Halogen family Questions in English

(A) When $KBr$ reacts with conc. $H_2SO_4$,the initial reaction produces $HBr$ gas: $KBr + H_2SO_4 \to KHSO_4 + HBr \uparrow$.
Since $HBr$ is a strong reducing agent,it is further oxidized by conc. $H_2SO_4$ to form reddish-brown bromine gas: $2HBr + H_2SO_4 \to Br_2 \uparrow + SO_2 + 2H_2O$.
Therefore,the reddish-brown gas evolved is $Br_2$.

(A) $1$. $AgNO_3$ is a well-known compound used in photography due to its light sensitivity.
$2$. When $AgNO_3$ reacts with a sodium salt $B$ $(NaBr)$,it forms a pale yellow precipitate $C$ $(AgBr)$.
$3$. The reaction is: $AgNO_3(aq) + NaBr(aq) \rightarrow AgBr(s) + NaNO_3(aq)$.
$4$. $AgBr$ is a pale yellow precipitate.
$5$. Sodium bromide $(NaBr)$ reacts with concentrated $H_2SO_4$ to produce brown vapours of bromine $(Br_2)$: $2NaBr + 3H_2SO_4 \rightarrow 2NaHSO_4 + SO_2 + Br_2 + 2H_2O$ (or simply $2NaBr + H_2SO_4 \rightarrow Na_2SO_4 + 2HBr$; $2HBr + H_2SO_4 \rightarrow Br_2 + SO_2 + 2H_2O$).
$6$. Thus,$A = AgNO_3$,$B = NaBr$,and $C = AgBr$.

(A) . Astatine $(At)$ is a radioactive element that belongs to Group $17$ (halogens) in the periodic table,similar to iodine $(I)$.
Due to its position in the same group,it exhibits chemical properties resembling those of iodine.

(B) Chlorine is more reactive than bromine and it displaces bromine from its salt solution.
When chlorine is added to a solution of potassium bromide,the bromine is displaced by the chlorine to form potassium chloride.
The chemical reaction is: $2 KBr + Cl_2 \rightarrow 2 KCl + Br_2$.
Since chlorine has a higher reduction potential than bromine,it acts as a stronger oxidizing agent and effectively replaces bromine.

(A) The displacement of a halide ion by a halogen depends on their standard reduction potentials.
According to the electrochemical series,the oxidizing power of halogens decreases in the order: $F_2 > Cl_2 > Br_2 > I_2$.
Since $Cl_2$ has a higher reduction potential than $Br_2$,it can oxidize bromide ions $(Br^-)$ to bromine $(Br_2)$ according to the reaction: $Cl_2(g) + 2Br^-(aq) \rightarrow 2Cl^-(aq) + Br_2(l)$.
Therefore,$Cl_2$ is the correct answer.

(B) The correct answer is $(B)$.
Fluorine $(F_2)$ is the strongest oxidizing agent among all halogens due to its high electronegativity and low bond dissociation energy.
Therefore,it can displace all other halogens ($Cl$,$Br$,and $I$) from their respective halide salts in aqueous solutions.

(D) . Its reduction to metallic silver. When $AgNO_3$ comes in contact with the skin,it undergoes photochemical reduction or reduction by organic matter present on the skin to form finely divided metallic silver $(Ag)$,which appears as a black stain. The reaction is: $AgNO_3 \xrightarrow{\text{light/organic matter}} Ag + \text{products}$.

(C) The reaction between potassium dichromate $(K_2Cr_2O_7)$,sodium chloride $(NaCl)$,and concentrated sulfuric acid $(H_2SO_4)$ is known as the chromyl chloride test.
The balanced chemical equation is:
$K_2Cr_2O_7 + 4NaCl + 6H_2SO_4 \to 2CrO_2Cl_2 + 2KHSO_4 + 4NaHSO_4 + 3H_2O$.
The product $CrO_2Cl_2$ is known as chromyl chloride,which is a deep red vapor.

