Halogen family Questions in English

(D) Reducing properties increase from fluorine to iodine,so it is oxidised by nitric acid.
$I_2 + 10 HNO_3 \longrightarrow 2 HIO_3 + 10 NO_2 + 4 H_2O$
Nitric acid oxidises iodine to iodic acid,$HIO_3$,and not periodic acid,$HIO_4$,which is of a higher oxidation state.
Hence,option $(D)$ is the correct.

(A) Fluorine $(F_2)$ is the strongest oxidizing agent among the halogens. Due to its high reactivity,it reacts vigorously with water to liberate oxygen gas $(O_2)$.
$2F_{2(g)} + 2H_2O_{(l)} \longrightarrow O_{2(g)} + 4HF_{(aq)}$

(B) $1$. For $X_2$: $Cl_2$ and $Br_2$ react with water to form $HX$ and $HOX$ (disproportionation reaction). $Cl_2 + H_2O \rightarrow HCl + HOCl$.
$2$. For $Y_2$: $F_2$ is a very strong oxidizing agent and reacts with water to oxidize it to $O_2$. $2F_2 + 2H_2O \rightarrow 4H^+ + 4F^- + O_2$.
$3$. For $Z_2$: $I_2$ is the least reactive and does not react with water spontaneously. Thus,$X_2 = Cl_2$,$Y_2 = F_2$,and $Z_2 = I_2$.

(B) The acidic strength of oxyacids depends on the oxidation state of the central atom and the stability of the conjugate base.
As the oxidation state of chlorine increases,the electronegativity of the chlorine atom increases,which pulls the electron density away from the $O-H$ bond,making the proton more acidic.
The oxidation states of chlorine in $HOCl$,$HClO_2$,$HClO_3$,and $HClO_4$ are $+1, +3, +5$,and $+7$ respectively.
Therefore,the acidic strength increases in the order: $HOCl < HClO_2 < HClO_3 < HClO_4$.

(B) The hydrolysis of interhalogen compounds like $BrF_5$ involves the reaction with water to form hydrohalic acid and an oxyacid of the halogen with a higher oxidation state.
For $BrF_5$,the balanced chemical equation is: $BrF_5 + 3H_2O \rightarrow HBrO_3 + 5HF$.
Here,$BrF_5$ reacts with $3$ moles of water to produce $1$ mole of bromic acid $(HBrO_3)$ and $5$ moles of hydrogen fluoride $(HF)$.
Therefore,the products are $HF$ and $HBrO_3$.

(B) In Whytlaw-Gray's method for the preparation of fluorine,the copper diaphragm is used to prevent the mixing of $H_2$ and $F_2$ gases liberated at the cathode and anode,respectively.
The reactions in the electrolytic cell are:
$KHF_2 \rightarrow KF + HF$
$KF \rightarrow K^+ + F^-$
At cathode:
$K^+ + e^- \rightarrow K$
$K + HF \rightarrow KF + H$
$2H \rightarrow H_2$
At anode:
$F^- \rightarrow F + e^-$
$2F \rightarrow F_2$

(A) The correct answer is $A$.
Heating potassium permanganate $(KMnO_4)$ results in the formation of potassium manganate $(K_2MnO_4)$,manganese dioxide $(MnO_2)$,and oxygen gas $(O_2)$.
The reaction is: $2KMnO_4 \xrightarrow{\Delta} K_2MnO_4 + MnO_2 + O_2$.
Since the statement in option $A$ claims it gives only potassium manganate and manganese dioxide,it is incorrect because oxygen is also produced.
Phosphine $(PH_3)$ is indeed used in smoke screens (Holme's signals).
Chlorine's bleaching action is due to oxidation in the presence of moisture.
Noble gases have weak van der Waals forces,leading to very low boiling points.

(B) In the reaction between $XeF_4$ and $SbF_5$,$XeF_4$ acts as a fluoride $(F^-)$ donor.
The reaction is: $XeF_4 + SbF_5 \rightarrow [XeF_3]^+ + [SbF_6]^-$.
For the $[XeF_3]^+$ cation,the central $Xe$ atom has $7 + 4 - 1 = 10$ valence electrons,leading to $sp^3d$ hybridization with two lone pairs,resulting in a $T$-shaped structure.
For the $[SbF_6]^-$ anion,the central $Sb$ atom has $5 + 6 + 1 = 12$ valence electrons,leading to $sp^3d^2$ hybridization with no lone pairs,resulting in an octahedral structure.

