the formation of a volatile iodide,$TiI_4$,which is subsequently decomposed to yield pure metal. Thus,$X_2$ is $I_2$.
$Y_2$ does not liberate $O_2$ from water and does not form $HY$ and $HOY$ with water. $I_2$ is the only halogen that does not react with water to produce $O_2$ or form $HX$ and $HOX$ species under standard conditions. Therefore,$Y_2$ is also $I_2$.
The reactions are:
$(i)$ $\text{Impure } Ti + 2I_2$ $\rightarrow TiI_4$ $\xrightarrow{\Delta} Ti + 2I_2$
(ii) $I_2 + H_2O \rightarrow \text{No reaction to form } O_2, HI, \text{ or } HOI$.
Thus,both $X_2$ and $Y_2$ are $I_2$. The correct option is $(C)$.

(C) Hydrofluoric acid $(HF)$ reacts with silicon dioxide $(SiO_2)$ to form silicon tetrafluoride $(SiF_4)$ and water,or hexafluorosilicic acid $(H_2SiF_6)$ in excess $HF$.
The reaction is: $SiO_2 + 4HF \rightarrow SiF_4 + 2H_2O$.
Because of this property,$HF$ is used to etch glass.
Hence,the correct option is $(C)$.

(A) The reaction of phosphorus with sulfuryl chloride $(SO_2Cl_2)$ is: $P_4 + 10 SO_2Cl_2 \longrightarrow 4 PCl_5 + 10 SO_2$ $(X = SO_2)$.
At $723 \ K$,the Deacon process for the manufacture of chlorine is: $4 HCl + O_2 \xrightarrow[CuCl_2]{723 \ K} 2 Cl_2 + 2 H_2O$ $(Y = Cl_2)$.
$Cl_2$ $(Y)$ reacts with $SO_2$ $(X)$ in the presence of water to form hydrochloric acid $(A)$ and sulfuric acid $(B)$: $SO_2 + 2 H_2O + Cl_2 \longrightarrow 2 HCl + H_2SO_4$.

(D) Deacon's process is used for the industrial preparation of chlorine gas $(Cl_2)$.
Thus,gas $X$ is $Cl_2$.
When chlorine reacts with iodine and water,it acts as an oxidizing agent and oxidizes iodine to iodic acid $(HIO_3)$.
The chemical reaction is:
$I_2 + 5Cl_2 + 6H_2O \rightarrow 2HIO_3 + 10HCl$

(A) When chlorine reacts with hot and concentrated $NaOH$,it undergoes a disproportionation reaction.
The balanced chemical equation is:
$3 Cl_2 + 6 NaOH \longrightarrow 5 NaCl + NaClO_3 + 3 H_2 O$
Thus,the products formed are sodium chloride $(NaCl)$,sodium chlorate $(NaClO_3)$,and water $(H_2 O)$.

(C) Chlorine heptoxide $(Cl_2O_7)$ reacts with water to form perchloric acid $(HClO_4)$,which is the strongest oxoacid of chlorine.
The chemical equation is: $Cl_2O_7 + H_2O \rightarrow 2HClO_4$.
In the formula $Cl_2O_7$,the ratio of chlorine atoms to oxygen atoms is $2: 7$.

(B) When chlorine reacts with $NaOH$,different products are formed depending upon the temperature and concentration of $NaOH$.
For cold and dilute $NaOH$:
$Cl_2 + 2 NaOH \longrightarrow NaCl + NaOCl + H_2 O$
Here,$A = NaCl$ and $B = NaOCl$.
For hot and concentrated $NaOH$:
$3 Cl_2 + 6 NaOH \longrightarrow 5 NaCl + NaClO_3 + 3 H_2 O$
Here,$C = NaCl$ and $D = NaClO_3$.
Therefore,the correct sequence is $A = NaCl, B = NaOCl, C = NaCl, D = NaClO_3$.

(D) The oxidising power of halogens in aqueous solution depends on the standard electrode potential $(E^{\circ})$,which is determined by the sum of enthalpy changes: sublimation,dissociation,electron gain,and hydration enthalpy.
For halogens,the reaction is: $X_{2(g/l/s)} + 2e^{-} \longrightarrow 2X^{-}_{(aq)}$.
Due to the very high hydration enthalpy of the fluoride ion $(F^{-})$,the overall reduction process for $F_2$ is the most spontaneous.
The hydration energy follows the order: $F^{-} > Cl^{-} > Br^{-} > I^{-}$.
Consequently,the ease of reduction follows the order: $F_2 > Cl_2 > Br_2 > I_2$.
Thus,the correct order of oxidising power is $F_2 > Cl_2 > Br_2 > I_2$.

(C) The physical properties of the given interhalogen compounds are as follows:
$1$. $IBr$ is a black solid.
$2$. $ClF_3$ is a colorless gas.
$3$. $BrF_3$ is a yellow-green liquid.
$4$. $ICl_3$ is an orange solid.
Therefore,the correct matching is $(a-iii), (b-iv), (c-ii), (d-i)$.

(B) The anhydride of an oxoacid is formed by removing water molecules from the acid such that the oxidation state of the central atom remains the same.
For $HOCl$,the reaction is $2HOCl \rightarrow Cl_2O + H_2O$.
Thus,the anhydride of $HOCl$ is $Cl_2O$.
The correct answer is $Cl_2O$,which corresponds to option $(B)$.

(C) When chlorine reacts with cold and dilute sodium hydroxide $(NaOH)$,it undergoes disproportionation to form sodium chloride $(NaCl)$ and sodium hypochlorite $(NaOCl)$.
The chemical equation is: $Cl_2 + 2NaOH \longrightarrow NaCl + NaOCl + H_2O$.