Halogen family Questions in English

(A) Fluorine is the most electronegative element and has the smallest atomic size among all halogens.
Due to its small size and high electronegativity,it cannot accommodate multiple oxygen atoms or exhibit positive oxidation states,thus it forms only one oxoacid,which is hypofluorous acid $(HOF)$.
Therefore,both Assertion $A$ and Reason $R$ are correct,and $R$ is the correct explanation of $A$.

(C) When bromine reacts with an excess of fluorine,the reaction is as follows:
$Br_2 + 5F_2 (\text{excess}) \longrightarrow 2BrF_5$
Thus,the interhalogen compound formed is $BrF_5$ (bromine pentafluoride).

(B) The interhalogens with a square pyramidal structure are those of the type $AX_{5}$,which undergo $sp^{3}d^{2}$ hybridization.
From the given list,the molecules with square pyramidal geometry are $BrF_{5}$,$IF_{5}$,and $ClF_{5}$.
Therefore,the total number of such interhalogens is $3$.

(C) Iodine $(I_{2})$ is a non-metal that acts as a reducing agent when it reacts with strong oxidizing agents like concentrated nitric acid $(HNO_{3})$.
In this reaction,$I_{2}$ is oxidized to iodic acid $(HIO_{3})$,while $HNO_{3}$ is reduced to nitrogen dioxide $(NO_{2})$.
The balanced chemical equation is:
$I_{2} + 10 HNO_{3(conc)} \rightarrow 2 HIO_{3} + 10 NO_{2} + 4 H_{2}O$

(B) $pK_a$ value of an acid is inversely proportional to its acidic strength,i.e.,the stronger the acid,the lower its $pK_a$ value.
Acidic strength increases with an increase in the oxidation number of the central chlorine atom.
The oxidation numbers of the central atom in the given acids are:

Acid	Oxidation Number of $Cl$
$HClO$	$+1$
$HClO_2$	$+3$
$HClO_3$	$+5$
$HClO_4$	$+7$

The acidic strength follows the order: $HClO < HClO_2 < HClO_3 < HClO_4$.
Therefore,the $pK_a$ order is the inverse of the acidic strength order: $HClO_4 < HClO_3 < HClO_2 < HClO$.

(C) The correct option is $C$.
When $MnO_2$ is treated with conc. $HCl$,it produces chlorine gas $(X)$ as follows:
$MnO_2 + 4HCl \rightarrow MnCl_2 + Cl_2(g) + 2H_2O$
Here,$X = Cl_2$.
Chlorine gas reacts with dry slaked lime $Ca(OH)_2$ to form bleaching powder,which is calcium oxychloride $Ca(OCl)Cl$ $(Y)$:
$Ca(OH)_2 + Cl_2 \rightarrow Ca(OCl)Cl + H_2O$
Here,$Y = Ca(OCl)Cl$.
Bleaching powder $(Y)$ reacts with dilute $HCl$ to regenerate chlorine gas $(X)$:
$Ca(OCl)Cl + 2HCl \rightarrow CaCl_2 + H_2O + Cl_2(g)$

(A) .
On heating $MnO_2$ with conc. $HCl$,$MnCl_2$ and $H_2O$ are produced along with the evolution of $Cl_2$ gas.
The balanced chemical equation is:
$MnO_2 + 4HCl \longrightarrow MnCl_2 + 2H_2O + Cl_2$

(A) The bond dissociation energy of $F_2$ is unexpectedly low due to the high inter-electronic repulsion between the lone pairs of the small $F$ atoms.
Therefore,the correct order of bond dissociation energy for halogens is $Cl_2 > Br_2 > F_2 > I_2$.
Thus,$Cl_2$ has the highest bond dissociation energy.

(A) The standard electrode potential $(E^{\circ})$ of a species is determined by the overall energy change involved in the process: $\frac{1}{2} F_2(g) + e^- \rightarrow F^-(aq)$.
This process involves three steps: $(1)$ Atomization (bond dissociation),$(2)$ Electron gain,and $(3)$ Hydration.
Fluorine has a very low bond dissociation enthalpy due to inter-electronic repulsions between lone pairs in the small $F_2$ molecule.
Additionally,the small size of the $F^-$ ion results in a very high hydration enthalpy,which makes the overall process highly exothermic.
Therefore,both the low bond dissociation enthalpy and the high hydration enthalpy contribute to the strong oxidizing power of fluorine.

