Halogen family Questions in English

(A) When an aqueous solution of $KClO_3$ is boiled with iodine $(I_2)$,iodine acts as a reducing agent and gets oxidized to iodate $(IO_3^-)$ while chlorate $(ClO_3^-)$ is reduced to chloride $(Cl^-)$.
The balanced chemical equation for this reaction is:
$2KClO_3 + I_2 \to 2KIO_3 + Cl_2$
Therefore,the product obtained is $KIO_3$.

(D) The reaction between $KMnO_4$ crystals and concentrated $HCl$ is a common laboratory method to produce chlorine gas.
$2KMnO_4 + 16HCl \to 2KCl + 2MnCl_2 + 5Cl_2 + 8H_2O$
Thus,the correct option is $(D)$.

(A) The oxidation states of chlorine in the given oxoacids are as follows:
$HClO_4$: $+7$
$HClO_3$: $+5$
$HClO_2$: $+3$
$HClO$: $+1$
As the oxidation number of the central halogen atom increases,the acidic strength of the oxoacid increases.
Therefore,the order of acidic strength is $HClO_4 > HClO_3 > HClO_2 > HClO$.
Thus,$HClO_4$ is the strongest acid.

(A) The halogens follow a trend in physical states: $F_2$ and $Cl_2$ are gases,$Br_2$ is a liquid,and $I_2$ is a solid at room temperature. Therefore,the statement that iodine is a solid is correct.
Chlorine is soluble in water (forming chlorine water).
Reactivity decreases down the group $(F_2 > Cl_2 > Br_2 > I_2)$,so chlorine is more reactive than bromine,and bromine is more reactive than iodine.

(C) When $KBr$ reacts with concentrated $H_2SO_4$,$HBr$ is initially formed:
$KBr + H_2SO_4 \rightarrow KHSO_4 + HBr$
$HBr$ is a strong reducing agent and reduces concentrated $H_2SO_4$ to $SO_2$,while $HBr$ itself gets oxidized to reddish-brown bromine $(Br_2)$ gas:
$2HBr + H_2SO_4 \rightarrow Br_2 + SO_2 + 2H_2O$
Thus,the reddish-brown gas evolved is $Br_2$.

(B) Sea weeds,particularly brown algae like $Laminaria$,are rich in iodine. They are a well-known commercial source for the extraction of iodine $(I_2)$.

(D) The reactivity of halogens depends on their oxidizing power and bond dissociation energy.
Fluorine $(F_2)$ has the highest oxidizing power and the lowest bond dissociation energy among the halogens,making it the most reactive.
The order of reactivity of halogens is: $I_2 < Br_2 < Cl_2 < F_2$.

(D) The correct answer is $(d)$. Both statements $(b)$ and $(c)$ are false.
$1$. The electron affinity of chlorine $(Cl)$ is higher than that of fluorine $(F)$ due to the small size and high inter-electronic repulsion in the $2p$ subshell of fluorine.
$2$. The boiling point of halogens increases down the group as the size and molecular mass increase,leading to stronger van der Waals forces. Thus,iodine $(I_2)$ has the maximum boiling point,not fluorine.

Element	$F$	$Cl$	$Br$	$I$
$E.A. \ (kJ/mol)$	$332.6$	$348.5$	$324.7$	$295.5$
Boiling point $(^oC)$	$-188.1$	$-34.6$	$59.5$	$185.2$

(D) The reducing power of halide ions depends on their ability to lose electrons,which is inversely proportional to the electronegativity of the corresponding halogen atom.
As we move down the group from $F$ to $I$,the size of the halide ion increases,and the hold of the nucleus on the valence electrons decreases.
Therefore,the ease of oxidation increases in the order: $F^{-} < Cl^{-} < Br^{-} < I^{-}$.
Thus,$I^{-}$ is the strongest reducing agent among the given options.

(D) Halogens belong to Group $17$ of the periodic table.
They have $7$ valence electrons in their outermost shell.
The general electronic configuration for the outermost shell of halogens is $ns^2 np^5$.

(A) The ability of a halogen to displace another depends on its reduction potential. Chlorine $(Cl_2)$ has a higher reduction potential than Bromine $(Br_2)$ and Iodine $(I_2)$,but lower than Fluorine $(F_2)$.
Therefore,$Cl_2$ can displace $Br^-$ ions from $NaBr$ solution according to the reaction:
$Cl_2 + 2NaBr \to 2NaCl + Br_2$
It cannot displace $F^-$ because $F_2$ is a stronger oxidizing agent than $Cl_2$.

(A) Concentrated nitric acid acts as a strong oxidizing agent and oxidizes iodine $(I_2)$ to iodic acid $(HIO_3)$.
The balanced chemical equation is:
$I_2 + 10HNO_3 \to 2HIO_3 + 10NO_2 + 4H_2O$
Therefore,the correct option is $(A)$.

