Halogen family Questions in English

(B) The reaction of nitrogen trichloride $(NCl_3)$ with water is as follows:
$NCl_3 + 4 H_2O \rightarrow NH_4OH + 3 HOCl$
$NCl_3$ is an explosive oily liquid that undergoes hydrolysis to form ammonium hydroxide and hypochlorous acid.

(A) 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 $(+1, +3, +5, +7)$,the electron-withdrawing effect of the oxygen atoms increases,which weakens the $O-H$ bond and facilitates the release of $H^+$ ions.
The oxidation states are: $HClO (+1)$,$HClO_2 (+3)$,$HClO_3 (+5)$,$HClO_4 (+7)$.
Additionally,the stability of the conjugate base anion increases as the negative charge is delocalized over more oxygen atoms: $ClO^- < ClO_2^- < ClO_3^- < ClO_4^-$.
Therefore,the correct order of acidic strength is: $HClO < HClO_2 < HClO_3 < HClO_4$.

(A) Compound $(A)$ is $NaCl$,which is a white crystalline solid having a rock salt crystal structure.
$(B)$ is $Cl_2$,a pungent-smelling,greenish-yellow gas produced by the reaction of $NaCl$ with conc. $H_2SO_4$ and $MnO_2$.
The reaction is: $2NaCl + 2H_2SO_4 + MnO_2 \rightarrow Na_2SO_4 + MnSO_4 + 2H_2O + Cl_2$.
Compound $(A)$ reacts with $AgNO_3$ to form a white precipitate of $(C)$,which is $AgCl$.
The reaction is: $AgNO_3 + NaCl \rightarrow NaNO_3 + AgCl(s)$ (white ppt).
Therefore,$(A) = NaCl$,$(B) = Cl_2$,and $(C) = AgCl$.

(C) Concentrated nitric acid $(HNO_3)$ acts as a strong oxidizing agent.
When iodine $(I_2)$ is treated with concentrated $HNO_3$,it gets oxidized to iodic acid $(HIO_3)$.
The balanced chemical equation for the reaction is:
$I_2 + 10 HNO_3 \rightarrow 2 HIO_3 + 10 NO_2 + 4 H_2O$
Thus,the product formed is $HIO_3$.

(B) Concentrated $H_2SO_4$ is a strong oxidizing agent. When it reacts with $NaBr$,it produces $HBr$,which is then immediately oxidized by the concentrated $H_2SO_4$ to $Br_2$. The reaction is: $2NaBr + 3H_2SO_4 \rightarrow 2NaHSO_4 + SO_2 + Br_2 + 2H_2O$.

(A) The stability of $Pb(IV)$ compounds decreases as the size of the halide ion increases due to the increasing reducing power of the halide ion.
$Pb^{4+}$ is a strong oxidizing agent and it oxidizes $I^-$ to $I_2$,$Br^-$ to $Br_2$,and $Cl^-$ to $Cl_2$,while $F^-$ is the least easily oxidized.
Therefore,$PbF_4$ is the most stable compound among the given options.

(D) The reactions are as follows:
$(I)$ $P_4 + 10SO_2Cl_2 \to 4PCl_5 + 10SO_2$
$(II)$ $P_4 + 8SOCl_2 \to 4PCl_3 + 4SO_2 + 2S_2Cl_2$
$(III)$ $PCl_5 + SO_2 \to POCl_3 + SO_2Cl_2$
$(IV)$ $PCl_5 + C_2H_5OH \to POCl_3 + C_2H_5Cl + HCl$
Thus,$POCl_3$ is prepared in reactions $(III)$ and $(IV)$.

(D) Hydrolysis of $SF_4$: $SF_4 + 3H_2O \rightarrow H_2SO_3 + 4HF$. Here,$H_2SO_3$ is an oxyacid and $HF$ is a hydra acid.
Hydrolysis of $PCl_3$: $PCl_3 + 3H_2O \rightarrow H_3PO_3 + 3HCl$. Here,$H_3PO_3$ is an oxyacid and $HCl$ is a hydra acid.
Hydrolysis of $P_4O_{10}$: $P_4O_{10} + 6H_2O \rightarrow 4H_3PO_4$. Here,only an oxyacid is formed.
Therefore,both $SF_4$ and $PCl_3$ produce both types of acids.