(B) Decomposes in sunlight.
The chemical reaction is: $2AgNO_3 \xrightarrow{h\nu} 2Ag + 2NO_2 + O_2$.
Silver nitrate is photosensitive and undergoes decomposition when exposed to light,hence it is stored in dark-coloured bottles.

(A) The reaction is: $K_2MnF_6 + 2SbF_5 \rightarrow 2KSbF_6 + MnF_3 + \frac{1}{2}F_2$.
In this reaction,the stronger Lewis acid $SbF_5$ displaces the weaker one,$MnF_4$,from its salt.
$MnF_4$ is unstable and readily decomposes to give $MnF_3$ and fluorine gas $(F_2)$.

(A) Chlorination is the most widely used method for disinfecting water supplies in cities. Chlorine $(Cl_2)$ acts as a powerful disinfectant that kills harmful microorganisms,making the water safe for consumption.

(B) The reaction proceeds as follows:
$4NaCl + K_2Cr_2O_7 + 6H_2SO_4 \to 2KHSO_4 + 4NaHSO_4 + 2CrO_2Cl_2 + 3H_2O$
The orange-red vapours evolved are of $CrO_2Cl_2$,which is known as chromyl chloride.
This is a characteristic test for the presence of chloride ions $(Cl^-)$.

(A) The reaction between $NaCl$,$MnO_2$,and concentrated $H_2SO_4$ is a standard laboratory method for the preparation of chlorine gas.
The chemical equation is: $4NaCl + MnO_2 + 4H_2SO_4 \rightarrow MnCl_2 + 4NaHSO_4 + 2H_2O + Cl_2 \uparrow$.
Chlorine gas $(Cl_2)$ is known for its characteristic greenish-yellow color.

(B) When $KI$ reacts with concentrated $H_2SO_4$,it first produces $HI$ (hydrogen iodide).
Since $HI$ is a strong reducing agent and concentrated $H_2SO_4$ is an oxidizing agent,the $HI$ is further oxidized to $I_2$ (iodine).
The reaction is: $2KI + 2H_2SO_4 \rightarrow K_2SO_4 + 2H_2O + I_2 + SO_2$.

(D) The acidic strength of oxoacids of chlorine depends on the oxidation state of the central chlorine atom.
As the oxidation state of $Cl$ increases,the acidic strength increases.
The oxidation states are:
$HClO (+1)$,$HClO_2 (+3)$,$HClO_3 (+5)$,$HClO_4 (+7)$.
Therefore,the correct order of acidic strength is $HClO < HClO_2 < HClO_3 < HClO_4$.

(B) The anhydride of an oxoacid is formed by removing water molecules from the acid.
For perchloric acid $(HClO_4)$,the dehydration reaction is:
$2HClO_4 \rightarrow Cl_2O_7 + H_2O$
Thus,the anhydride of $HClO_4$ is $Cl_2O_7$.

(A) Among the halides,the reducing power increases down the group because the size of the atom increases and the bond dissociation energy decreases.
$HI$ is a strong reducing agent because the $H-I$ bond is the weakest among $HCl, HBr,$ and $HI$,making it easy to release $H^+$ and electrons.

(C) As we move down the halogen group,the atomic size increases due to the addition of new shells.
Consequently,the effective nuclear charge experienced by the valence electrons decreases,making it easier to remove an electron.
Therefore,the ionization enthalpy decreases down the group.

(B) The active ingredient in bleaching powder responsible for its bleaching action is calcium hypochlorite,$Ca(OCl)_2$.
When bleaching powder reacts with water or acids,it releases nascent oxygen or hypochlorite ions,which perform the oxidation of colored substances to colorless ones.

(B) The reaction between $KCl$,solid $K_2Cr_2O_7$,and concentrated $H_2SO_4$ is known as the chromyl chloride test.
The chemical equation for this reaction is:
$K_2Cr_2O_7 + 6H_2SO_4 + 4KCl \rightarrow 2CrO_2Cl_2 + 6KHSO_4 + 3H_2O$
The red-brown gas evolved is chromyl chloride $(CrO_2Cl_2)$.