(A) The reaction $2 Cu^{2+} + 4 X^{-} \rightarrow Cu_2 X_{2(s)} + X_2$ represents the reduction of $Cu^{2+}$ to $Cu^+$ by halide ions.
This reaction occurs specifically with iodide ions $(I^-)$ because $CuI_2$ is unstable and decomposes to form $Cu_2I_2$ and $I_2$.
Chlorine and bromine are not strong enough reducing agents to reduce $Cu^{2+}$ to $Cu^+$ under these conditions.
Therefore,the correct answer is $I^-$ (Iodine).

(C) Statement $I$ is correct: According to Fajan's rule,the cation with higher charge $(Sb^{+5})$ has greater polarising power than the one with lower charge $(Sb^{+3})$,making $SbCl_5$ more covalent than $SbCl_3$.
Statement $II$ is incorrect: Generally,the lower oxides of halogens are more stable than the higher oxides. Higher oxides of halogens are often unstable and tend to decompose or explode.

(C) The central atom $Cl$ in $ClF_5$ has $7$ valence electrons. It forms $5$ bonds with $F$ atoms and has $1$ lone pair of electrons.
Total electron pairs = $5 + 1 = 6$,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.
$ClF_5$ exists as a colourless liquid at room temperature.

(D) Among the halide ions,only $I^{-}$ can be oxidized to $I_2$ by oxygen in an acidic medium because it has the highest standard oxidation potential.
The balanced chemical equation is:
$4I^{-}_{(aq)} + 4H^{+}_{(aq)} + O_{2(g)} \rightarrow 2I_{2(s)} + 2H_2O_{(l)}$

(D) Fluorine is the most electronegative element in the periodic table and has no $d$-orbitals in its valence shell. Due to its high electronegativity and lack of $d$-orbitals,it only exhibits an oxidation state of $-1$ in its compounds. Other halogens like $Cl$,$Br$,and $I$ exhibit variable oxidation states (e.g.,$-1, +1, +3, +5, +7$) due to the presence of vacant $d$-orbitals.

(B) disproportionation reaction is a type of redox reaction where the same element is simultaneously oxidized and reduced.
$F_2$ is the most electronegative element and cannot exhibit a positive oxidation state,so it cannot undergo disproportionation.
$Cl_2$,$Br_2$,and $I_2$ can exist in various positive oxidation states (such as $+1, +3, +5, +7$) and thus undergo disproportionation in alkaline media.
Therefore,$Cl_2$,$Br_2$,and $I_2$ can undergo disproportionation reactions.

(B) Statement $[A]$ is correct: $HClO_4$ is a much stronger acid than $HClO$ because the perchlorate anion $(ClO_4^-)$ is highly resonance-stabilized,whereas the hypochlorite anion $(ClO^-)$ is not.
Statement $[B]$ is incorrect: The reaction between $Cl_2$ and $H_2O$ produces $HCl$ and $HOCl$ $(HClO)$,not $HClO_4$.
Statement $[C]$ is correct: In $HClO_4$,the central $Cl$ atom is $sp^3$ hybridized ($4$ sigma bonds). In $HClO$,the central $Cl$ atom is also $sp^3$ hybridized ($2$ lone pairs and $2$ sigma bonds).
Statement $[D]$ is correct: Since $HClO_4$ is a very strong acid,its conjugate base $ClO_4^-$ is an extremely weak base,much weaker than $H_2O$.

(C) The colour of halogens arises due to the absorption of light in the visible region,which promotes an electron from the highest occupied molecular orbital $(HOMO)$ to the lowest unoccupied molecular orbital $(LUMO)$.
For $X_2$ molecules,the $HOMO$ is the $\pi^*$ orbital and the $LUMO$ is the $\sigma^*$ orbital.
As we move down the group from $F_2$ to $I_2$,the size of the atoms increases,leading to a decrease in the energy gap between the $\pi^*$ and $\sigma^*$ orbitals.
This decrease in the $HOMO-LUMO$ gap allows for the absorption of light of lower energy (longer wavelength) as we go down the group,resulting in the observed shift in colour from yellow to violet.
Thus,both statements $B$ and $C$ describe the same physical phenomenon.