(D) Aqua regia is a mixture of concentrated $HCl$ and concentrated $HNO_3$ in a $3:1$ ratio.
Gold dissolves in aqua regia due to the oxidizing action of $HNO_3$ and the complexing ability of $Cl^-$ ions.
The reaction proceeds as follows:
$Au(s) + 3NO_3^-(aq) + 6H^+(aq) \rightarrow Au^{3+}(aq) + 3NO_2(g) + 3H_2O(l)$
$Au^{3+}(aq) + 4Cl^-(aq) \rightarrow [AuCl_4]^-(aq)$
The overall reaction is:
$Au(s) + 3HNO_3(aq) + 4HCl(aq) \rightarrow H[AuCl_4](aq) + 3NO_2(g) + 3H_2O(l)$
Thus,the final product formed is chloroauric acid,$HAuCl_4$.

(B) The bond energy of $F-F$ is exceptionally low due to the high inter-electronic repulsion between the lone pairs of the small $F$ atoms. The bond energy order is $Br-F > F-F$. Therefore,the correct statement is $Br-F > F-F$.

(B) The structure of polythionic acids $H_2S_nO_6$ contains a chain of $n$ sulfur atoms,which involves $(n-1)$ $S-S$ bonds. This statement is $True$.
$(b)$ $F_2$ is a strong oxidizing agent. It reacts with $H_2O$ to produce $HF$ and $O_2$,and also forms $O_3$ in small amounts. This statement is $True$.
$(c)$ Hydrolysis of $XeF_6$ is not a disproportionation reaction; it is a series of partial hydrolysis steps leading to $XeOF_4$,$XeO_2F_2$,and finally $XeO_3$. This statement is $False$.
$(d)$ $Al$ reacts with $NaOH$ to form a soluble complex,sodium tetrahydroxoaluminate,$Na[Al(OH)_4]$,not a precipitate of $Al(OH)_3$. This statement is $False$.
Therefore,the correct sequence is $True, True, False, False$.

(D) The reactivity of halogens as oxidizing agents decreases down the group: $F_2 > Cl_2 > Br_2 > I_2$.
Reaction $(i)$: $F_2$ is a stronger oxidizing agent than $Cl_2, Br_2,$ and $I_2$,so it can displace them from their halide ions. This reaction is correct.
Reaction $(ii)$: $Cl_2$ is a stronger oxidizing agent than $Br_2$ and $I_2$,so it can displace them from their halide ions. This reaction is correct.
Reaction $(iii)$: $I_2$ is a weaker oxidizing agent than $Br_2$,so it cannot displace $Br_2$ from $Br^-$. This reaction is incorrect.
Therefore,only reactions $(i)$ and $(ii)$ are correct.

(B) The greenish yellow gas is $Cl_2$.
When $Cl_2$ reacts with hot and concentrated potassium hydroxide $(KOH)$,it forms potassium chlorate $(KClO_3)$,which is a halate.
$KClO_3$ is widely used in the manufacturing of fireworks and safety matches due to its oxidizing properties.
The chemical reaction is: $3Cl_2 + 6KOH \to KClO_3 + 5KCl + 3H_2O$.

(C) The feasibility of displacement reactions among halogens depends on their standard reduction potentials. $A$ halogen with a higher reduction potential can displace a halogen with a lower reduction potential from its salt solution.
The order of oxidizing power is $F_2 > Cl_2 > Br_2 > I_2$.
Therefore,$F_2$ can displace $Cl^-$,$Br^-$,and $I^-$; $Cl_2$ can displace $Br^-$ and $I^-$; and $Br_2$ can displace $I^-$.
In option $C$,$Br_2$ is trying to displace $F^-$. Since $Br_2$ is a weaker oxidizing agent than $F_2$,this reaction is not feasible.

(B) Concentrated $H_2SO_4$ is a strong oxidizing agent.
When $NaBr$ reacts with concentrated $H_2SO_4$,it first produces $HBr$ $(NaBr + H_2SO_4 \to NaHSO_4 + HBr)$.
However,the produced $HBr$ is then oxidized by the concentrated $H_2SO_4$ to $Br_2$ $(2HBr + H_2SO_4 \to Br_2 + SO_2 + 2H_2O)$.
Therefore,it cannot be used to prepare pure $HBr$.

(A) When chlorine gas $(Cl_2)$ reacts with cold and dilute sodium hydroxide $(NaOH)$,it undergoes a disproportionation reaction.
The balanced chemical equation is:
$Cl_2 + 2NaOH (cold, dil.) \to NaCl + NaOCl + H_2O$
Thus,the products formed are sodium chloride $(NaCl)$ and sodium hypochlorite $(NaOCl)$.

