Halogen family Questions in English

(D) Interhalogens are compounds formed by the combination of two different halogens with the general formula $XX'_n$,where $n = 1, 3, 5, 7$.
In these compounds,the central atom $X$ is the larger,less electronegative halogen,and $X'$ is the smaller,more electronegative halogen.
For $ICl_4$,the formula does not follow the general $XX'_n$ rule where $n$ must be an odd integer $(1, 3, 5, 7)$.
Therefore,$ICl_4$ does not exist as a stable interhalogen compound.

(D) Disproportionation is a reaction in which an element in one oxidation state is simultaneously oxidised and reduced.
$F_2$ is a strong oxidising agent and it oxidises water to $O_2$ or $OF_2$.
$Cl_2$ reacts with water to undergo disproportionation: $Cl_2 + H_2O \rightarrow HCl + HOCl$.
$Br_2$ and $I_2$ also undergo similar disproportionation reactions in water,although to a lesser extent than $Cl_2$.
Therefore,all three halogens $(Cl_2, Br_2, I_2)$ can undergo disproportionation in water,while $F_2$ does not.

(D) $-$ Bond dissociation energy of $F_2$ is less than that of $Cl_2$ due to interelectronic repulsion.
$(b)$ $-$ $Cl$ has higher electron affinity $(EA)$ than $F$ because of the small size and high electron density of $F$.
$(c)$ $-$ $HF$ is a weaker acid than $HCl$ because the $H-F$ bond is much stronger than the $H-Cl$ bond.
$(d)$ $-$ Boiling point increases down the group in halogens due to an increase in the magnitude of van der Waals forces with increasing molecular size.

(C) Due to the larger size of the iodine atom,it can accommodate up to seven small fluorine atoms around it.
In contrast,due to the smaller sizes of chlorine and bromine atoms,they cannot accommodate seven fluorine atoms.
Thus,the steric factor dominates in the case of chlorine and bromine.

(B) Hydrofluoric acid $(HF)$ reacts with silica $(SiO_2)$ present in glass to form silicon tetrafluoride $(SiF_4)$ and water.
This reaction is used for etching or making permanent markings on glass surfaces.

(C) The oxy acids of chlorine are $HOCl$ $(n=1)$,$HClO_2$ $(n=2)$,$HClO_3$ $(n=3)$,and $HClO_4$ $(n=4)$.
$1$. Acidic character increases as the oxidation state of $Cl$ increases (i.e.,as $n$ increases) because the stability of the conjugate base increases.
$2$. Oxidising power increases as the stability of the acid decreases,which occurs as $n$ decreases.
$3$. Thermal stability increases with the increase in the oxidation state of the central atom,so it increases as $n$ increases.
$4$. The $Cl-O$ bond order is calculated based on the number of resonance structures; it decreases as $n$ decreases. However,the statement in option $C$ is incorrect because thermal stability actually increases with the increasing value of $n$.

(C) $1.$ Ionization energy decreases down the group as the atomic size increases.
$2.$ Ionic radius increases down the group due to the addition of new shells.
$3.$ Bond energy of $X_2$ molecules generally decreases down the group (except for $F_2$ due to inter-electronic repulsion).
$4.$ Enthalpy of vaporization increases down the group due to the increase in magnitude of van der Waals forces with increasing molecular size.
Therefore,properties $(II)$ and $(IV)$ increase with increasing atomic number.

(B) The reaction of chlorine with barium hydroxide is: $6Cl_2 + 6Ba(OH)_2 \to Ba(ClO_3)_2 (X) + 5BaCl_2 + 6H_2O$.
When $X$ $(Ba(ClO_3)_2)$ reacts with $H_2SO_4$,it forms chloric acid $(Y)$:
$Ba(ClO_3)_2 + H_2SO_4 \to 2HClO_3 (Y) + BaSO_4 \downarrow$.
Chloric acid $(HClO_3)$ decomposes upon heating above $365 \ K$ to produce chlorine dioxide $(Z)$:
$2HClO_3 \xrightarrow{\Delta > 365 \ K} 2ClO_2 (Z) + H_2O + \frac{1}{2} O_2$.
Thus,$Y$ is $HClO_3$ and $Z$ is $ClO_2$.

