Halogen family Questions in English

(D) The reaction of iodine $(I_2)$ with water at room temperature $(R.T.)$ is given as 'No reaction'.
Since there is no reaction,no oxy acids are formed.
Therefore,the product is not an oxy acid,and it does not have an $-ic$ or $-ous$ suffix.
Thus,the correct option is $D$.

(A) $F_2$ is a very strong oxidizing agent. It oxidizes water to oxygen at room temperature: $2F_2(g) + 2H_2O(l) \to 4HF(aq) + O_2(g)$.
In this reaction,$F_2$ is reduced to $F^-$,but it does not undergo disproportionation because fluorine cannot exhibit a positive oxidation state.
Other halogens like $Cl_2$ and $Br_2$ undergo disproportionation in water to form hydrohalic acid and hypohalous acid: $X_2 + H_2O \to HX + HOX$.

(A) The hydrolysis of $ClF_5$,$XeF_2$,and $SF_4$ produces $HF$ as one of the products. $HF$ reacts with the silica $(SiO_2)$ of glass vessels to form $SiF_4$ or $H_2SiF_6$.
In the case of $BF_3$,the hydrolysis reaction is:
$4BF_3 + 3H_2O \rightarrow H_3BO_3 + 3H[BF_4]$
The product $H[BF_4]$ (fluoroboric acid) does not react with silica $(SiO_2)$.

(C) The reaction of bromine with cold dilute sodium hydroxide is: $Br_2 + 2NaOH \longrightarrow NaBr + NaBrO + H_2O$.
Here,$Y$ is $NaBr$ and $Z$ is $NaBrO$.
$NaBr$ gives a yellow precipitate of $AgBr$ with $AgNO_3$.
$NaBrO$ (sodium hypobromite) acts as an oxidizing agent.
It can oxidize $Cr^{3+}$ to $Cr^{6+}$,$Fe^{2+}$ to $Fe^{3+}$,and $Sn^{2+}$ to $Sn^{4+}$.
However,$Al^{3+}$ is already in its highest stable oxidation state $(+3)$,so it cannot be further oxidized by $NaBrO$.

(D) $I_2$ is oxidized to $HIO_3$ by strong oxidizing agents like concentrated $HNO_3$,concentrated $H_2SO_4$,and $Cl_2$ water.
$1$. $I_2 + 10HNO_3 \rightarrow 2HIO_3 + 10NO_2 + 4H_2O$
$2$. $I_2 + 5H_2SO_4 \rightarrow 2HIO_3 + 5SO_2 + 4H_2O$
$3$. $I_2 + 5Cl_2 + 6H_2O \rightarrow 2HIO_3 + 10HCl$
Concentrated $H_3PO_4$ is a non-oxidizing acid and cannot oxidize $I_2$ to $HIO_3$.

(D) The reaction $I_2 + H_2O \rightarrow HI + HOI$ is non-spontaneous at room temperature $(R.T.)$ because the standard Gibbs free energy change $\Delta G^o$ is positive.
Conversely,the reactions involving $F_2$,$Cl_2$,and $Br_2$ with water are spontaneous $(\Delta G^o < 0)$ under standard conditions.

(D) $1$. $SOCl_2$ decomposes upon heating to release $SO_2$ and $Cl_2$ gas.
$2$. $PbCl_4$ is thermally unstable and decomposes to give $PbCl_2$ and $Cl_2$ gas.
$3$. $FeCl_3$ decomposes at high temperatures to give $FeCl_2$ and $Cl_2$ gas.
$4$. $Hg_2Cl_2$ (calomel) undergoes disproportionation upon heating to give $Hg$ and $HgCl_2$,but it does not release $Cl_2$ gas.