(A) $(A). (CH_3)_2 SiCl_2 + 2 H_2O \rightarrow (CH_3)_2 Si(OH)_2 + 2 HCl$. The product $(CH_3)_2 Si(OH)_2$ undergoes polymerization to form silicones. Thus,it matches $(p)$ and $(s)$.
$(B). 3 XeF_4 + 6 H_2O \rightarrow XeO_3 + 2 Xe + 12 HF + 1.5 O_2$. This reaction involves hydrogen halide $(HF)$ formation $(p)$,redox reaction $(q)$,reaction with glass (due to $HF$ produced) $(r)$,and $O_2$ formation $(t)$. Thus,it matches $(p, q, r, t)$.
$(C). Cl_2 + H_2O \rightarrow HCl + HOCl$. Further,$2 HOCl \rightarrow 2 HCl + O_2$. This involves hydrogen halide $(HCl)$ formation $(p)$,redox reaction $(q)$,and $O_2$ formation $(t)$. Thus,it matches $(p, q, t)$.
$(D). VCl_5 + H_2O \rightarrow VOCl_3 + 2 HCl$. This involves hydrogen halide $(HCl)$ formation $(p)$. Thus,it matches $(p)$.

(A) $PCl_5$ reacts with compounds containing oxygen to form $POCl_3$ under specific conditions. Let us analyze each compound:
$1$. $O_2$: Does not react with $PCl_5$.
$2$. $CO_2$: Does not react with $PCl_5$.
$3$. $SO_2$: $PCl_5 + SO_2 \rightarrow POCl_3 + SOCl_2$. (Reacts)
$4$. $H_2O$: $PCl_5 + H_2O \rightarrow POCl_3 + 2HCl$. (Reacts with limited water)
$5$. $H_2SO_4$: $2PCl_5 + H_2SO_4 \rightarrow 2POCl_3 + SO_2Cl_2 + 2HCl$. (Reacts)
$6$. $P_4O_{10}$: $6PCl_5 + P_4O_{10} \rightarrow 10POCl_3$. (Reacts)
Thus,the compounds that react to give $POCl_3$ are $SO_2, H_2O, H_2SO_4$,and $P_4O_{10}$. The total number of such compounds is $4$.

(D) $(1)$ The reaction of gold with aqua regia produces $NO$ (nitric oxide),not $NO_2$. Thus,statement $(1)$ is incorrect.
$(2)$ Aqua regia is indeed prepared by mixing concentrated $HCl$ and concentrated $HNO_3$ in a $3:1$ volume ratio. Thus,statement $(2)$ is correct.
$(3)$ The reaction is $Au + 4H^+ + NO_3^- + 4Cl^- \rightarrow AuCl_4^- + NO + 2H_2O$. The complex ion $AuCl_4^-$ contains gold in the $+3$ oxidation state. Thus,statement $(3)$ is correct.
$(4)$ The yellow-orange colour of aqua regia is due to the formation of nitrosyl chloride $(NOCl)$ and chlorine $(Cl_2)$. Thus,statement $(4)$ is correct.

(B) Hypochlorite ion: $ClO^{-}$
Chlorate ion: $ClO_{3}^{-}$
Perchlorate ion: $ClO_{4}^{-}$
$A.$ Acidic strength order: $HClO < HClO_{3} < HClO_{4}$. Therefore,the conjugate base strength order is $ClO^{-} > ClO_{3}^{-} > ClO_{4}^{-}$. Thus,$A$ is correct.
$B.$ Hypochlorite ion $(ClO^{-})$: Linear shape. Chlorate ion $(ClO_{3}^{-})$: Trigonal pyramidal shape due to the lone pair on $Cl$. Perchlorate ion $(ClO_{4}^{-})$: Tetrahedral shape. Only the chlorate ion's shape is significantly distorted by a lone pair. Thus,$B$ is correct.
$C.$ Disproportionation reactions:
$I.$ $3ClO^{-} \rightarrow 2Cl^{-} + ClO_{3}^{-}$
$II.$ $4ClO_{3}^{-} \rightarrow 3ClO_{4}^{-} + Cl^{-}$
The products are different. Thus,$C$ is incorrect.
$D.$ Hypochlorite is a strong oxidizing agent and oxidizes sulfite $(SO_{3}^{2-})$ to sulfate $(SO_{4}^{2-})$: $ClO^{-} + SO_{3}^{2-} \rightarrow Cl^{-} + SO_{4}^{2-}$. Thus,$D$ is correct.
Correct statements are $A, B, D$.

