Halogen family Questions in English

(C) Aqua regia is a freshly prepared mixture of concentrated nitric acid $(HNO_3)$ and concentrated hydrochloric acid $(HCl)$ in the ratio of $1:3$ by volume.
Therefore,the halogen acid used is $HCl$.

(A) . Bromine is the only non-metal that exists in a liquid state at room temperature $(25 ^\circ C)$.

(A) The bleaching action of chlorine is due to the process of oxidation. $Cl_2$ reacts with water to produce nascent oxygen,which is a strong oxidizing agent. The reaction is: $Cl_2 + H_2O \to 2HCl + [O]$ (nascent oxygen). This nascent oxygen oxidizes the colored substance to a colorless substance.

(B) The reactivity of halogens towards hydrogen decreases down the group as the electronegativity of the halogen decreases.
Among the halogens $(F, Cl, Br, I)$,iodine $(I)$ has the lowest electronegativity and the largest atomic size,making it the least reactive towards hydrogen.

(B) Iodine $(I_2)$ is a non-polar molecule and has very low solubility in water.
However,it dissolves readily in an aqueous solution of potassium iodide $(KI)$ due to the formation of a soluble complex ion,triiodide $(I_3^-)$.
The reaction is: $KI + I_2 \to KI_3$.

(A) The correct order of acidic strength is $HF < HCl < HBr < HI$.
As we move down the group in the periodic table,the size of the halogen atom increases,which leads to a decrease in the bond dissociation enthalpy of the $H-X$ bond.
Since the $H-X$ bond becomes weaker down the group,it becomes easier to release the $H^+$ ion,thereby increasing the acidic strength.

(B) Caliche is crude Chile saltpetre $(NaNO_3)$ which contains about $0.02\%$ iodine as sodium iodate $(NaIO_3)$.

(A) When chlorine $(Cl_2)$ reacts with cold and dilute sodium hydroxide $(NaOH)$,it undergoes a disproportionation reaction.
In this reaction,chlorine is simultaneously oxidized and reduced to form chloride $(Cl^{-})$ and hypochlorite $(ClO^{-})$ ions.
The balanced chemical equation is: $2NaOH + Cl_2 \rightarrow NaCl + NaOCl + H_2O$.
The ionic form of this reaction is: $Cl_2 + 2OH^{-} \rightarrow Cl^{-} + ClO^{-} + H_2O$.

(C) Among the halogens,fluorine is the most electronegative and smallest in size,which limits its ability to form stable oxyacids. While $HOF$ (hypofluorous acid) exists,it is extremely unstable. In the context of standard chemistry,it is generally stated that all halogens except fluorine form a series of stable oxyacids.

(D) Iodine $(I_2)$ is the least electronegative and the weakest oxidizing agent among the halogens $(F_2, Cl_2, Br_2, I_2)$.
Therefore,it cannot displace other halogens from their respective salt solutions $(NaF, NaBr, NaCl)$.
Thus,the reactions $I_2 + NaF \xrightarrow{} \text{No reaction}$,$I_2 + NaBr \xrightarrow{} \text{No reaction}$,and $I_2 + NaCl \xrightarrow{} \text{No reaction}$ do not occur.
The correct option is $D$.

(A) 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 ease of releasing $H^+$ ions increases in the order: $HF < HCl < HBr < HI$.
Thus,$HF$ is the weakest acid among the given hydrohalic acids.

(D) The reaction between potassium dichromate $(K_2Cr_2O_7)$ and concentrated hydrochloric acid $(HCl)$ results in the liberation of chlorine gas $(Cl_2)$.
The balanced chemical equation is:
$K_2Cr_2O_7 + 14HCl \to 2KCl + 2CrCl_3 + 7H_2O + 3Cl_2$

(D) $F$ (Fluorine) is the most electronegative element in the periodic table. Due to its high electronegativity and the absence of $d$-orbitals in its valence shell,it cannot expand its octet or share electrons in a way that results in a positive oxidation state. Therefore,it always exhibits a $-1$ oxidation state in its compounds.

(A) The acid strength of oxyacids increases with an increase in the oxidation state of the central atom.
The oxidation states of chlorine in $HClO$,$HClO_2$,$HClO_3$,and $HClO_4$ are $+1$,$+3$,$+5$,and $+7$ respectively.
As the oxidation state increases,the electron-withdrawing power of the chlorine atom increases,which stabilizes the conjugate base $(ClO_n^-)$ by dispersing the negative charge.
Therefore,the correct order of acid strength is $HClO < HClO_2 < HClO_3 < HClO_4$.