(B) The correct answer is $(B)$.
$Cl^-$ ions have a weak reducing nature and can only be oxidized by strong oxidizing agents.
$(A)$ $2HCl + PbO_2 \to PbO + Cl_2 \uparrow + H_2O$ (Oxidation occurs)
$(B)$ $HCl + \text{conc. } H_2SO_4 \to \text{No reaction}$ (Concentrated $H_2SO_4$ acts as a dehydrating agent but does not oxidize $HCl$ to $Cl_2$)
$(C)$ $4HCl + MnO_2 \to MnCl_2 + Cl_2 \uparrow + 2H_2O$ (Oxidation occurs)
$(D)$ $14HCl + K_2Cr_2O_7 \to 2CrCl_3 + 2KCl + 3Cl_2 \uparrow + 7H_2O$ (Oxidation occurs)

(D) The reaction is: $Cl_{2(g)} + H_2O \xrightarrow{R.T.} \underbrace{HCl}_{(P)} + \underbrace{HClO}_{(Q)}$.
Molecular weight of $HCl$ $(P)$ is $36.5 \ g/mol$ and $HClO$ $(Q)$ is $52.5 \ g/mol$. Since $36.5 < 52.5$,$P$ is $HCl$ and $Q$ is $HClO$.
$(A)$ $HCl$ reacts with $CrO_3$ to form chromyl chloride $(CrO_2Cl_2)$,which gives deep red vapours. This statement is correct.
$(B)$ $HClO$ is an oxidizing agent and exhibits bleaching properties due to the release of nascent oxygen. This statement is correct.
$(C)$ $MnO_2$ oxidizes $HCl$ to $Cl_2$ gas upon heating: $MnO_2 + 4HCl \to MnCl_2 + Cl_2 + 2H_2O$. This statement is correct.
$(D)$ $HCl$ does not react with $H_2S$,whereas $HClO$ (being an oxidizing agent) reacts with $H_2S$ to form sulfur. Thus,the statement that $P$ reacts with $H_2S$ while $Q$ does not is incorrect.

(C) $NO_2$ reacts with water to form $HNO_2$ and $HNO_3$.
$P_4$ does not react with water but reacts with $NaOH$ to form $PH_3$ and $NaH_2PO_2$.
$Al$ does not react with water (due to the formation of a protective oxide layer) but reacts with $NaOH$ to form $NaAlO_2$ and $H_2$.
$I_2$ does not react with water but reacts with $NaOH$ to form $NaI$ and $NaOI$.
Therefore,the species that do not react with water but react with $NaOH$ are $(ii), (iii),$ and $(iv)$.

(C) The process involves two main steps:
$1$. The absorption of $Br_2$ in a hot aqueous solution of $Na_2CO_3$:
$3Br_2 + 3Na_2CO_3 \to 5NaBr + NaBrO_3 + 3CO_2 \uparrow$
$2$. The recovery of pure $Br_2$ by treating the resulting mixture with concentrated $H_2SO_4$:
$5NaBr + NaBrO_3 + 3H_2SO_4 \xrightarrow{\Delta} 3Br_2 + 3Na_2SO_4 + 3H_2O$
Thus,$H_2SO_4$ is used to obtain pure $Br_2$.

(A) The reaction between $NaCl$,$K_2Cr_2O_7$,and concentrated $H_2SO_4$ is known as the Chromyl Chloride test.
The reaction is: $4NaCl + K_2Cr_2O_7 + 6H_2SO_4 \rightarrow 2CrO_2Cl_2 + 2KHSO_4 + 4NaHSO_4 + 3H_2O$.
Here,$X$ is Chromyl chloride $(CrO_2Cl_2)$.
In $CrO_2Cl_2$,let the oxidation state of $Cr$ be $x$.
$x + 2(-2) + 2(-1) = 0$ $\Rightarrow x - 4 - 2 = 0$ $\Rightarrow x = +6$.
Thus,the oxidation state of the central atom $(Cr)$ in $X$ is $+6$.