(D) The given compounds are interhalogen or polyhalide structures that exist in ionic forms in the solid state.
$I_4O_9$ exists as $I^{3+}(IO_3^-)_3$,which dissociates as $I_4O_9 \rightleftharpoons I^{3+} + 3IO_3^-$.
$I_2O_4$ exists as $IO^+IO_3^-$,which dissociates as $I_2O_4 \rightleftharpoons IO^+ + IO_3^-$.
$CsBr_3$ is a polyhalide salt that dissociates as $CsBr_3 \rightleftharpoons Cs^+ + Br_3^-$.
Since all the given representations are correct,the incorrectly matched option is none of these.

(C) The correct order for boiling points of hydrogen halides is $HF > HI > HBr > HCl$.
$HF$ has the highest boiling point due to strong intermolecular hydrogen bonding.
For the remaining hydrogen halides $(HCl, HBr, HI)$,the boiling point increases with increasing molecular mass due to stronger van der Waals forces of attraction.
Therefore,the order given in option $C$ is incorrect.

(B) Aqua-regia is a mixture of concentrated $HCl$ and concentrated $HNO_3$ in a $3:1$ molar ratio.
When mercury $(Hg)$ is heated with aqua-regia,it reacts to form mercury$(II)$ chloride $(HgCl_2)$.
The chemical reaction is: $3Hg + 2HNO_3 + 6HCl \to 3HgCl_2 + 2NO + 4H_2O$.

(B) Photographic films or plates are coated with a light-sensitive emulsion containing $AgBr$ (silver bromide) suspended in gelatin. When light hits the film,$AgBr$ undergoes a photochemical reaction to form metallic silver,which creates the image. Therefore,$AgBr$ is the essential ingredient.

(C) Aqua regia is a mixture of concentrated $HCl$ and concentrated $HNO_3$ in a $3:1$ molar ratio.
When gold $(Au)$ reacts with aqua regia,it dissolves to form chloroauric acid $(HAuCl_4)$.
The chemical reaction is as follows:
$Au + 4HCl + HNO_3 \rightarrow HAuCl_4 + NO + 2H_2O$.

(D) The reaction between iodine $(I_2)$ and concentrated nitric acid $(HNO_3)$ is an oxidation reaction where iodine is oxidized to iodic acid $(HIO_3)$.
The balanced chemical equation is:
$I_2 + 10HNO_3 \to 2HIO_3 + 10NO_2 + 4H_2O$
Thus,the product formed is iodic acid $(HIO_3)$.

(B) Oxygen does not react directly with $Cl$ (chlorine). It reacts with $P$ to form $P_4O_{10}$,with $Na$ to form $Na_2O_2$,and with $S$ to form $SO_2$.

(A) For a given halogen atom,the oxidizing power of its oxyacids decreases as the oxidation state of the halogen increases.
In $HOCl$,the oxidation state of $Cl$ is $+1$.
In $HClO_2$,the oxidation state of $Cl$ is $+3$.
In $HClO_3$,the oxidation state of $Cl$ is $+5$.
In $HClO_4$,the oxidation state of $Cl$ is $+7$.
Since the oxidizing power is inversely proportional to the oxidation state of the halogen,$HOCl$ is the strongest oxidizing agent.

(D) Iodine $(I_{2})$ is sparingly soluble in water but dissolves readily in an aqueous solution of potassium iodide $(KI)$ to form the triiodide ion complex,potassium triiodide $(KI_{3})$.
The chemical reaction is: $I_{2(s)} + KI_{(aq)} \to KI_{3(aq)}$

(B) When iodine $(I_2)$ is dissolved in non-polar solvents like carbon tetrachloride $(CCl_4)$,it forms a violet-colored solution.

(D) $Cl_2$ gas is dried by passing it over concentrated $H_2SO_4$.
$CaO$,$NaOH$,and $KOH$ react with chlorine gas,so they cannot be used as drying agents.

(A) The reactivity of halogens decreases as we move down the group in the periodic table.
Therefore,the correct order of reactivity is $F > Cl > Br > I$.