(A) $1.$ The reaction of $Cl_2$ with cold dilute $NaOH$ is: $Cl_2 + 2NaOH \rightarrow NaCl + NaOCl + H_2O$. Here,$NaOCl$ is the salt of hypochlorous acid $(P)$.
The reaction of $Cl_2$ with hot concentrated $NaOH$ is: $3Cl_2 + 6NaOH \rightarrow 5NaCl + NaClO_3 + 3H_2O$. Here,$NaClO_3$ is the salt of chloric acid $(Q)$.
$2.$ The reaction of $Cl_2$ with $SO_2$ in the presence of charcoal gives sulfuryl chloride: $SO_2 + Cl_2 \xrightarrow{\text{charcoal}} SO_2Cl_2 (R)$.
$SO_2Cl_2$ reacts with white phosphorus $(P_4)$ to give phosphorus pentachloride: $10SO_2Cl_2 + P_4 \rightarrow 4PCl_5 (S) + 10SO_2$.
Hydrolysis of $PCl_5$ gives phosphoric acid: $PCl_5 + 4H_2O \rightarrow H_3PO_4 (T) + 5HCl$.

(B) Let us analyze the structures of the oxoacids of chlorine:
$(i) HClO$: $H-O-Cl$ ($3$ lone pairs on $Cl$)
$(ii) HClO_2$: $H-O-Cl=O$ ($2$ lone pairs on $Cl$,$1$ $Cl=O$ bond)
$(iii) HClO_3$: $H-O-Cl(=O)_2$ ($1$ lone pair on $Cl$,$2$ $Cl=O$ bonds)
$(iv) HClO_4$: $H-O-Cl(=O)_3$ ($0$ lone pairs on $Cl$,$3$ $Cl=O$ bonds)
Evaluating the statements:
$(A)$ Number of $Cl=O$ bonds in $(ii)$ is $1$ and in $(iii)$ is $2$. Total = $3$. Statement $(A)$ is incorrect.
$(B)$ Number of lone pairs on $Cl$ in $(ii)$ is $2$ and in $(iii)$ is $1$. Total = $3$. Statement $(B)$ is correct.
$(C)$ In $HClO_4$,$Cl$ is bonded to $4$ oxygen atoms (one $-OH$ and three $=O$). Steric number = $4+0 = 4$,so hybridization is $sp^3$. Statement $(C)$ is correct.
$(D)$ Acidic strength increases with the number of oxygen atoms attached to the central atom. Thus,$HClO_4$ is the strongest acid,not $HClO$. Statement $(D)$ is incorrect.
Therefore,the correct statements are $(B)$ and $(C)$.

(C) The reaction of $HClO_3$ with $HCl$ produces $ClO_2$ gas,which is paramagnetic due to the presence of an unpaired electron.
$2HClO_3 + 2HCl \rightarrow 2ClO_2 + Cl_2 + 2H_2O$
When $ClO_2$ reacts with $O_3$,it forms $Cl_2O_6$.
$2ClO_2 + 2O_3 \rightarrow Cl_2O_6 + 2O_2$.

(D) The oxidising capacity of a species is directly proportional to its standard reduction potential $(E^{\ominus})$.
Higher the positive value of standard reduction potential,stronger is the oxidising agent.
Comparing the given values:
$Sn^{4+}/Sn^{2+} = +1.15 \ V$
$Tl^{+}/Tl = +1.26 \ V$
$Al^{3+}/Al = -1.66 \ V$
$Pb^{4+}/Pb^{2+} = +1.67 \ V$
Since $Pb^{4+}/Pb^{2+}$ has the highest positive reduction potential $(+1.67 \ V)$,it has the strongest oxidising capacity.

