Halogen family Questions in English

(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.

(C) The reactions are as follows:
$(A)$ $2F_2 + 4NaOH \longrightarrow 4NaF + O_2 + 2H_2O$ (Forms $O_2$)
$(B)$ $2F_2 + 2H_2O \longrightarrow 4HF + O_2$ (Forms $O_2$)
$(C)$ $Cl_2 + 2NaOH \text{ (cold, dil)} \longrightarrow NaCl + NaClO + H_2O$ (Does not form $O_2$)
$(D)$ $CaOCl_2 + H_2SO_4 \longrightarrow CaSO_4 + H_2O + Cl_2 + \frac{1}{2}O_2$ (Forms $O_2$)
Therefore,the pair that does not form oxygen is $Cl_2$ and cold,dilute $NaOH$.

(D) Fluorine is the most electronegative element and acts as a strong oxidizing agent. When $F_2$ is passed through a hot and concentrated $KOH$ solution,it oxidizes water to oxygen gas.
The balanced chemical equation is:
$2F_2 + 4KOH \longrightarrow 4KF + 2H_2O + O_2$

(C) When chlorine is passed through an aqueous solution of sodium thiosulfate (hypo),it acts as an oxidizing agent. The reaction is as follows:
$Na_2S_2O_3 + H_2O + Cl_2 \longrightarrow Na_2SO_4 + 2HCl + S \downarrow$
Thus,the products formed are sodium sulfate $(Na_2SO_4)$,hydrochloric acid $(HCl)$,and sulfur $(S)$.

(A) Hypo is sodium thiosulfate,$Na_2S_2O_3 \cdot 5H_2O$. When moist chlorine gas reacts with sodium thiosulfate,it acts as an oxidizing agent. The reaction is:
$Na_2S_2O_3 + 5H_2O + 4Cl_2 \rightarrow 2NaHSO_4 + 8HCl$.
However,in the context of standard chemistry problems regarding the reaction of hypo with chlorine,it is often represented as:
$Na_2S_2O_3 + H_2O + Cl_2 \rightarrow Na_2SO_4 + S + 2HCl$.
Thus,the products formed are $Na_2SO_4, S, \text{and } HCl$.

(B) When chlorine gas reacts with hot and concentrated sodium hydroxide $(NaOH)$,it undergoes a disproportionation reaction to form sodium chloride $(NaCl)$ and sodium chlorate $(NaClO_3)$.
The balanced chemical equation is:
$3Cl_2 + 6NaOH \rightarrow 5NaCl + NaClO_3 + 3H_2O$
Therefore,apart from sodium chloride,sodium chlorate $(NaClO_3)$ is formed.

(A) As the atomic size increases down the group,the $A-A$ bond length increases.
Since bond energy is inversely proportional to bond length,the bond energy decreases as the size of the halogen atom increases.
Therefore,the order of bond energy for the given halogens is $Cl_2 > Br_2 > I_2$.

(D) The bond dissociation energy of halogen molecules generally decreases down the group due to the increase in atomic size and bond length.
However,$F_2$ is an exception due to the high inter-electronic repulsion between the lone pairs of the small fluorine atoms.
The correct order for bond dissociation energies is $Cl_2 > Br_2 > F_2 > I_2$.
Among the given options,the order $Cl_2 > Br_2 > I_2$ is correct.

(D) The reaction $2 F_{2} + 2 H_{2}O \rightarrow 4 HF + O_{2}$ occurs because $F_{2}$ is a strong oxidizing agent.
$F$ is the most electronegative element in the periodic table,so the statement that $F$ is less electronegative than $O$ is false.
$F_{2}$ is a stronger oxidizing agent than $O_{2}$ because of its high electronegativity and the low bond dissociation energy of the $F-F$ bond compared to the $O=O$ bond.
The $H-F$ bond is stronger than the $H-O$ bond due to the high electronegativity of $F$.

(D) Anhydrous $FeCl_3$ is prepared by the reaction of dry $Cl_2$ gas with heated iron metal.
The reaction is: $2 Fe + 3 Cl_2 \xrightarrow{\Delta} 2 FeCl_3$.
If $HCl$ is used,it often results in the formation of hydrated ferric chloride due to the presence of water.

(B) The acidic strength of hydrohalic acids $(HX)$ depends on the bond dissociation enthalpy of the $H-X$ bond.
As we move down the group from $F$ to $I$,the atomic size increases,which leads to a decrease in the bond dissociation enthalpy of the $H-X$ bond.
Therefore,the ease of releasing $H^+$ ions increases,making the acidity increase in the order: $HF < HCl < HBr < HI$.

(B) Chlorine bleach is $NaOCl$. The hypochlorite ion in chlorine bleach dissociates to give nascent oxygen as shown below.
$OCl^{-} \longrightarrow [O] + Cl^{-}$
Therefore,the reactive species in chlorine bleach is $OCl^{-}$.
So,the option $(B)$ is correct.

(D) $Cl_{2}O_{7}$ is the anhydride of $HClO_{4}$.
An acid anhydride is a compound that reacts with water to form an acid.
The reaction is:
$Cl_{2}O_{7} + H_{2}O \longrightarrow 2HClO_{4}$ (Perchloric acid)
$Cl_{2}O_{7}$ is the oxide of chlorine in its highest oxidation state $(+7)$,which corresponds to perchloric acid $(HClO_{4})$.

(C) Fluorine is a powerful oxidizing agent. It reacts vigorously with water at room temperature to produce oxygen gas and hydrofluoric acid $(HF)$. In an aqueous medium,$HF$ dissociates into $H^{+}$ and $F^{-}$ ions. The chemical equation for the reaction is:
$2F_2 + 2H_2O \rightarrow 4H^{+} + 4F^{-} + O_2$

(D) The reaction of chlorine gas with red hot calcium oxide is given by the following chemical equation:
$2 CaO + 2 Cl_2 \xrightarrow{\text{Red hot}} 2 CaCl_2 + O_2 \uparrow$
Thus,the products formed are calcium chloride and oxygen.

(A) The thermal decomposition of chloric acid $(HClO_{3})$ is given by the following reaction:
$3 HClO_{3} \stackrel{\Delta}{\longrightarrow} HClO_{4} + Cl_{2} + 2 O_{2} + H_{2}O$
Thus,the products formed are perchloric acid $(HClO_{4})$,chlorine $(Cl_{2})$,oxygen $(O_{2})$,and water $(H_{2}O)$.

(C) The preparation of $Cl_2$ gas from concentrated $HCl$ at room temperature is achieved by using strong oxidizing agents like $KMnO_4$.
The chemical reaction is: $2KMnO_4 + 16HCl \rightarrow 2KCl + 2MnCl_2 + 8H_2O + 5Cl_2$.
While $MnO_2$ also oxidizes $HCl$ to $Cl_2$,it typically requires heating,whereas $KMnO_4$ can perform this reaction at room temperature.

Correct order of
($i$) Boiling point: $HF > HI > HBr > HCl$
($ii$) Melting point: $HI > HF > HBr > HCl$

(A) Statement $I$ is true; the standard reduction potential values $(E^0)$ decrease from $F_2$ to $I_2$,hence the order of oxidising power is $F_2 > Cl_2 > Br_2 > I_2$.
Statement $II$ is also true; the "layer test" uses the fact that a stronger oxidising agent $(Cl_2)$ can displace a weaker halide from its solution.
The reaction $Cl_2 + 2Br^- \rightarrow 2Cl^- + Br_2$ is a displacement redox reaction where bromide is oxidised to $Br_2$.
Both statements are correct.