(D) The reaction of manganese dioxide with concentrated hydrochloric acid is:
$MnO_2 + 4HCl \rightarrow MnCl_2 + Cl_2 + 2H_2O$
The greenish yellow gas $X$ is chlorine $(Cl_2)$.
When chlorine reacts with excess ammonia,the reaction is:
$8NH_3 + 3Cl_2 \rightarrow 6NH_4Cl + N_2$
Here,$Y$ is $NH_4Cl$ and $Z$ is $N_2$.

(B) The oxidizing power of a species is directly related to its standard reduction potential. Higher reduction potential indicates a stronger oxidizing agent.
For the given oxoanions of halogens,the reduction potential values are determined by the stability of the species and the electronegativity of the central atom.
The standard reduction potentials for the reduction of these species to their respective lower oxidation states follow the order $BrO_4^{-} > IO_4^{-} > ClO_4^{-}$.
Therefore,the correct order of oxidizing power is $BrO_4^{-} > IO_4^{-} > ClO_4^{-}$.

(B) The reaction $ClF_3 + H_2O \longrightarrow HCl + HOF + F_2$ is chemically incorrect. The hydrolysis of $ClF_3$ typically yields $HF$,$HCl$,and $HClO_2$ or $HClO_3$ depending on conditions,but it does not produce $HOF$ and $F_2$ in this manner.
Option $(a)$ is a standard displacement reaction where $Cl_2$ oxidizes $Br^-$.
Option $(c)$ is a standard disproportionation reaction of $Cl_2$ in cold dilute $NaOH$.
Option $(d)$ is a standard acid-base reaction where $SO_2$ is evolved.

(D) The ability of a halogen to act as an oxidizing agent decreases down the group $(F_2 > Cl_2 > Br_2 > I_2)$.
$F_2$ is the strongest oxidizing agent and can oxidize $Cl^{-}$,$Br^{-}$,and $I^{-}$ to their respective halogens.
However,$Cl_2$ is a weaker oxidizing agent than $F_2$ and cannot oxidize the fluoride ion $(F^{-})$ to fluorine gas $(F_2)$.
Therefore,the reaction $Cl_2 + 2F^{-} \longrightarrow 2Cl^{-} + F_2$ does not occur.

(D) When ozone $(O_3)$ reacts with potassium iodide $(KI)$ in an aqueous medium,it acts as an oxidizing agent.
The chemical reaction is:
$2 KI_{(aq)} + H_2O_{(l)} + O_{3(g)} \rightarrow 2 KOH_{(aq)} + I_{2(s)} + O_{2(g)}$
The ionic form of this reaction is:
$2 I^{-}_{(aq)} + H_2O_{(l)} + O_{3(g)} \rightarrow 2 OH^{-}_{(aq)} + I_{2(s)} + O_{2(g)}$
Thus,the product formed is iodine $(I_2)$.

(A) Let us evaluate each reaction:
$(I)$ $2 NaOH + SO_2 \longrightarrow Na_2SO_3 + H_2O$. This is a correct acid-base reaction.
$(II)$ $XeF_4 + O_2F_2 \stackrel{143 K}{\longrightarrow} XeF_6 + O_2$. This is a correct reaction for the fluorination of xenon fluorides.
$(III)$ $PCl_5 + 4 H_2O \longrightarrow H_3PO_4 + 5 HCl$. This is a correct hydrolysis reaction of phosphorus pentachloride.
$(IV)$ $2 NaNO_2 + 2 HCl \longrightarrow 2 NaCl + NO + NO_2 + H_2O$. This is a correct reaction where nitrous acid $(HNO_2)$ formed in situ decomposes into $NO$,$NO_2$,and $H_2O$.
All reactions $(I, II, III, IV)$ are correct. However,based on standard textbook patterns for this specific question format,the most accurate set is $I, II, III, IV$.

(A) When moist chlorine gas reacts with hypo $(Na_2S_2O_3)$,it acts as an oxidizing agent. The reaction produces sodium sulphate,hydrochloric acid,and colloidal sulphur.
The balanced chemical equation is:
$Na_2S_2O_3 + H_2O + Cl_2 \rightarrow Na_2SO_4 + 2HCl + S$

(B) The chemical formula of bleaching powder is $CaOCl_2$.
It is prepared by the action of chlorine gas on dry slaked lime,$Ca(OH)_2$.
The reaction is:
$Ca(OH)_2 + Cl_2 \xrightarrow{\Delta} CaOCl_2 + H_2O$

(C) Noble metals such as gold and platinum are soluble in aqua-regia,which is a mixture of concentrated hydrochloric acid $(HCl)$ and concentrated nitric acid $(HNO_3)$ in a molar ratio of $3:1$.
This means $3$ parts of $HCl$ and $1$ part of $HNO_3$ are mixed.
Therefore,the correct mixture is $1$ part conc. $HNO_3$ and $3$ parts conc. $HCl$.
Hence,option $C$ is correct.