(C) Bleaching powder,$CaOCl_2$,reacts with moisture or dilute acids to release nascent oxygen or hypochlorous acid $(HClO)$.
$CaOCl_2 + H_2O \rightarrow Ca(OH)_2 + Cl_2$
$Cl_2 + H_2O \rightarrow HCl + HClO$
$HClO \rightarrow HCl + [O]$
The nascent oxygen $[O]$ or the hypochlorous acid $(HClO)$ is responsible for the bleaching action by oxidizing colored substances to colorless substances.

(A) The reaction of fluorine with dilute $NaOH$ is as follows:
$2F_2 + 2NaOH (dil.) \to 2NaF + OF_2 + H_2O$
Thus,fluorine reacts with dilute $NaOH$ to produce oxygen difluoride $(OF_2)$.

(A) $MnO_2$ acts as an oxidizing agent. It can oxidize $Cl^-$,$Br^-$,and $I^-$ to their respective halogens ($Cl_2$,$Br_2$,$I_2$).
However,$F^-$ (fluoride ion) cannot be oxidized by $MnO_2$ because fluorine is the strongest oxidizing agent among halogens,and its oxidation potential is extremely high.

(D) . The reaction between concentrated $HCl$ and potassium permanganate $(KMnO_4)$ is a standard laboratory method for the preparation of chlorine gas at room temperature.
The balanced chemical equation is: $2KMnO_4 + 16HCl \to 2KCl + 2MnCl_2 + 8H_2O + 5Cl_2$.

(B) Concentrated $H_2SO_4$ is a strong oxidizing agent.
When $NaBr$ reacts with concentrated $H_2SO_4$,it first produces $HBr$.
However,the concentrated $H_2SO_4$ then oxidizes the produced $HBr$ to $Br_2$ gas.
The reaction is: $2NaBr + 3H_2SO_4 \rightarrow 2NaHSO_4 + SO_2 + Br_2 + 2H_2O$.

(A) The displacement reaction depends on the oxidizing power of halogens. The oxidizing power decreases down the group: $F_2 > Cl_2 > Br_2 > I_2$.
$Cl_2$ can displace $Br^-$ and $I^-$ from their salts because it is a stronger oxidizing agent than $Br_2$ and $I_2$.
However,$Cl_2$ cannot displace $F^-$ from $NaF$ because $F_2$ is a stronger oxidizing agent than $Cl_2$.
Therefore,the reaction $Cl_2 + NaF \to \text{No reaction}$ occurs.

(D) When a fluoride salt (e.g.,$NaF$) is heated with concentrated $H_2SO_4$,$HF$ gas is produced: $2NaF + H_2SO_4 \rightarrow Na_2SO_4 + 2HF$.
Unlike chloride,bromide,or iodide salts,fluoride salts do not undergo oxidation by $MnO_2$ in the presence of $H_2SO_4$ to release the free halogen $(F_2)$.
This is because the standard reduction potential of $F_2$ is very high,making it the strongest oxidizing agent,and it cannot be produced by chemical oxidation of $F^-$ ions.
Therefore,no $F_2$ gas is evolved.

(A) Fluorine is a strong oxidizing agent. When it reacts with hot and concentrated $NaOH$,it oxidizes water to oxygen. The reaction is: $2F_2 + 4NaOH \to 4NaF + 2H_2O + O_2$.

(A) Aqua regia is a freshly prepared 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 like gold and platinum.

(B) Fluorine is the most electronegative element and has the smallest size among halogens.
Crucially,it lacks $d$-orbitals in its valence shell $(n=2)$.
Due to the absence of vacant $d$-orbitals,it cannot expand its octet or promote electrons to higher energy levels to exhibit positive oxidation states like $+3, +5,$ or $+7$.
Therefore,it only shows a $-1$ oxidation state.

(A) Fluorine $(F)$ is the most electronegative element in the periodic table and has no $d$-orbitals in its valence shell. Due to this,it can only exhibit an oxidation state of $-1$ and does not show variable oxidation states like other halogens $(Cl, Br, I)$.

(A) The correct answer is $(A)$.
Solid $NaF$ is used to purify fluorine gas by removing $HF$ fumes through the formation of $NaHF_2$ according to the reaction: $NaF + HF \rightarrow NaHF_2$.