(D) The dissolution of iodine $(I_2)$ in an aqueous solution of potassium iodide $(KI)$ occurs due to the formation of the triiodide ion $(I_3^-)$.
The chemical reaction is: $KI + I_2 \to KI_3$.

(A) $H_2SO_4$ is an oxidizing agent. In the reaction $2KBr + H_2SO_4 \to K_2SO_4 + 2HBr$,the produced $HBr$ is further oxidized by $H_2SO_4$ to $Br_2$ $(2HBr + H_2SO_4 \to Br_2 + SO_2 + 2H_2O)$. Therefore,this reaction cannot be used to produce pure $HBr$.

(B) Fluorine $(F_2)$ is the most reactive halogen due to its low bond dissociation energy and high electronegativity.
It reacts with hydrogen $(H_2)$ even in the dark and at very low temperatures.
The reaction is: $H_2 + F_2 \to 2HF$.

(A) The high reactivity of fluorine is primarily due to the low bond dissociation energy of the $F-F$ bond.
This is caused by the strong inter-electronic repulsions between the non-bonding electrons of the two small fluorine atoms,which weakens the bond.

(A) Hydrofluoric acid $(HF)$ reacts with silica $(SiO_2)$ present in glass to form soluble hydrofluorosilicic acid $(H_2SiF_6)$.
Because of this chemical reaction,it cannot be stored in glass bottles.
The reaction is as follows:
$SiO_2 + 4HF \rightarrow SiF_4 + 2H_2O$
$SiF_4 + 2HF \rightarrow H_2SiF_6$

(D) The correct answer is $D$.
Fluorine $(F_2)$ is the strongest oxidizing agent among the halogens due to its high electronegativity and high hydration energy of the fluoride ion.
Iodine $(I_2)$ is the weakest oxidizing agent among the halogens.
Therefore,the pair '$D$. The strongest oxidizing halogen—Iodine' is incorrectly matched.

(D) As the atomic number increases,the electronegativity of the halogens decreases.
Consequently,their tendency to gain electrons decreases down the group.

(A) Halogens $(F_2, Cl_2, Br_2, I_2)$ are all diatomic molecules in their elemental state. They have a general valence shell configuration of $ns^2 np^5$,which allows them to gain one electron to form univalent halide ions $(X^-)$. While most halogens exhibit multiple oxidation states,the statement that they are all diatomic and form univalent ions is the most fundamental and universally correct characteristic among the given options.

(A) $F_2$ is the strongest oxidizing agent in the periodic table.
The oxidizing power of halogens is determined by their standard reduction potential $(E^\circ)$,which is influenced by factors such as enthalpy of atomization,electron gain enthalpy,and hydration enthalpy.
The standard reduction potentials ($E^\circ$ in $V$) for halogens are:
$F_2: +2.87 \ V$
$Cl_2: +1.36 \ V$
$Br_2: +1.07 \ V$
$I_2: +0.54 \ V$
Due to the low bond dissociation enthalpy of $F-F$ and high hydration enthalpy of the $F^-$ ion,$F_2$ exhibits the highest reduction potential,making it the strongest oxidizing agent.

(A) The correct answer is $A$. Fluorine $(F_2)$ is the strongest oxidizing agent among the halogens.
It can displace other halogens from their halide salts because it has the highest reduction potential.
The displacement reactions are as follows:
$F_2 + 2Cl^- \to Cl_2 + 2F^-$
$F_2 + 2Br^- \to Br_2 + 2F^-$
$F_2 + 2I^- \to I_2 + 2F^-$
Thus,$F_2$ displaces $Cl$,$Br$,and $I$ from their respective compounds.

(A) $Fluorine$ is the most electronegative element in the periodic table.
Due to its very high electronegativity,it cannot lose electrons to form positive oxidation states and always exhibits a $-1$ oxidation state in its compounds.

(B) The isolation of fluorine is difficult due to several reasons:
$1$. Fluorine is the strongest oxidizing agent,making it difficult to oxidize $F^-$ ions to $F_2$ using chemical oxidants.
$2$. The potential required for the discharge of fluoride ions is very high,not the lowest.
$3$. Fluorine is highly reactive and attacks most glass vessels,forming $SiF_4$.
$4$. It has a high affinity for hydrogen,but this is not the primary reason for the difficulty in its isolation compared to its reactivity with apparatus.
$5$. Electrolysis of aqueous $HF$ does not yield $F_2$ because water is oxidized more easily than $F^-$.
Therefore,the statement that fluorine reacts with most glass vessels is a correct observation regarding the difficulties encountered.