(B) The bond dissociation energy of halogens generally decreases down the group due to the increase in atomic size and bond length. However,$F_2$ is an exception because of the very small size of the fluorine atom,which leads to significant inter-electronic repulsion between the lone pairs of the two fluorine atoms. This makes the $F-F$ bond weaker than the $Cl-Cl$ bond. The order of bond dissociation energy is $Cl_2 > Br_2 > F_2 > I_2$. Therefore,$Cl_2$ has the highest bond dissociation energy.

(D) The electron gain enthalpy of fluorine is less negative than that of chlorine because of the small size of the fluorine atom.
Due to the small size,the electron density in the $2p$ subshell of fluorine is very high.
When an incoming electron approaches,it experiences significant inter-electronic repulsion from the existing electrons.
In contrast,chlorine has a larger atomic size,which reduces the inter-electronic repulsion,making it easier for an incoming electron to be accommodated.

(B) In $IF_5$,the iodine atom is in the $+5$ oxidation state. Iodine can exhibit a maximum oxidation state of $+7$ in its compounds. Therefore,$IF_5$ can react with additional fluorine to form $IF_7$ according to the reaction: $IF_5 + F_2 \rightarrow IF_7$. In contrast,$SF_6$ has sulfur in its maximum oxidation state of $+6$,and $KF$ and $MgF_2$ contain elements in their stable oxidation states that cannot be further oxidized by fluorine.

(A) Fluorine is the strongest oxidizing agent among the halogens. When it reacts with water,it oxidizes water to oxygen $(O_2)$ and ozone $(O_3)$.
The chemical reaction is as follows:
$2F_2(g) + 2H_2O(l) \rightarrow 4HF(aq) + O_2(g)$
$3F_2(g) + 3H_2O(l) \rightarrow 6HF(aq) + O_3(g)$
Thus,the products formed are $HF$,$O_2$,and $O_3$.

(A) Bleaching powder is chemically represented as $Ca(OCl)Cl$.
It is a mixed salt of calcium,containing two different anions: chloride $(Cl^-)$ and hypochlorite $(OCl^-)$.
These anions are derived from their respective acids: $HCl$ (hydrochloric acid) and $HClO$ (hypochlorous acid).
Therefore,it is a mixed calcium salt of $HCl$ and $HClO$.

(D) $HCl$ is a gas at room temperature $(R.T.)$. Hydrochloric acid is an aqueous solution of $HCl$,which is a liquid,but pure $HCl$ exists as a gas under standard conditions. Therefore,the statement that $HCl$ is liquid at room temperature is false.

(D) $1$. Electrolysis of brine ($NaCl$ solution) produces $Cl_2$ gas at the anode: $2NaCl(aq) + 2H_2O(l) \rightarrow 2NaOH(aq) + H_2(g) + Cl_2(g)$.
$2$. Deacon's process involves the oxidation of $HCl$ by atmospheric oxygen in the presence of $CuCl_2$ catalyst to produce $Cl_2$: $4HCl + O_2 \xrightarrow{CuCl_2} 2Cl_2 + 2H_2O$.
$3$. Heating $MnO_2$ with concentrated $HCl$ is a standard laboratory method to produce $Cl_2$: $MnO_2 + 4HCl \rightarrow MnCl_2 + 2H_2O + Cl_2$.
$4$. Heating $KCl$ with concentrated $H_2SO_4$ produces $HCl$ gas,not $Cl_2$: $KCl + H_2SO_4 \rightarrow KHSO_4 + HCl(g)$. Therefore,this method cannot produce $Cl_2$.

(C) The given order $HClO < HClO_2 < HClO_3 < HClO_4$ represents the increasing order of acidic strength and the stability of the corresponding oxoanions $(ClO^- < ClO_2^- < ClO_3^- < ClO_4^-)$.
As the number of oxygen atoms increases,the stability of the conjugate base increases due to the dispersal of negative charge,which in turn increases the acidic strength.
However,the oxidising power of these oxoacids decreases as the oxidation state of chlorine increases,because the stability of the acid increases.
Therefore,the order for oxidising nature is $HClO > HClO_2 > HClO_3 > HClO_4$.