(C) Fluorine is prepared by the electrolysis of a fused mixture of $KHF_2$ and $HF$ (Moissan's method).
The reactions are as follows:
$KHF_2 \to KF + HF$
$KF \to K^+ + F^-$
At cathode: $2K^+ + 2e^- \to 2K$; $2K + 2HF \to 2KF + H_2$
At anode: $2F^- \to F_2 + 2e^-$

(D) The oxidizing power of a halogen is determined by the standard electrode potential $(E^\circ)$,which depends on three factors: enthalpy of dissociation,enthalpy of hydration,and enthalpy of sublimation.
$1$. Fluorine has the lowest enthalpy of dissociation ($F-F$ bond is weak).
$2$. Fluorine has the highest enthalpy of hydration due to its small size.
$3$. These factors combined make the reduction potential of fluorine the highest,making it the strongest oxidizing agent.
Therefore,the correct answer is $D$.

(B) $SO_2$ bleaches flowers by reduction:
$2H_2O + SO_2 \to H_2SO_4 + 2[H]$
$2[H] + \text{Coloured flower} \xrightarrow{\text{Reduction}} \text{Colourless reduced flower}$
$Cl_2$ bleaches flowers by oxidation:
$Cl_2 + H_2O \to 2HCl + [O]$
$[O] + \text{Coloured flower} \xrightarrow{\text{Oxidation}} \text{Colourless oxidised flower}$
Therefore,the correct pair is $SO_2$ and $Cl_2$.

(A) Fluorine does not form oxyacids because it is the most electronegative element and cannot exhibit positive oxidation states,which are required for the formation of oxyacids.

(C) Chlorine can be prepared by the oxidation of hydrochloric acid $(HCl)$ using oxidizing agents like manganese dioxide $(MnO_2)$ or potassium permanganate $(KMnO_4)$.
$1$. Using $MnO_2$: $MnO_2 + 4HCl \to MnCl_2 + 2H_2O + Cl_2$
$2$. Using $KMnO_4$: $2KMnO_4 + 16HCl \to 2KCl + 2MnCl_2 + 8H_2O + 5Cl_2$
Thus,both $MnO_2$ and $KMnO_4$ are used in the preparation of chlorine.

(B) The thermal decomposition or reaction of $KClO_3$ under specific conditions leads to the formation of chlorine dioxide $(ClO_2)$.
Specifically,the reaction with concentrated sulfuric acid is: $3KClO_3 + 3H_2SO_4 \xrightarrow{\Delta} 3KHSO_4 + HClO_4 + 2ClO_2 + H_2O$.

(B) The correct answer is $B$.
For the halogens,the electronegativity follows the order $F > Cl > Br > I$.
As the atomic size increases down the group,the attraction for the shared pair of electrons decreases,leading to a decrease in electronegativity.

(C) The correct answer is $C$.
As we move down the group from $F$ to $I$,the number of electron shells increases.
This increase in the number of shells leads to an increase in the atomic and ionic radius.

(B) The reducing character of halogens increases down the group from $F$ to $I$.
Because $I_2$ has the strongest reducing property among the halogens,it can be easily oxidized by strong oxidizing agents like concentrated nitric acid $(HNO_3)$.
The chemical reaction is:
$I_2 + 10HNO_3 \to 2HIO_3 + 10NO_2 + 4H_2O$

(A) The reaction $S + 2X_2 \to SX_4$ represents the formation of sulphur tetrahalides.
Sulphur reacts with fluorine to form $SF_4$ and with chlorine to form $SCl_4$.
Bromine is not strong enough to oxidize sulphur to the $+4$ oxidation state to form $SBr_4$ under standard conditions,and iodine is even less reactive.
Therefore,only $F$ and $Cl$ show this reaction.

(D) The reactivity of halogens decreases down the group $(F_2 > Cl_2 > Br_2 > I_2)$.
An element with higher electronegativity and higher oxidation potential can displace an element with lower electronegativity from its salt solution.
Since $I_2$ is the least reactive among the given halogens,it cannot displace $Cl^-$,$Br^-$,or $F^-$ from their respective salt solutions ($KCl$,$KBr$,$KF$).
Therefore,no reaction occurs.

(A) Fluorine is the most electronegative element and acts as a strong oxidizing agent. When it reacts with cold and dilute $NaOH$,it forms sodium fluoride $(NaF)$ and oxygen difluoride $(OF_2)$.
The balanced chemical equation is:
$2NaOH(aq) + 2F_2(g) \to 2NaF(aq) + OF_2(g) + H_2O(l)$

(C) Paramagnetism is observed in species that contain one or more unpaired electrons.
In $ClO_2$,the chlorine atom has an odd number of valence electrons,resulting in an unpaired electron in the molecule.
Therefore,$ClO_2$ is paramagnetic.
$CO_2$,$SO_2$,and $SiO_2$ have all their electrons paired,making them diamagnetic.