(C) The trends for the first four elements of group-$17$ $(F, Cl, Br, I)$ are as follows:
$1$. Covalent radius: $F < Cl < Br < I$ (Regular trend).
$2$. Electron affinity: $Cl > F > Br > I$ (Irregular,as $F$ has lower electron affinity than $Cl$ due to small size and inter-electronic repulsion).
$3$. Ionic radius: $F^- < Cl^- < Br^- < I^-$ (Regular trend).
$4$. First ionization energy: $F > Cl > Br > I$ (Regular trend).
Thus,only electron affinity shows an irregularity in the expected periodic trend.

(C) The given order $HClO < HClO_2 < HClO_3 < HClO_4$ represents the increasing order of oxidation state of chlorine $(+1, +3, +5, +7)$.
As the oxidation state increases,the acidic strength and thermal stability of the oxoacids increase.
Similarly,the stability of the corresponding anions $(ClO^- < ClO_2^- < ClO_3^- < ClO_4^-)$ also increases due to the dispersal of negative charge.
However,the oxidising power decreases as the oxidation state increases because the chlorine atom is already in a higher oxidation state.
Therefore,the correct order for oxidising nature is $HClO > HClO_2 > HClO_3 > HClO_4$.

(C) The correct order of bond dissociation energy for halogens is $Cl_{2} > Br_{2} > F_{2} > I_{2}$.
$F_{2}$ has a lower bond energy than $Cl_{2}$ and $Br_{2}$ due to high inter-electronic repulsion between the lone pairs of the small $F$ atoms.
Therefore,the order given in option $C$ $(F_{2} > Cl_{2} < Br_{2} > I_{2})$ is incorrect.

(D) Hydrolysis of a compound requires the presence of vacant $d$-orbitals on the central atom to accept the lone pair of electrons from the oxygen atom of water,or the presence of a highly polar bond.
$NF_3$ does not undergo hydrolysis because nitrogen lacks vacant $d$-orbitals and the $N-F$ bond is very strong,making it kinetically inert.
$CCl_4$ does not undergo hydrolysis because carbon lacks vacant $d$-orbitals to accommodate the incoming water molecule,and the $C-Cl$ bond is sterically hindered.
$NCl_3$ undergoes hydrolysis to form $NH_3$ and $HOCl$ because the $N-Cl$ bond is polar and the chlorine atom can stabilize the transition state.
Therefore,both $NF_3$ and $CCl_4$ do not undergo hydrolysis.

(C) The reaction between $K_2Cr_2O_7$,concentrated $H_2SO_4$,and $NaCl$ is known as the Chromyl Chloride test.
The chemical equation is: $K_2Cr_2O_7 + 4NaCl + 6H_2SO_4 \rightarrow 2CrO_2Cl_2 + 2KHSO_4 + 4NaHSO_4 + 3H_2O$.
The red-orange vapours produced are of Chromyl chloride $(CrO_2Cl_2)$.

(A) Atom $X$ belongs to group $17$ and is not in the second period,so $X$ could be $Cl$,$Br$,or $I$.
The general formula for the most acidic oxyacid of halogens is $HXO_4$ (e.g.,$HClO_4$,$HBrO_4$,$HIO_4$).
Statement $A$ claims the most acidic oxyacid is $HXO_3$,which is incorrect.
The highest oxidation state for group $17$ elements is $+7$.
Halogens can form interhalogen compounds with other elements of their group.
Oxides of halogens are acidic in nature,so their aqueous solutions have a $pH < 7$.

(A) Fluorine $(F_2)$ is a very strong oxidizing agent and it oxidizes water to oxygen $(O_2)$ or ozone $(O_3)$ instead of undergoing disproportionation.
$2F_2 + 2H_2O \rightarrow 4HF + O_2$
Iodine $(I_2)$ is a very weak oxidizing agent and it is not strong enough to undergo disproportionation in water,as the reaction is non-spontaneous.
Chlorine $(Cl_2)$ and Bromine $(Br_2)$ undergo disproportionation reactions in water to form hypohalous acids and hydrohalic acids.
Therefore,the pair that does not undergo disproportionation with water is $F_2$ and $I_2$.