(B) When $Hg_2Cl_2$ is heated with $NaCl$,the sublimate formed is $HgCl_2$.
In this process,$Hg_2Cl_2$ reacts with $NaCl$ to form $HgCl_2$ which sublimes.
The chemical reaction is:
$Hg_2Cl_2 + 2NaCl \rightarrow 2HgCl + 2NaCl$ (This is not the standard industrial process).
Actually,the reaction is:
$HgSO_4 + 2NaCl \rightarrow HgCl_2 + Na_2SO_4$.
$HgCl_2$ is the volatile component that sublimes.

(D) Gold $(Au)$ is a noble metal and is insoluble in common mineral acids like $HCl$,$H_2SO_4$,or $HNO_3$ individually.
It dissolves in aqua regia,which is a mixture of concentrated $HCl$ and concentrated $HNO_3$ in a $3:1$ molar ratio.
The reaction is: $Au + 4H^+ + NO_3^- + 4Cl^- \rightarrow [AuCl_4]^- + NO + 2H_2O$.

(B) $AgNO_3$ (Silver nitrate) is photosensitive and undergoes decomposition when exposed to light.
To prevent this photochemical decomposition,it is stored in dark-colored (brown) bottles.
The decomposition reaction is:
$2AgNO_3 \xrightarrow{hv} 2Ag + 2NO_2 + O_2$

(D) Historically,$Cl_2$ (chlorine gas) has been used in the bleaching process to remove lignin from wood pulp to produce white paper. The oxidation of lignin by chlorine breaks down the complex organic structure,allowing it to be washed away.

(B) The correct option is $(B)$.
Bond energy of $Cl_2$ is the highest among all halogen molecules.
The bond dissociation energies for $F_2$,$Cl_2$,$Br_2$,and $I_2$ are approximately $37 \, kcal \, mol^{-1}$,$58 \, kcal \, mol^{-1}$,$46 \, kcal \, mol^{-1}$,and $36 \, kcal \, mol^{-1}$ respectively.
Due to the small size of the fluorine atom,the lone pair-lone pair repulsions are very high in $F_2$,which makes its bond weaker than that of $Cl_2$.

(D) . $F^-$ is the strongest reducing agent among halides,but $F_2$ is the strongest oxidizing agent. The oxidation of $F^-$ to $F_2$ is not possible by any chemical oxidizing agent because the standard oxidation potential of $F^-$ is very low (highly negative). Therefore,$F^-$ can be oxidized to $F_2$ only by electrolysis.

(B) $Cl_2$ acts as a bleaching agent due to oxidation,which is permanent.
$SO_2$ acts as a bleaching agent due to reduction,which is temporary because the bleached substance gets re-oxidized by atmospheric oxygen to regain its original color.

(D) . The reaction is $\text{Cl}_2 + \text{H}_2\text{O} \xrightarrow{h\nu} \text{HCl} + \text{HOCl}$.
Initially,chlorine water is $\text{Greenish yellow}$ due to the presence of dissolved $\text{Cl}_2$ gas.
Upon exposure to sunlight $(h\nu)$,$\text{Cl}_2$ reacts with water to form $\text{HCl}$ and $\text{HOCl}$,which are colourless.
Therefore,the solution changes from $\text{Greenish yellow}$ to $\text{Colourless}$.

(D) The bond dissociation energy order for halogens is $Cl_2 > Br_2 > F_2 > I_2$ because of the small size of the $F$ atom,which leads to strong inter-electronic repulsion between the non-bonding electrons. Thus,option $(a)$ is incorrect.
For electron gain enthalpy,the order is $Cl > F > Br > I$. Due to the small size of $F$,the incoming electron experiences more repulsion,making its electron gain enthalpy less negative than that of $Cl$. Thus,option $(c)$ is also incorrect.
Therefore,both $(a)$ and $(c)$ represent incorrect trends.