(B) The reaction of silica $(SiO_2)$ with hydrogen fluoride $(HF)$ proceeds in two steps.
First,silica reacts with $HF$ to form silicon tetrafluoride:
$SiO_2 + 4 HF \rightarrow SiF_4 + 2 H_2O$
Then,the $SiF_4$ formed reacts further with excess $HF$ to produce hexafluorosilicic acid:
$SiF_4 + 2 HF \rightarrow H_2SiF_6$
Thus,the final product is $H_2SiF_6$.

(B) Pseudohalogens are polyatomic analogues of halogen ions. $ICN$ (iodine cyanide) is considered a pseudohalogen analogue of an interhalogen because it behaves similarly to interhalogens like $ICl$ or $IBr$ in its chemical properties.

(A) $HF$ (hydrofluoric acid) reacts with silica $(SiO_2)$ present in glass to form silicon tetrafluoride $(SiF_4)$ and water,which causes etching of the glass.
Therefore,$HF$ is stored in containers made of $Cu$,$Pb$,or wax,which are resistant to its corrosive action.

(B) The oxyacids of chlorine are $HOCl$,$HClO_2$,$HClO_3$,and $HClO_4$.
In all these oxyacids,the central chlorine atom is $sp^3$ hybridized.
Also,all these oxyacids are monobasic,meaning they contain only one $OH$ group attached to the chlorine atom,which can release one $H^+$ ion.
Therefore,they show similarity in their hybridisation state and basicity.

(D) $1$. Bond dissociation energy: Due to the small size of the $F$ atom,the lone pair-lone pair repulsion is high,making the $F-F$ bond weaker than the $Cl-Cl$ bond. Thus,$F-F < Cl-Cl$ is correct.
$2$. Thermal stability: As the bond length increases from $HF$ to $HI$,the bond dissociation energy decreases,making $HF$ the most stable. Thus,$HI < HBr < HCl < HF$ is correct.
$3$. Reducing power: As the bond dissociation energy decreases from $HF$ to $HI$,the ease of releasing $H^+$ ions increases,making $HI$ the strongest reducing agent. Thus,$HI > HBr > HCl > HF$ is correct.
Therefore,all statements are correct.

(D) The chromyl chloride test is used for the detection of chloride ions $(Cl^-)$.
In this test,a chloride salt is heated with potassium dichromate $(K_2Cr_2O_7)$ and concentrated sulfuric acid $(H_2SO_4)$.
The reaction produces chromyl chloride $(CrO_2Cl_2)$ vapors.
In $K_2Cr_2O_7$,the oxidation state of $Cr$ is $+6$.
In $CrO_2Cl_2$,the oxidation state of $Cr$ is calculated as: $x + 2(-2) + 2(-1) = 0$,which gives $x - 4 - 2 = 0$,so $x = +6$.
Since the oxidation state of $Cr$ is $+6$ in both the reactant $(K_2Cr_2O_7)$ and the product $(CrO_2Cl_2)$,it remains constant.

(B) $1$. Ozone $(O_3)$ is a strong oxidizing agent and oxidizes $SO_2$ to $SO_3$: $O_3 + SO_2 \rightarrow O_2 + SO_3$. This statement is correct.
$2$. Diamond has a three-dimensional network structure where all carbon atoms are $sp^3$ hybridized and satisfy their valency; it does not have dangling bonds. Graphite and fullerene also do not possess dangling bonds in their stable forms. Thus,the statement that all three have dangling bonds is incorrect.
$3$. Chlorine reacts with excess ammonia to form nitrogen and ammonium chloride: $8NH_3 + 3Cl_2 \rightarrow N_2 + 6NH_4Cl$. This statement is correct.
$4$. Bromine reacts with hot and concentrated $NaOH$ to form sodium bromide and sodium bromate: $3Br_2 + 6NaOH \rightarrow 5NaBr + NaBrO_3 + 3H_2O$. This statement is correct.