(B) Correct: Due to the small size of the fluorine atom,the incoming electron experiences strong inter-electronic repulsion,making its electron gain enthalpy less negative than that of chlorine.
$(b)$ Incorrect: Iodine is a solid at room temperature,while fluorine and chlorine are gases and bromine is a liquid. Thus,it has a different physical state.
$(c)$ Correct: Fluorine is the most electronegative element and only shows a $-1$ oxidation state.
$(d)$ Correct: Chlorine and bromine undergo disproportionation reactions with water to form hydrohalic acid and hypohalous acid.

(C) The reaction is: $2 NaCl + MnO_2 + 2 H_2SO_4 \rightarrow Na_2SO_4 + MnSO_4 + 2 H_2O + Cl_2$.
The gas $X$ is chlorine $(Cl_2)$,which is a greenish-yellow gas with a pungent,suffocating smell.
Chlorine gas acts as a strong oxidizing agent in the presence of moisture and exhibits bleaching properties due to the formation of nascent oxygen.
Therefore,the statement that '$X$ does not show bleaching property' is incorrect.

(C) The acidic strength of hydrohalic acids 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,$HI$ is the strongest acid among the given options.

(D) The bond dissociation enthalpy depends on the bond length and the repulsion between lone pairs of electrons on the bonded atoms.
In $F-F$,the atoms are very small,which leads to significant inter-electronic repulsion between the lone pairs of the two fluorine atoms.
This repulsion weakens the $F-F$ bond,making its bond dissociation enthalpy lower than that of $Cl-Cl$.
Therefore,the correct decreasing order of bond dissociation enthalpy is $Cl-Cl > Br-Br > F-F > I-I$.

(D) The bond dissociation enthalpy of halogens generally decreases down the group due to an increase in atomic size.
However,$F_2$ is an exception because of the small size of the fluorine atom.
In $F_2$,the lone pairs on the fluorine atoms experience significant inter-electronic repulsion,which weakens the $F-F$ bond.
Therefore,$F_2$ has a lower bond dissociation enthalpy than $Cl_2$ and $Br_2$,but $I_2$ has the lowest bond dissociation enthalpy due to the largest atomic size and longest bond length.

(A) The bond dissociation enthalpy of halogens generally decreases down the group due to an increase in atomic size.
However,$F_2$ has an exceptionally low bond dissociation enthalpy due to the high inter-electronic repulsion between the lone pairs of the small $F$ atoms.
The correct order of bond dissociation enthalpy is: $I_2 < F_2 < Br_2 < Cl_2$.
Therefore,$Cl_2$ has the highest bond dissociation enthalpy.

(A) $Group \ 17$ elements are known as halogens. Among the given options,$At$ (Astatine) belongs to $group \ 17$.
$Zn$ (Zinc) is in $group \ 12$.
$As$ (Arsenic) is in $group \ 15$.
$Te$ (Tellurium) is in $group \ 16$.

(A) The general outer electronic configuration $ns^2 np^5$ corresponds to the halogen group (Group $17$).
Iodine $(I)$ belongs to the halogen group and has the outer electronic configuration $5s^2 5p^5$,which fits the general form $ns^2 np^5$ where $n=5$.
Therefore,the correct element is $I$.

(B) Among the given elements,$Se$ (Selenium),$I$ (Iodine),and $S$ (Sulfur) are solids at room temperature.
$Br$ (Bromine) is a non-metallic element that exists in a liquid state at room temperature.

(B) The chlorinating property of $PCl_{5}$ is demonstrated by its ability to provide chlorine atoms to other substances.
When $PCl_{5}$ reacts with metals like $Sn$,it acts as a chlorinating agent by transferring chlorine atoms to the metal,resulting in the formation of metal chlorides and $PCl_{3}$.
Specifically,the reaction $2 PCl_{5} + Sn \longrightarrow SnCl_{4} + 2 PCl_{3}$ shows $PCl_{5}$ acting as a chlorinating agent for $Sn$.