(C) $1$. $PH_5$ and $BiCl_5$ do not exist due to the inert pair effect and steric hindrance.
$2$. $SO_2$ contains $p\pi -d\pi$ bonding between sulfur and oxygen atoms.
$3$. $SeF_4$ has a see-saw shape (due to $sp^3d$ hybridization with one lone pair),while $CH_4$ has a tetrahedral shape ($sp^3$ hybridization). Thus,they do not have the same shape.
$4$. $I_3^+$ has a bent geometry due to the presence of two lone pairs on the central iodine atom.

(B) Calcium hypochlorite,with the formula $Ca(OCl)_2$,is the primary active ingredient in bleaching powder.
It is responsible for the bleaching action and disinfection properties of the compound.
The chemical reaction for the preparation of bleaching powder is:
$3Ca(OH)_2 + 2Cl_2 \xrightarrow{\text{below } 35 \ ^\circ C} Ca(OCl)_2 \cdot CaCl_2 \cdot Ca(OH)_2 \cdot 2H_2O$.

(A) $A-iii$: $XX'$ has linear geometry as only two atoms are present.
$B-i$: $XX'_3$ has $T$-shaped geometry as $3$ bond pairs and $2$ lone pairs of electrons are present. The electron pair geometry is trigonal bipyramidal.
$C-iv$: $XX'_5$ has square-pyramidal geometry as $5$ bond pairs and $1$ lone pair of electrons are present. The electron pair geometry is octahedral.
$D-ii$: $XX'_7$ has pentagonal bipyramidal geometry as $7$ bond pairs and $0$ lone pairs of electrons are present.

(D) The correct order of bond dissociation enthalpy for halogen molecules is $Cl_2 > Br_2 > F_2 > I_2$.
Although bond dissociation energy generally decreases down the group due to an increase in atomic size,$F_2$ is an exception.
In $F_2$,the small size of the fluorine atoms leads to strong inter-electronic repulsion between the lone pairs of the two atoms,which significantly weakens the $F-F$ bond.
Therefore,the bond dissociation enthalpy of $F_2$ is lower than that of $Cl_2$ and $Br_2$.

(D) The acidic strength of oxoacids of chlorine increases with an increase in the oxidation state of the central chlorine atom.
The oxidation states of $Cl$ in the given acids are:
$HClO: +1$
$HClO_2: +3$
$HClO_3: +5$
$HClO_4: +7$
Since the acidic strength is directly proportional to the oxidation state,the correct order is $HClO < HClO_2 < HClO_3 < HClO_4$.

(D) The strength of an oxidising agent is determined by its ability to gain electrons,which is related to its reduction potential.
Fluorine $(F_2)$ has the highest electronegativity and the highest standard reduction potential among the halogens.
As we move down the group,the electronegativity decreases,making it harder to gain electrons.
Therefore,$F_2$ is the strongest oxidising agent as it is reduced most readily to the $F^-$ ion.

(A) The order given in option $A$ is incorrect.
The correct order of bond dissociation energy is $Cl_2 > Br_2 > F_2 > I_2$.
This is because the $F-F$ bond is weaker than the $Cl-Cl$ and $Br-Br$ bonds due to the small size of the fluorine atom,which leads to significant interelectronic repulsion between the lone pairs of the two fluorine atoms.

(D) Let us analyze each reaction:
$1$. $3Br_2 6NaOH (\text{hot and conc.}) \rightarrow 5NaBr NaBrO_3 3H_2O$. Thus,option $A$ is incorrect as it does not mention $NaBrO_3$.
$2$. $SO_2 O_3 \rightarrow SO_3 O_2$. This is correct.
$3$. $Si 2NaOH H_2O \rightarrow Na_2SiO_3 2H_2$. This is correct.
$4$. With excess $NH_3$,$Cl_2$ reacts as: $3Cl_2 8NH_3 \rightarrow N_2 6NH_4Cl$. The statement in option $D$ claims $HCl$ is formed,but $HCl$ reacts with excess $NH_3$ to form $NH_4Cl$. Therefore,both $A$ and $D$ are technically incorrect based on standard chemical descriptions,but $D$ is the most commonly cited incorrect statement in this context due to the product being $NH_4Cl$ instead of $HCl$.