(D) Fluorine is a very strong oxidizing agent and reacts with water to produce oxygen and ozone along with hydrogen fluoride.
The chemical reactions are as follows:
$2F_2(g) + 2H_2O(l) \to 4HF(aq) + O_2(g)$
$3F_2(g) + 3H_2O(l) \to 6HF(aq) + O_3(g)$
Thus,the reaction of fluorine with water yields $HF, O_2$ and $O_3$.

(D) Chlorine gas is dried over Conc. $H_2SO_4$.
It cannot be dried over $CaO$,$NaOH$,or $KOH$ because these substances are basic and react with acidic $Cl_2$ gas.

(A) When cold and dilute $NaOH$ reacts with $Cl_2$,it undergoes a disproportionation reaction to form sodium hypochlorite $(NaClO)$ and sodium chloride $(NaCl)$.
The balanced chemical equation is: $2NaOH + Cl_2 \to NaClO + NaCl + H_2O$.

(D) Chlorine acts as a strong oxidizing agent and is widely used as a bleaching agent for paper and textiles. It is also used in the sterilization of water and in the preparation of various antiseptics. However,it is not used for the extraction of metals like $Ag$ and $Cu$ from their ores,as this process typically involves hydrometallurgy or pyrometallurgy techniques not involving chlorine gas.

(B) $Cl_2 + H_2O \to 2HCl + [O]$ (Nascent oxygen)
$\text{Coloured matter} + [O] \xrightarrow{\text{Bleaching}} \text{Colourless matter}$ (Oxidation)
Chlorine acts as a bleaching agent only in the presence of moisture because it produces nascent oxygen,which is responsible for the bleaching action.

(B) $CaO$ reacts with $Cl_2$ to form bleaching powder: $CaO + Cl_2 \to CaOCl_2$.
$NaHCO_3$ does not react with $Cl_2$ under normal conditions.
$CO_2$ reacts with both $CaO$ and $NaHCO_3$ (in the presence of water).
Therefore,the correct gas is $Cl_2$.

(C) The reaction of chlorine gas with dry slaked lime $(Ca(OH)_2)$ at room temperature produces bleaching powder $(CaOCl_2)$.
The chemical equation is:
$Ca(OH)_2 + Cl_2 \to CaOCl_2 + H_2O$
Thus,the main product is $CaOCl_2$.

(C) Bromine is present in sea water in the form of bromide salts like $KBr$ or $MgBr_2$.
To extract bromine,chlorine gas $(Cl_2)$ is bubbled through the concentrated sea water (bittern).
Since chlorine is a stronger oxidizing agent than bromine,it displaces bromine from the bromide ions according to the following reaction:
$2Br^- (aq) + Cl_2 (g) \rightarrow 2Cl^- (aq) + Br_2 (l)$

(B) In the extraction of bromine from sea water,the concentrated mother liquor containing bromide ions $(Br^-)$ is treated with chlorine gas $(Cl_2)$.
Chlorine is a stronger oxidizing agent than bromine and displaces bromine from the bromide solution.
The chemical reaction is: $MgBr_2 + Cl_2 \to MgCl_2 + Br_2$.

(B) When iodine $(I_2)$ is dissolved in non-polar solvents like carbon tetrachloride $(CCl_4)$,it forms a violet-coloured solution.
This is due to the presence of molecular iodine in the solvent.

(B) $KI$ reacts with concentrated $H_2SO_4$ to produce $HI$ and $KHSO_4$.
$KI + H_2SO_4 \xrightarrow{\Delta} KHSO_4 + HI$
$HI$ is a strong reducing agent,so it reduces $H_2SO_4$ to $SO_2$ and gets oxidized to $I_2$.
$2HI + H_2SO_4 \to I_2 + SO_2 + 2H_2O$
Thus,the final product is $I_2$ (violet vapours).

(C) $HI$ cannot be prepared by the reaction of $KI$ with concentrated $H_2SO_4$ because $H_2SO_4$ acts as a strong oxidizing agent.
Initially,the reaction produces $HI$:
$KI + H_2SO_4 \to KHSO_4 + HI$
However,the produced $HI$ is a strong reducing agent,and concentrated $H_2SO_4$ oxidizes $HI$ to $I_2$:
$2HI + H_2SO_4 \to I_2 + SO_2 + 2H_2O$
Therefore,$I_2$ is obtained instead of $HI$.

(B) The reaction between sodium chloride,concentrated $H_2SO_4$,and potassium dichromate is known as the chromyl chloride test.
The balanced chemical equation is: $4NaCl + K_2Cr_2O_7 + 3H_2SO_4 \to K_2SO_4 + 2Na_2SO_4 + 2CrO_2Cl_2 + 3H_2O$.
The product $CrO_2Cl_2$ is known as Chromyl chloride.