(A) The ease of reduction is determined by the standard reduction potential $(E^\circ)$.
Fluorine $(F_2)$ has the highest standard reduction potential among all halogens.
Therefore,$F_2$ is the strongest oxidizing agent and is most easily reduced to fluoride ions $(F^-)$.

(B) The balanced chemical equation for the reaction of fluorine with hot and concentrated $KOH$ is:
$2F_2 + 4KOH \to 4KF + 2H_2O + O_2$
From the stoichiometry of the balanced equation,for $2 \ mol$ of $F_2$ and $4 \ mol$ of $KOH$,the products formed are $4 \ mol$ of $KF$,$2 \ mol$ of $H_2O$,and $1 \ mol$ of $O_2$.
Dividing the coefficients by $2$ to match the $1 \ mol$ of $F_2$ given in the question:
$1F_2 + 2KOH \to 2KF + 1H_2O + 0.5O_2$
Thus,the molar ratio of $KF : H_2O : O_2$ is $2 : 1 : 0.5$.

(A) The bond dissociation enthalpy of $F_2$ is lower than that of $Cl_2$ due to the strong inter-electronic repulsion between the lone pairs of the small $F$ atoms.

(A) Bleaching powder $(CaOCl_2)$ reacts with dilute hydrochloric acid $(HCl)$ to produce calcium chloride $(CaCl_2)$,water $(H_2O)$,and chlorine gas $(Cl_2)$.
The chemical equation is: $CaOCl_2 + 2HCl \to CaCl_2 + H_2O + Cl_2$.

(A) All halogens are diatomic and form monovalent ions (e.g.,$F^-$,$Cl^-$,$Br^-$,$I^-$).

(C) Halogen elements have one electron less than the noble gas configuration. Therefore,they have a very strong tendency to accept an electron and easily form anions.

(D) The oxidizing power of oxoacids of halogens decreases as the oxidation state of the central halogen atom increases.
In $HOCl$,the oxidation state of $Cl$ is $+1$.
In $HClO_2$,the oxidation state of $Cl$ is $+3$.
In $HClO_3$,the oxidation state of $Cl$ is $+5$.
In $HClO_4$,the oxidation state of $Cl$ is $+7$.
Since $HClO_4$ has the highest oxidation state $(+7)$,it is the least oxidizing in nature.

(D) The reaction of potassium dichromate $(K_2Cr_2O_7)$ with concentrated hydrochloric acid $(HCl)$ produces chlorine gas $(Cl_2)$:
$K_2Cr_2O_7 + 14HCl \to 2KCl + 2CrCl_3 + 7H_2O + 3Cl_2$

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

(D) Fluorine $(F_2)$ is the strongest oxidizing agent among the halogens. When it reacts with a hot concentrated solution of potassium hydroxide $(KOH)$,it oxidizes water to oxygen $(O_2)$ and gets reduced to fluoride ions. The balanced chemical equation is: $2F_2 + 4KOH \to 4KF + O_2 + 2H_2O$.

(B) The mixture of concentrated $HNO_3$ and concentrated $HCl$ in a $3:1$ molar ratio is known as aqua regia.
The chemical reaction is: $HNO_3 + 3HCl \to NOCl + 2H_2O + 2Cl$.
The product $NOCl$ (nitrosyl chloride) is formed along with nascent chlorine.

(D) The enrichment of ${}_{92}U^{235}$ involves the conversion of uranium into uranium hexafluoride $(UF_6)$.
$ClF_3$ (chlorine trifluoride) is a powerful fluorinating agent used to convert uranium metal or uranium oxides into $UF_6$ at relatively low temperatures.
Therefore,the correct substance used is $ClF_3$.

(A) The acidic strength of oxyacids of the same halogen increases with the increase in the oxidation state of the halogen atom.
Calculating the oxidation states:
$HClO$: $+1$
$HClO_2$: $+3$
$HClO_3$: $+5$
$HClO_4$: $+7$
Therefore,the correct order of acidic strength is: $HClO < HClO_2 < HClO_3 < HClO_4$.