(C) The decreasing order of acid strength of chlorine oxoacids is: $HClO_4 > HClO_3 > HClO_2 > HOCl$.
Reasoning: Consider the stability of the conjugate bases ($ClO_4^-$,$ClO_3^-$,$ClO_2^-$,$ClO^-$).
The negative charge is more delocalized on $ClO_4^-$ due to resonance involving more oxygen atoms,making $ClO_4^-$ the most stable conjugate base and thus $HClO_4$ the strongest acid.
As the number of oxygen atoms around the $Cl$-atom increases,the oxidation state of the $Cl$-atom increases,which increases the electron-withdrawing effect,thereby facilitating the release of the $H^+$ ion.

(B) $ICl$ is an interhalogen compound. It contains two atoms of different electronegativities,making the bond polar.
In contrast,the molecules of chlorine $(Cl_2)$,bromine $(Br_2)$,and iodine $(I_2)$ consist of atoms with the same electronegativity,resulting in non-polar bonds.
Due to the polarity of the bond,$ICl$ is more reactive than the corresponding homonuclear halogen molecules.

(C) When chlorine gas reacts with cold and dilute aqueous $NaOH$,it undergoes a disproportionation reaction.
The chemical equation is:
$Cl_{2} + 2NaOH \text{ (cold/dilute)} \rightarrow NaCl + NaOCl + H_{2}O$
In this reaction,$Cl_{2}$ is reduced to $Cl^{-}$ (in $NaCl$) and oxidized to $ClO^{-}$ (in $NaOCl$).
Therefore,the products are $Cl^{-}$ and $ClO^{-}$.

(D) Aqua regia is a mixture of concentrated $HCl$ and concentrated $HNO_3$ in a $3:1$ ratio. It is a powerful oxidizing agent capable of dissolving noble metals like gold and platinum,as well as various metal sulfides. $HgS$,$NiS$,and $CoS$ are all soluble in aqua regia.

(B) $1$. $H_3PO_3$ (phosphorous acid) has two $P-OH$ bonds and one $P-H$ bond,making it dibasic and a reducing agent. So,option $A$ is incorrect.
$2$. Pyrosilicate ion is $Si_2O_7^{6-}$,which is formed by sharing one oxygen atom between two $SiO_4^{4-}$ tetrahedra. This is correct.
$3$. Caro's acid $(H_2SO_5)$ contains a peroxo bond,but oleum $(H_2S_2O_7)$ does not. So,option $C$ is incorrect.
$4$. In $BrF_5$ (square pyramidal geometry),the apical $Br-F$ bond is shorter than the equatorial $Br-F$ bonds due to the repulsion from the lone pair. So,option $D$ is incorrect.

(A) The oxidizing strength of a halogen in aqueous solution is determined by the standard electrode potential $(E^{\circ})$,which depends on the energy changes involved in the process: $\frac{1}{2} X_2(g)$ $\rightarrow X(g)$ $\rightarrow X^-(g)$ $\rightarrow X^-(aq)$.
This process involves three main energy terms:
$1$. Bond dissociation enthalpy $(\frac{1}{2} \Delta_{diss} H^{\circ})$: Energy required to break the $X-X$ bond.
$2$. Electron gain enthalpy $(\Delta_{eg} H^{\circ})$: Energy released when an electron is added to the atom.
$3$. Hydration enthalpy $(\Delta_{hyd} H^{\circ})$: Energy released when the ion is hydrated in water.
Fluorine $(F_2)$ has a low bond dissociation enthalpy,high negative electron gain enthalpy,and very high negative hydration enthalpy compared to chlorine $(Cl_2)$. The combination of these three factors makes $F_2$ a much stronger oxidizing agent in water.