(A) The thermal stability 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 compounds are:
$HOCl$: $+1$
$HOClO$: $+3$
$HOClO_2$: $+5$
$HOClO_3$: $+7$
As the oxidation state increases,the number of oxygen atoms bonded to the chlorine atom increases,which increases the stability of the molecule due to resonance and the electronegative effect of oxygen atoms.
Therefore,$HOClO_3$ (perchloric acid) is the most thermally stable compound among the given options.

(D) Silica $(SiO_2)$ is an acidic oxide and is generally inert to most acids. However,it reacts with hydrofluoric acid $(HF)$ to form silicon tetrafluoride $(SiF_4)$ and water. The reaction is: $SiO_2 + 4HF \rightarrow SiF_4 + 2H_2O$. Further,$SiF_4$ can react with excess $HF$ to form hexafluorosilicic acid $(H_2SiF_6)$: $SiF_4 + 2HF \rightarrow H_2SiF_6$. Thus,silica is soluble in $HF$.

(C) Marine plants,such as brown algae (kelp),are known to accumulate high concentrations of iodine from seawater. Therefore,they serve as an important commercial source of iodine.

(C) Industrially,bromine is obtained from concentrated sea water or brine (common salt solution) by displacement with chlorine gas.
$2Br^- + Cl_2 \rightarrow 2Cl^- + Br_2$
Therefore,the correct option is $C$.

(D) Iodine $(I_{2})$ is sparingly soluble in water but dissolves readily in an aqueous solution of potassium iodide $(KI)$ due to the formation of the triiodide ion $(I_{3}^{-})$.
The reaction is: $I_{2} + KI \rightarrow KI_{3}$.
Thus,the resulting species is $KI_{3}$.

(D) The oxidizing power of halogens depends on their standard reduction potential $(E^\circ)$.
Fluorine $(F_2)$ has the highest standard reduction potential $(+2.87 \ V)$,making it the strongest oxidizing agent.
The order of standard reduction potentials is $I_2 < Br_2 < Cl_2 < F_2$.
Therefore,the correct order of oxidizing power is $I_2 < Br_2 < Cl_2 < F_2$.

(B) Iodine $(I_2)$ is a non-polar molecule.
When it is dissolved in a non-polar solvent like carbon tetrachloride $(CCl_4)$,it retains its molecular state and exhibits a characteristic violet color.
In contrast,when dissolved in polar solvents like water (in the presence of $KI$),it forms a polyiodide complex $(I_3^-)$,which appears brown.

(D) $Cl_2$ gas is acidic in nature. To dry it,we must pass it through a drying agent that is also acidic or neutral,so that it does not react with the gas.
$CaO$,$NaOH$,and $KOH$ are basic in nature and will react with $Cl_2$ gas.
Concentrated $H_2SO_4$ is an acidic drying agent and does not react with $Cl_2$,making it suitable for drying the gas.

(B) The balanced chemical equation for the reaction of fluorine with hot and concentrated $KOH$ is:
$2F_2 + 4KOH \rightarrow 4KF + 2H_2O + O_2$
Dividing the entire equation by $2$ to match the condition of $1 \ mol$ of $F_2$:
$F_2 + 2KOH \rightarrow 2KF + H_2O + 0.5O_2$
From the balanced equation,the molar ratio of $KF : H_2O : O_2$ is $2 : 1 : 0.5$.

(A) Bleaching powder is chemically $CaOCl_2$. When it reacts with concentrated $HCl$,it releases $Cl_2$ gas.
The chemical reaction is: $CaOCl_2 + 2HCl \rightarrow CaCl_2 + H_2O + Cl_2 \uparrow$.