(D) $A)$ The reaction of hydrochloric acid with sodium sulphide produces hydrogen sulphide gas: $2HCl(aq) + Na_2S(s) \rightarrow 2NaCl(aq) + H_2S(g) \uparrow$.
$B)$ The reaction of concentrated sulphuric acid with a mixture of sodium chloride and manganese dioxide produces chlorine gas: $MnO_2(s) + 2NaCl(s) + 3H_2SO_4(conc.) \rightarrow 2NaHSO_4(aq) + MnSO_4(aq) + 2H_2O(l) + Cl_2(g) \uparrow$.

(D) When gold $(Au)$ is dissolved in aqua-regia,it reacts to form chloroauric acid $(HAuCl_4)$.
The chemical reaction is as follows:
$Au + 3HCl + HNO_3 \rightarrow HAuCl_4 + NO + 2H_2O$
In an aqueous solution,chloroauric acid dissociates to form the tetrachloroaurate$(III)$ ion:
$HAuCl_4 \rightarrow H^{+} + \left[AuCl_4\right]^{-}$
Therefore,the final chemical form of gold in the solution is the $\left[AuCl_4\right]^{-}$ ion.

(A) Silica $(SiO_2)$ present in glass reacts with hydrofluoric acid $(HF)$ to produce silicon tetrafluoride $(SiF_4)$ and water.
$SiO_2 + 4HF \longrightarrow SiF_4 + 2H_2O$

(B) mixture of concentrated $HCl$ and $HNO_3$ in a $3:1$ ratio is known as aqua-regia.
The chemical reaction between them is as follows:
$3HCl + HNO_3 \longrightarrow NOCl + 2H_2O + Cl_2$
As shown in the reaction,$NOCl$ (nitrosyl chloride) is produced.
Therefore,option $(B)$ is the correct answer.

(C) The hydrolysis of nitrogen trichloride $(NCl_3)$ involves the reaction with water to produce ammonia $(NH_3)$ and hypochlorous acid $(HOCl)$.
The balanced chemical equation is:
$NCl_3 + 3H_2O \longrightarrow NH_3 + 3HOCl$
Therefore,$X$ is $HOCl$.

(C) The reaction of white phosphorus $(P_4)$ with sulfuryl chloride $(SO_2Cl_2)$ is a chlorination reaction.
The balanced chemical equation is:
$P_4 + 10 SO_2Cl_2 \longrightarrow 4 PCl_5 + 10 SO_2$
Here,$x$ is $PCl_5$ and $y$ is $SO_2$.
Therefore,the correct option is $(C)$.

(C) The reaction of $SO_2$ with $Cl_2$ in the presence of charcoal (catalyst) gives sulphuryl chloride $(A)$: $SO_2 + Cl_2 \xrightarrow{\text{charcoal}} SO_2Cl_2$ $(A)$.
Sulphuryl chloride $(A)$ reacts with white phosphorus $(P_4)$ to give $SO_2$ and phosphorus pentachloride $(B)$: $10SO_2Cl_2 + P_4 \rightarrow 4PCl_5$ $(B)$ $+ 10SO_2$.
Compound $(B)$ is $PCl_5$.
In the solid state,$PCl_5$ exists as an ionic solid $[PCl_4]^+[PCl_6]^-$.
Therefore,the correct statement is that '$B$' in the solid state exists as an ionic solid.

(B) $SF_6$ is an octahedral molecule where the sulfur atom is sterically protected by six fluorine atoms,making it chemically inert and highly stable.
$SF_6$ is indeed a gas at room temperature.
While both statements are factually correct,the fact that it is a gas is not the reason for its chemical stability.
Therefore,$A$ is true,$R$ is true,but $R$ is not the correct explanation for $A$.

(B) Deacon's process is an industrial method for the production of chlorine gas $(Cl_2)$.
In this process,hydrogen chloride $(HCl)$ is oxidized by atmospheric oxygen $(O_2)$ in the presence of a catalyst,copper$(II)$ chloride $(CuCl_2)$,at a temperature of approximately $723 \ K$.
The chemical equation for this reaction is: $4 HCl + O_2 \xrightarrow[723 \ K]{CuCl_2} 2 Cl_2 + 2 H_2 O$.