(D) Iodine exhibits oxidation states of $-1$ (e.g.,$HI$),$+1$ (e.g.,$ICl$),$+3$ (e.g.,$ICl_3$),$+5$ (e.g.,$HIO_3$),and $+7$ (e.g.,$IF_7$).

(C) $H_2SO_4$ acts as a strong oxidizing agent.
It oxidizes $HI$ to $I_2$ according to the reaction: $2HI + H_2SO_4 \rightarrow I_2 + SO_2 + 2H_2O$.
Therefore,$HI$ cannot be prepared by this method.

(A) The reducing power of hydrogen halides 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 $H-I$ bond is the weakest and can easily release $H$ atoms,making $HI$ the strongest reducing agent among the given options.

(B) Euchlorine is a mixture of $Cl_2$ and $ClO_2$ gases,obtained by the action of hydrochloric acid on potassium chlorate $(KClO_3)$.

(D) The halogen oxide $I_4O_9$ is ionic in nature. It is composed of tripositive iodine and iodate ions,represented as $[I^{3+}][(IO_3^-)_3]$.

(B) Halogens,except for fluorine,exhibit variable oxidation states ranging from $-1$ to $+7$. Fluorine shows only $-1$ oxidation state due to the absence of $d$-orbitals. Therefore,the statement that halogens show only $-1$ oxidation state is incorrect.

(D) The bleaching action of chlorine is due to the process of oxidation in the presence of moisture.
$i) \ H_2O + Cl_2 \to 2HCl + [O]$
$ii) \ \text{Coloured substance} + [O] \to \text{Colourless substance}$

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

(B) Interhalogen compounds are generally covalent and diamagnetic in nature. They are more reactive than halogens (except fluorine). Most of them are volatile solids or liquids at $298 \ K$ except $ClF$,which is a gas. Therefore,the statement that they are highly volatile is generally true,but the statement that they are not explosive is incorrect because some interhalogen compounds are known to be explosive.

(D) The basic character of halogens is related to their ability to donate electrons or their electronegativity.
As we move down the group from $F$ to $I$,the electronegativity decreases.
Since $I$ (Iodine) has the lowest electronegativity among the given halogens,it has the least tendency to hold onto its electrons,making it the most basic in character compared to the others.

(D) $SF_6$ (Sulfur hexafluoride) is an inert,non-toxic,and highly stable gas. Due to its high dielectric strength and excellent insulating properties,it is widely used as a gaseous insulator in high-voltage electrical equipment such as high-voltage generators and circuit breakers.

(B) Iodine forms several oxides such as $I_2O_4$,$I_2O_5$,and $I_2O_7$.
Among these,$I_2O_5$ is the most stable and well-known covalent oxide of iodine.
It is a white crystalline solid and is used in the estimation of carbon monoxide.

(D) As the atomic number of halogens increases,the size of the atom increases due to the addition of new shells.
This leads to a decrease in the effective nuclear charge felt by the incoming electron.
Consequently,the tendency to gain an electron decreases,meaning the electron gain enthalpy becomes less negative,or the halogen gains an electron less easily.

(C) When chlorine gas is passed over dry slaked lime $(Ca(OH)_2)$,it forms bleaching powder $(CaOCl_2)$.
The chemical reaction is:
$Ca(OH)_2 + Cl_2 \to CaOCl_2 + H_2O$
Here,$Ca(OH)_2$ is slaked lime and $CaOCl_2$ is bleaching powder.

(D) Fluorine reacts with dilute $NaOH$ at low temperatures to form oxygen difluoride $(OF_2)$:
$2F_2 + 2NaOH \text{ (cold, dilute)} \to 2NaF + OF_2 + H_2O$
Fluorine reacts with concentrated $NaOH$ at higher temperatures to form oxygen $(O_2)$:
$2F_2 + 4NaOH \text{ (hot, conc.)} \to 4NaF + O_2 + 2H_2O$
Therefore,the products are $OF_2$ and $O_2$ (or $NaF$ and $O_2$ depending on the specific focus of the option). Given the options provided,$D$ is the most appropriate choice as it lists the primary products formed in the respective reactions.