(A) $1$. Halogens $(F_2, Cl_2, Br_2, I_2)$ exist as diatomic molecules $(X_2)$.
$2$. They have a valence shell configuration of $ns^2 np^5$,making them monovalent (valency = $1$) as they gain one electron to complete their octet.
$3$. While fluorine only shows an oxidation state of $-1$,other halogens $(Cl, Br, I)$ show variable oxidation states $(-1, +1, +3, +5, +7)$.
$4$. Option $A$ is the most accurate general description of their elemental state and valency.

(A) The oxidizing power of a halogen depends on the standard electrode potential $(E^\circ)$.
$F_2$ has a very high positive standard electrode potential $(+2.87 \ V)$ compared to $Br_2$ $(+1.09 \ V)$.
This is primarily due to the low bond dissociation enthalpy of the $F-F$ bond and the high hydration enthalpy of the $F^-$ ion.
However,the small size of the fluorine atom leads to high electron-electron repulsion in the $F^-$ ion,which is a significant factor in its thermodynamic properties,but the overall oxidizing strength is governed by the sum of atomization,ionization/electron gain,and hydration energies.
Among the given options,the small size of the fluorine atom is the fundamental reason for its unique properties,including its high oxidizing power.

(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: $HF < HCl < HBr < HI$.
Since $HF$ has the strongest $H-X$ bond due to its small size,it is the weakest acid among the given options.

(D) Chlorine is prepared by the oxidation of hydrochloric acid $(HCl)$ using oxidizing agents.
Common laboratory methods include:
$1$. $MnO_2 + 4HCl \rightarrow MnCl_2 + Cl_2 + 2H_2O$
$2$. $2KMnO_4 + 16HCl \rightarrow 2KCl + 2MnCl_2 + 8H_2O + 5Cl_2$
Both $MnO_2$ and $KMnO_4$ are effective oxidizing agents used to produce chlorine gas from $HCl$.

(D) Fluorine $(F_2)$ is the most electronegative element and acts as a strong oxidizing agent. When $F_2$ is passed through a hot and concentrated solution of $KOH$,it oxidizes water to oxygen $(O_2)$ and itself gets reduced to fluoride ions $(F^-)$.
The chemical reaction is: $2F_2 + 4KOH \rightarrow 4KF + 2H_2O + O_2$.
Therefore,the correct option is $D$.

(D) The correct order of electron gain enthalpy is $Cl > F > Br > I$. Due to the small size of the $F$ atom,there is strong inter-electronic repulsion,making the addition of an electron less favorable compared to $Cl$. Thus,$F < Cl$.
Also,the bond dissociation energy order is $Cl_2 > Br_2 > F_2 > I_2$. The $F-F$ bond is weaker than the $Cl-Cl$ bond due to high inter-electronic repulsion between the lone pairs of the small $F$ atoms.
Therefore,both options $B$ and $D$ are inconsistent with the properties indicated. However,in standard multiple-choice questions of this type,$D$ is often the primary focus due to the anomalous weakness of the $F-F$ bond.

(B) The mixture of concentrated $HCl$ and $HNO_3$ in a $3:1$ molar ratio is known as aqua regia.
The chemical reaction is: $HNO_3 + 3HCl \rightarrow NOCl + Cl_2 + 2H_2O$.
The product $NOCl$ (nitrosyl chloride) is formed in this reaction.

(D) The enrichment of ${}^{235}U$ requires the conversion of uranium into $UF_6$ (uranium hexafluoride),which is a volatile compound.
$ClF_3$ (chlorine trifluoride) is used as a fluorinating agent in this process to convert uranium metal or uranium oxides into $UF_6$.

(A) Iodine is a halogen with the valence shell configuration $ns^2 np^5$.
It can exhibit oxidation states of $-1, +1, +3, +5,$ and $+7$.
Since the provided options are incomplete based on standard chemical knowledge,the most comprehensive set including the common states is selected.