(C) Fluorine is the most electronegative element in the periodic table,which is a true statement.
It exhibits only $-1$ oxidation state because it lacks $d$-orbitals and is the most electronegative,making it unable to show positive oxidation states,which is a true statement.
Fluorine has the lowest bond dissociation enthalpy among all halogens due to the high inter-electronic repulsion between the lone pairs of the small $F$ atoms,making the statement 'It has high bond dissociation enthalpy among all halogens' false.
Fluorine forms only one oxoacid,which is $HOF$ (hypofluorous acid),which is a true statement.

(C) The boiling point of hydrides of halogens depends on two main factors: intermolecular hydrogen bonding and van der Waals forces.
$HF$ has the highest boiling point due to strong intermolecular hydrogen bonding.
For the remaining hydrides ($HCl$,$HBr$,$HI$),the boiling point increases with increasing molecular mass due to the increase in the magnitude of van der Waals forces.
Thus,the order is $HI > HBr > HCl$.
Combining these,the overall decreasing order of boiling point is $HF > HI > HBr > HCl$.

(D) Interhalogen compounds are represented by the general formula $XX^\prime_n$,where $X$ is the less electronegative halogen and $X^\prime$ is the more electronegative halogen,and $n = 1, 3, 5, 7$.
$1$. They are covalent in nature due to the small difference in electronegativity between the two halogens.
$2$. They are diamagnetic because all electrons are paired.
$3$. They are generally more reactive than individual halogens (except fluorine) because the $X-X^\prime$ bond is weaker than the $X-X$ or $X^\prime-X^\prime$ bonds.
$4$. The statement "If these are represented as $XX^\prime$,the number of $X^\prime$ atoms is always even" is false,as $n$ can be $1, 3, 5, 7$,which are odd numbers.

(B) Fluorine $(F)$ is the most electronegative element and has no $d$-orbitals,so it forms only one oxoacid,which is hypofluorous acid $(HOF)$.
Chlorine $(Cl)$,Bromine $(Br)$,and Iodine $(I)$ can exhibit higher oxidation states due to the presence of vacant $d$-orbitals.
Chlorine forms four types of oxoacids: hypochlorous acid $(HOCl)$,chlorous acid $(HOClO)$,chloric acid $(HOClO_2)$,and perchloric acid $(HOClO_3)$.
Among the halogens,Chlorine forms the maximum number of stable oxoacids.

(B) The oxoacids of chlorine are as follows:
$1$. Hypochlorous acid ($HOCl$ or $HClO$): Oxidation state of $Cl$ is $+1$.
$2$. Halous acid $(HClO_2)$: Oxidation state of $Cl$ is $+3$.
$3$. Halic acid $(HClO_3)$: Oxidation state of $Cl$ is $+5$.
$4$. Perhalic acid $(HClO_4)$: Oxidation state of $Cl$ is $+7$.
Therefore,the molecular formula of halous acid of chlorine is $HClO_2$.

(B) Interhalogen compounds of the type $XX'_3$ (e.g.,$ClF_3$,$BrF_3$) have a central halogen atom in the $+3$ oxidation state.
According to $VSEPR$ theory,the central atom has $3$ bond pairs and $2$ lone pairs of electrons,resulting in a total of $5$ electron pairs.
This corresponds to $sp^3d$ hybridization with a trigonal bipyramidal electron geometry.
Due to the presence of $2$ lone pairs in the equatorial positions to minimize repulsion,the molecule adopts a bent '$T$' shape.

(A) Among the given interhalogen compounds,$ICl$ (iodine monochloride) exists as a ruby-red liquid at room temperature.
$ClF_3$ is a gas.
$BrF_5$ is a liquid,but $ICl$ is the most commonly cited example for this property in standard textbooks.
$IF_7$ is a gas.

(A) Fluorine is the most electronegative element and has a small atomic size. Due to the absence of $d$-orbitals and its high electronegativity,it cannot exhibit positive oxidation states except in the case of $HOF$ (hypofluorous acid).
In $HOF$,the oxidation state of fluorine is $-1$,while oxygen is in the $-2$ state and hydrogen is in the $+1$ state. Other oxoacids like $HFO_2$,$HFO_3$,and $HFO_4$ do not exist because fluorine cannot expand its octet or accommodate positive oxidation states required for these structures.