(B) Let us analyze each reaction:
$1$. $P_4 + 8SOCl_2 \rightarrow 4PCl_3 + 4SO_2 + 2S_2Cl_2$. Here,$PCl_3$ is formed.
$2$. $P_4 + 10SO_2Cl_2 \rightarrow 4PCl_5 + 10SO_2$. This reaction produces $PCl_5$,not $PCl_3$.
$3$. $PH_3$ reacts with $CaOCl_2$ to form $PCl_3$ as a product.
$4$. $2Ag + PCl_5 \rightarrow 2AgCl + PCl_3$. Here,$PCl_3$ is formed.
Therefore,the reaction between $P_4$ and $SO_2Cl_2$ does not form $PCl_3$.

(B) Aqua regia is a mixture of concentrated hydrochloric acid $(HCl)$ and concentrated nitric acid $(HNO_3)$ in a molar ratio of $3:1$.
It is a highly corrosive,fuming liquid that is capable of dissolving noble metals such as gold $(Au)$ and platinum $(Pt)$,which are otherwise resistant to attack by individual acids.

(B) The reaction between bromine and fluorine depends on the molar ratio of the reactants.
When $Br_2$ reacts with an excess of $F_2$,the product formed is $BrF_5$.
However,when $F_2$ is diluted with an inert gas like $N_2$,the concentration of $F_2$ is effectively reduced.
Under these conditions,the reaction favors the formation of the lower interhalogen compound,which is $BrF_3$ $(Br_2 + 3F_2 \rightarrow 2BrF_3)$.

(D) Photography relies on the photosensitivity of silver halides,particularly $AgBr$,which decomposes upon exposure to light.
Sodium thiosulphate $(Na_2S_2O_3)$ is used as a 'fixing' agent to dissolve and remove the unexposed $AgBr$ from the film,preventing further darkening.
$AuCl_3$ (gold chloride) is used in the process of toning to improve the image quality and stability.
Therefore,all the given statements are correct.

(D) The reaction of ionic bromides (like $NaBr$) with $MnO_2$ and conc. $H_2SO_4$ produces $Br_2$ (red-orange gas),which is $(C)$.
The reaction of ionic iodides (like $KI$) with $MnO_2$ and conc. $H_2SO_4$ produces $I_2$ (violet gas),but the question specifies $(E)$ has the same colour as $(C)$,implying $(E)$ is also $Br_2$ or a similar red-orange species.
However,in standard qualitative analysis,$Br_2$ and $I_2$ are distinct. If $(C)$ is $Br_2$ and $(E)$ is $Br_2$,they are identical. If the question implies $(C)$ is $Br_2$ and $(E)$ is $I_2$,they differ in colour.
Assuming the standard context where $(C)$ is $Br_2$ and $(E)$ is $I_2$,they differ in colour,atomicity (both are diatomic $X_2$),and their reaction with $NaOH$. Since $Br_2$ and $I_2$ both react with $NaOH$,the resulting solutions have different colours ($Br_2$ gives a colorless solution of $NaBr/NaBrO_3$,while $I_2$ gives a colorless solution of $NaI/NaIO_3$).
Given the options,$(2)$ and $(3)$ are the most likely differences if $(C)$ and $(E)$ are different halogens.

(C) The reaction described is the formation of polyhalide ions,specifically the triiodide ion $(I_3^-)$.
$KI(aq) + I_2(s) \rightarrow KI_3(aq)$
In this reaction,$I^-$ acts as a Lewis base (donor) and $I_2$ acts as a Lewis acid (acceptor) to form the $I_3^-$ ion,which gives a characteristic brown-coloured solution.
This property is specific to iodine among the given halogens.

(D) When chlorine water is added to a solution containing iodide ions $(I^-)$,iodine $(I_2)$ is liberated.
$2KI + Cl_2 \rightarrow 2KCl + I_2$
This liberated $I_2$ dissolves in the chloroform $(CHCl_3)$ layer,imparting a violet colour.
Upon adding excess chlorine water,the liberated $I_2$ is further oxidized to iodate $(IO_3^-)$,which is colourless.
$I_2 + 5Cl_2 + 6H_2O \rightarrow 2HIO_3 + 10HCl$
Thus,the disappearance of the violet colour confirms the presence of iodide ions.