(B) The correct answer is $B$.
$1$. Chlorine $(Cl_2)$ oxidises ferrous salts $(Fe^{2+})$ to ferric salts $(Fe^{3+})$ in acidic medium: $2Fe^{2+} + Cl_2 \rightarrow 2Fe^{3+} + 2Cl^-$. This statement is correct.
$2$. Chlorine oxidises iodine $(I_2)$ to iodic acid $(HIO_3)$,not periodic acid $(HIO_4)$,in the presence of water: $I_2 + 5Cl_2 + 6H_2O \rightarrow 2HIO_3 + 10HCl$. Thus,statement $B$ is incorrect.
$3$. Chlorine acts as a bleaching agent due to the oxidation of coloured substances by nascent oxygen: $Cl_2 + H_2O \rightarrow 2HCl + [O]$. This statement is correct.
$4$. Chlorine is manufactured by Deacon's process,which involves the oxidation of hydrogen chloride gas by atmospheric oxygen in the presence of $CuCl_2$ as a catalyst: $4HCl + O_2 \xrightarrow{CuCl_2} 2Cl_2 + 2H_2O$. This statement is correct.

(D) The given reaction is the industrial method for the preparation of chlorine gas,known as the Deacon's process.
In this process,$HCl$ is oxidized by atmospheric oxygen in the presence of $CuCl_2$ as a catalyst at $723 \ K$.
$4 HCl + O_2 \xrightarrow[723 \ K]{CuCl_2} 2 Cl_2 + 2 H_2 O$

(B, C) Statement $(I)$ is incorrect because both $Au$ and $Pt$ dissolve in aqua regia to form soluble compounds $HAuCl_4$ and $H_2PtCl_6$.
Statement $(II)$ is correct because in perchloric acid $(HClO_4)$,chlorine exhibits an oxidation state of $+7$.
Statement $(III)$ is correct because $HCl$ has the lowest boiling point among hydrogen halides due to the absence of hydrogen bonding and lower molecular mass compared to $HBr$ and $HI$.
Statement $(IV)$ is incorrect because the stability of halogen oxides decreases in the order $I > Cl > Br$.

(B) Deacon's method involves the preparation of $Cl_2$ from $HCl$ gas by oxidation at about $723 \ K$ in the presence of a catalyst like $CuCl_2$.
The chemical reaction is:
$4 HCl + O_2 \xrightarrow[723 \ K]{CuCl_2} 2 H_2 O + 2 Cl_2$

(A) $Cl_2$ reacts with $CO$ to form phosgene $(COCl_2)$: $CO + Cl_2 \rightarrow COCl_2$.
$I_2O_5$ is a strong oxidizing agent used for the quantitative estimation of carbon monoxide $(CO)$: $I_2O_5 + 5CO \rightarrow I_2 + 5CO_2$.
$O_3$ (Ozone) acts as a powerful disinfectant and bleaching agent due to its ability to release nascent oxygen.
Therefore,all three pairs are correctly matched.

(D) The thermal stability of $OF_2$ at room temperature is due to the small difference between the electronegativities of $O$ and $F$.
The order of stability of the oxides of halogens is $I > Cl > Br$.
$I_2O_5$ is a very good oxidizing agent and is used for the estimation of $CO$.
$ClO_2$ is a well-known bleaching agent.
Thus,statements $A$,$C$,and $D$ are correct.

(C) Interhalogen compounds are generally more reactive than halogens (except $F_2$) because the $X-X'$ bond is weaker than the $X-X$ bond.
They are diamagnetic in nature as all electrons are paired.
During the hydrolysis of interhalogen compounds,the less electronegative halogen forms an oxyacid,and the more electronegative halogen forms a hydrohalic acid.
For $ICl$,the reaction is: $ICl + H_2O \rightarrow HCl + HOI$.
Therefore,the statement that the products are $HI + HOCl$ is incorrect.

(D) Fluorine is present mainly as insoluble fluorides such as fluorspar $(CaF_2)$,cryolite $(Na_3AlF_6)$,and fluoroapatite $(3Ca_3(PO_4)_2 \cdot CaF_2)$.
Small quantities are also present in soil,river water,plants,and the bones and teeth of animals.
Carnallite is a double salt with the formula $KCl \cdot MgCl_2 \cdot 6H_2O$.
It does not contain fluorine,therefore it is not a mineral of fluorine.

(C) The acidic character of hypohalous acids $(HOX)$ depends on the electronegativity of the halogen atom $(X)$.
As the electronegativity of the halogen increases,the $O-H$ bond becomes more polarized,facilitating the release of the $H^+$ ion.
The electronegativity order of halogens is $Cl > Br > I$.
Therefore,the acidic strength order is $HClO > HBrO > HIO$.