(A) The anhydride of an acid is formed by the removal of water molecules from the acid.
Perchloric acid is $HClO_4$.
When two molecules of $HClO_4$ undergo dehydration,they form $Cl_2O_7$ and $H_2O$.
The reaction is: $2HClO_4 \xrightarrow{\Delta} Cl_2O_7 + H_2O$.
Therefore,$Cl_2O_7$ is the anhydride of perchloric acid.

(C) $I_2$ reacts with $I^-$ ions present in the $KI$ solution to form the triiodide complex ion,$I_3^-$.
The reaction is: $I_2 + I^- \rightarrow I_3^-$.
Thus,$I_2$ dissolves in $KI$ solution due to the formation of $KI_3$ (potassium triiodide).

(B) The correct answer is $B$.
The bond dissociation energy of halogens follows the order: $Cl_2 > Br_2 > F_2 > I_2$.
Although $F_2$ is the smallest,the lone pair-lone pair repulsion between the small fluorine atoms makes its bond weaker than that of $Cl_2$.

Element	Bond Energy $(kJ \ mol^{-1})$
$F_2$	$158.8$
$Cl_2$	$242.6$
$Br_2$	$192.8$
$I_2$	$151.0$

(B) The correct answer is $(B)$.
$I_2O_5$ is the only stable and true covalent oxide of iodine.
$I_2O_4$ is known to be an ionic compound,specifically iodyl iodate,which can be represented as $[IO]^+ [IO_3]^-$.
$I_2O_7$ is not a stable oxide of iodine.
Therefore,$I_2O_5$ is the standard covalent oxide used in the estimation of carbon monoxide.

(C) The reaction between concentrated $HCl$ and potassium chlorate $(KClO_3)$ upon heating produces chlorine gas $(Cl_2)$ and chlorine dioxide $(ClO_2)$.
The balanced chemical equation is: $3KClO_3 + 8HCl \rightarrow 3KCl + 4H_2O + 3ClO_2 + Cl_2$.

(A) Ozone $(O_3)$ is a strong oxidizing agent. When it reacts with dry iodine $(I_2)$,it oxidizes iodine to form iodine tetroxide $(I_4O_9)$. However,the reaction with dry iodine specifically yields $I_4O_9$. The balanced chemical equation is: $6I_2 + 9O_3 \to 3I_4O_9 + 9O_2$. Note: In some contexts,the product is represented as $I_4O_9$. Given the options,$I_4O_9$ is the correct chemical species formed.

(A) When solid $NaCl$ is heated with $MnO_2$ and concentrated $H_2SO_4$,a greenish-yellow gas with a suffocating odor is evolved.
The reaction proceeds in two steps:
$NaCl + H_2SO_4 \to NaHSO_4 + HCl$
$MnO_2 + 4HCl \to MnCl_2 + 2H_2O + Cl_2$
Thus,the gas liberated is $Cl_2$.

(D) The correct answer is $D$. This is the Chromyl chloride test.
$4NaCl + K_2Cr_2O_7 + 3H_2SO_4 \xrightarrow{\Delta} K_2SO_4 + 2Na_2SO_4 + 2CrO_2Cl_2 + 3H_2O$
In this reaction,$CrO_2Cl_2$ (Chromyl chloride) is formed,which appears as deep red vapours.
When these vapours are passed into $NaOH$ solution,they form a yellow solution of sodium chromate:
$CrO_2Cl_2 + 4NaOH \to Na_2CrO_4 + 2NaCl + 2H_2O$
Since all statements $A$,$B$,and $C$ are correct,the correct option is $D$.

(B) The reaction is known as the chromyl chloride test for chloride ions.
$4NaCl + K_2Cr_2O_7 + 6H_2SO_4 \rightarrow 2KHSO_4 + 4NaHSO_4 + 2CrO_2Cl_2 + 3H_2O$
In this reaction,$CrO_2Cl_2$ (Chromyl chloride) is formed,which appears as orange-red vapours.

(B) When potassium iodide $(KI)$ is heated with concentrated $H_2SO_4$,the initial reaction produces hydrogen iodide $(HI)$:
$2KI + 2H_2SO_4 \text{ (conc.)} \to 2KHSO_4 + 2HI$
Since $HI$ is a strong reducing agent,it is further oxidized by concentrated $H_2SO_4$ to iodine $(I_2)$:
$2HI + H_2SO_4 \text{ (conc.)} \to I_2 + SO_2 + 2H_2O$
Thus,the final product formed is iodine $(I_2)$.