(D) $HF$ is a very weak reducing agent because the $F^-$ ion has a very high hydration energy and the $F-F$ bond is very strong.
Consequently,$HF$ does not act as a reducing agent and cannot be oxidized by any of the given oxidizing agents.

(D) The acidic strength of hydrohalic acids increases down the group as the bond dissociation enthalpy decreases.
As the size of the halogen atom increases from $F$ to $I$,the $H-X$ bond length increases and the bond strength decreases.
Therefore,the order of acidic strength is $HF < HCl < HBr < HI$.
Thus,$HI$ is the strongest acid.

(D) An acid anhydride is an oxide that forms an acid when reacted with water.
To find the anhydride of $HClO_4$,we remove water $(H_2O)$ from two molecules of the acid:
$2HClO_4 \to H_2O + Cl_2O_7$.
Thus,$Cl_2O_7$ is the anhydride of perchloric acid $(HClO_4)$.

(B) As we move down the group,the magnitude of $Van \ der \ Waals$ forces increases due to an increase in molecular size and atomic mass.
Consequently,the physical state changes from gas to liquid and then to solid.
$F_2$ and $Cl_2$ are gases,$Br_2$ is a liquid,and $I_2$ is a solid at room temperature.

(C) Concentrated $HNO_3$ acts as a strong oxidizing agent and oxidizes iodine $(I_2)$ to iodic acid $(HIO_3)$.
The balanced chemical equation for the reaction is:
$I_2 + 10HNO_3 \to 2HIO_3 + 10NO_2 + 4H_2O$
Therefore,the correct option is $C$.

(A) The compound $IF_5$ can react further with fluorine to form $IF_7$ because iodine can expand its octet and accommodate more fluorine atoms.
$IF_5 + F_2 \to IF_7$
Other compounds like $NaF$,$CaF_2$,and $SF_6$ are stable and do not react further with fluorine under normal conditions.

(A) Pseudohalide ions are monovalent negative ions composed of two or more electronegative atoms that exhibit chemical properties similar to those of halide ions $(X^-)$.
These ions can form dimeric molecules known as pseudohalogens,which behave similarly to halogens $(X_2)$.
Among the given options,the cyanide ion $(CN^-)$ is a well-known pseudohalide,and its corresponding pseudohalogen is cyanogen $(CN)_2$.

Pseudohalide	Pseudohalogen
$CN^-$ (Cyanide)	$(CN)_2$ (Cyanogen)
$SCN^-$ (Thiocyanate)	$(SCN)_2$ (Thiocyanogen)

(B) Fluorine is a very strong oxidizing agent due to its high electronegativity and low bond dissociation enthalpy.
It reacts with water to oxidize it to oxygen:
$2F_2(g) + 2H_2O(l) \to 4HF(aq) + O_2(g)$

(D) Fluorine $(F_2)$ is prepared by the electrolytic oxidation of fluoride ions. Specifically,it is obtained by the electrolysis of a mixture of potassium hydrogen fluoride $(KHF_2)$ and anhydrous hydrogen fluoride $(HF)$.
Other elements like $Ca$ are also prepared by electrolysis,but in the context of the $p$-block elements chapter,$F_2$ is the standard example of an element prepared by electrolysis.

(D) The ability of a halogen to displace another halogen from its salt solution depends on its oxidizing power. The oxidizing power of halogens decreases down the group: $F_2 > Cl_2 > Br_2 > I_2$.
Since $Cl_2$ is a stronger oxidizing agent than $Br_2$,it can oxidize $Br^-$ ions to $Br_2$ and get reduced to $Cl^-$ ions.
The reaction is: $Cl_2 + 2KBr \rightarrow 2KCl + Br_2$.
Therefore,$Cl_2$ will liberate $Br_2$ from $KBr$.

(D) Chlorine dioxide $(ClO_2)$ is best prepared by passing dry chlorine gas over hot silver chlorate $(AgClO_3)$.
The chemical reaction is as follows:
$2 AgClO_3 + Cl_2 \rightarrow 2 AgCl + 2 ClO_2 + O_2$

(B) The mixture of concentrated $HCl$ and $HNO_3$ in a $3:1$ ratio is known as aqua regia.
The chemical reaction is: $3HCl + HNO_3 \rightarrow NOCl + 2H_2O + Cl_2$.
Thus,the mixture contains nitrosyl chloride $(NOCl)$.

(A) . The statement "$HF$ is a stronger acid than $HCl$" is false.
$HF$ is a weak acid in aqueous solution due to strong intermolecular hydrogen bonding and high bond dissociation enthalpy of the $H-F$ bond,whereas $HCl$ is a strong acid.