(A) The colours of the halogens are as follows:
$F_2$ is pale yellow.
$Cl_2$ is greenish yellow.
$Br_2$ is reddish brown (often simplified to red in some contexts).
$I_2$ is violet.
Matching these with the given options:
$A(F_2) - III$ (Yellow)
$B(Cl_2) - IV$ (Greenish yellow)
$C(Br_2) - I$ (Red)
$D(I_2) - II$ (Violet)
Therefore,the correct sequence is $A-III, B-IV, C-I, D-II$.

(A) The reaction of $Br_2$ with excess $F_2$ is given by:
$Br_2 + 5F_2 \rightarrow 2BrF_5$
Thus,the product $P$ is $BrF_5$.
In $BrF_5$,the central $Br$ atom has $7$ valence electrons. It forms $5$ bonds with $F$ atoms and has $1$ lone pair.
According to $VSEPR$ theory,the steric number is $5 + 1 = 6$,which corresponds to $sp^3d^2$ hybridization.
The geometry is octahedral,but due to the presence of one lone pair,the shape is square pyramidal.

(B) The balanced chemical equation for the reaction of chlorine with excess ammonia is:
$8 NH_3 + 3 Cl_2 \longrightarrow N_2 + 6 NH_4Cl$
From the stoichiometry of the reaction,$3$ moles of $Cl_2$ react with $8$ moles of $NH_3$.
Therefore,$1$ mole of $Cl_2$ will react with $\frac{8}{3}$ moles of $NH_3$.
Thus,$Z = \frac{8}{3}$.

(A) The bond dissociation energy of the $F_2$ molecule is unexpectedly lower than that of the $Cl_2$ molecule.
This is because the fluorine atom has a very small size,which causes significant inter-electronic repulsion between the lone pairs of electrons on the two fluorine atoms.
This repulsion weakens the $F-F$ bond,making it easier to break.
In contrast,chlorine has a larger atomic size,providing more space for the lone pairs,which reduces the repulsion and results in a stronger bond.
Therefore,both the assertion and the reason are true,and the reason correctly explains the assertion.

(A) The reaction of chlorine with cold and dilute sodium hydroxide is a disproportionation reaction.
$Cl_2 + 2 NaOH \text{ (cold, dilute)} \longrightarrow NaCl + NaOCl + H_2O$
In this reaction,$Cl_2$ is simultaneously oxidized to hypochlorite $(NaOCl)$ and reduced to chloride $(NaCl)$.

(B) In Whytlaw-Gray's method for the preparation of fluorine,the copper diaphragm is used to prevent the mixing of $H_2$ and $F_2$ gases liberated at the cathode and anode,respectively.
The reactions in the electrolytic cell are as follows:
$KHF_2 \rightarrow KF + HF$
$KF \rightarrow K^+ + F^-$
At the cathode:
$K^+ + e^- \rightarrow K$
$K + HF \rightarrow KF + H$
$2H \rightarrow H_2$
At the anode:
$F^- \rightarrow F + e^-$
$2F \rightarrow F_2$

(B) The reaction of fluorine with dilute $NaOH$ is given by:
$2F_2 + 2NaOH \rightarrow 2NaF + OF_2 + H_2O$
Thus,the gaseous product $A$ is oxygen difluoride $(OF_2)$.
In $OF_2$,the central oxygen atom is $sp^3$ hybridized.
It has two bond pairs and two lone pairs of electrons.
Due to the strong repulsion between the two lone pairs of electrons on the oxygen atom,the bond angle is compressed from the ideal tetrahedral angle of $109^{\circ} 28^{\prime}$ to $103^{\circ}$.

(D) Perchloric acid $(HClO_4)$ is an oxoacid of chlorine where the oxidation state of chlorine is $+7$. It does not contain a peroxy linkage ($-O-O-$ bond).
In contrast,perphosphoric acid $(H_3PO_5)$,pernitric acid $(HNO_4)$,and perdisulphuric acid $(H_2S_2O_8)$ all contain at least one peroxy linkage ($-O-O-$ bond) in their structures.

(D) $I$. Bleaching powder $(CaOCl_2)$ reacts with ethanol to produce chloroform $(CHCl_3)$. This is a standard laboratory preparation method.
$II$. Bleaching powder decomposes in the presence of cobalt chloride $(CoCl_2)$ catalyst to release oxygen gas: $2CaOCl_2 \xrightarrow{CoCl_2} 2CaCl_2 + O_2$.
$III$. Fluorine is prepared by the electrolysis of a fused mixture of potassium hydrogen fluoride $(KHF_2)$ and anhydrous hydrogen fluoride $(HF)$. Aqueous $KHF_2$ cannot be used because water would be oxidized to oxygen instead of fluoride ions being oxidized to fluorine.
Thus,statements $I$ and $II$ are correct.