(B) Aqua-regia is a mixture of concentrated $HCl$ and $HNO_3$ in a $3:1$ molar ratio. \\ The reaction of platinum $(Pt)$ with aqua-regia produces chloroplatinic acid $(H_2PtCl_6)$ and nitrogen dioxide $(NO_2)$. \\ The balanced chemical equation is: $Pt + 4HNO_3 + 6HCl \rightarrow H_2PtCl_6 + 4NO_2 + 4H_2O$.

(A) When $NaCl$ reacts with concentrated $H_2SO_4$,it produces $HCl$ gas.
$HCl$ is a diatomic molecule with $18$ electrons (even number),making it diamagnetic.
$NaBr$ and $NaI$ react with concentrated $H_2SO_4$ to produce $Br_2$ and $I_2$ respectively,which are diatomic but are not the primary gaseous products expected in this context compared to the hydrogen halide gas.
Thus,$NaCl$ is the correct species that yields a diatomic gas $(HCl)$ with diamagnetic behaviour.

(D) The stability of oxoanions increases with the number of oxygen atoms due to resonance stabilization. Thus,$ClO_4^- > ClO_3^- > ClO_2^- > ClO^-$ is correct.
$(b)$ The bond energy order for halogens is $Cl_2 > Br_2 > F_2 > I_2$. $F_2$ has an anomalously low bond energy due to inter-electronic repulsions between lone pairs. Thus,this is correct.
$(c)$ The bond angle order is $OF_2 (103^\circ) < OH_2 (104.5^\circ) < OCl_2 (111^\circ)$. This is correct.
$(d)$ Reducing power increases as the size of the halide ion increases,making it easier to lose electrons. Thus,$I^- > Br^- > Cl^-$ is correct.
Since all statements are correct,the answer is 'All of these'.

(D) mixed anhydride is an oxide that reacts with water to produce two different acids.
$ClO_2$ (which exists as $Cl_2O_4$) reacts with water to form $HClO_2$ and $HClO_3$:
$2ClO_2 + H_2O \rightarrow HClO_2 + HClO_3$
$ClO_3$ (which exists as $Cl_2O_6$) reacts with water to form $HClO_3$ and $HClO_4$:
$2ClO_3 + H_2O \rightarrow HClO_3 + HClO_4$
Therefore,both $ClO_2$ and $ClO_3$ are mixed anhydrides.

(D) The reaction of chlorine with sodium hydroxide depends on the concentration and temperature of the $NaOH$ solution.
$1$. With cold and dilute $NaOH$:
$Cl_2 + 2NaOH (cold \& dil.) \rightarrow NaCl + NaOCl + H_2O$
Here,$A$ is $NaOCl$ (sodium hypochlorite).
$2$. With hot and concentrated $NaOH$:
$3Cl_2 + 6NaOH (hot \& conc.) \rightarrow 5NaCl + NaClO_3 + 3H_2O$
Here,$B$ is $NaClO_3$ (sodium chlorate).
Therefore,$A$ is $NaOCl$ and $B$ is $NaClO_3$.

(C) $1$. $2KClO_3 \xrightarrow{\Delta} 2KCl + 3O_2$ (Produces $O_2$)
$2$. $2XeF_2 + 2H_2O \longrightarrow 2Xe + 4HF + O_2$ (Produces $O_2$)
$3$. $6XeF_4 + 12H_2O \longrightarrow 4Xe + 2XeO_3 + 24HF + 3O_2$ (Produces $O_2$)
$4$. $2PbO_2 \xrightarrow{\Delta} 2PbO + O_2$ (Produces $O_2$)
$5$. $XeF_6 + 3H_2O \longrightarrow XeO_3 + 6HF$ (Does not produce $O_2$)
Total processes yielding $O_2$ are $1, 2, 3,$ and $4$. Hence,the total count is $4$.

(D) $O_2F_2$ acts as a fluorinating agent,for example,it fluorinates plutonium to $PuF_6$ at low temperatures.
$OF_2$ also acts as a fluorinating agent.
$ClF_3$ is a very strong fluorinating agent used in the nuclear industry for the production of $UF_6$.
Therefore,all the given substances act as fluorinating agents.

(C) The reaction $I_2 + H_2O \to HI + HOI$ is incorrect because the equilibrium lies far to the left.
The hydrolysis of iodine is not spontaneous as the value of $\Delta G$ is positive $(+ve)$.

(C) The acidic strength of oxoacids is directly proportional to the electronegativity of the central atom when the oxidation state is the same.
For the series $HClO_4$,$HBrO_4$,and $HIO_4$,the central atoms are $Cl$,$Br$,and $I$ respectively.
Since electronegativity decreases down the group $(Cl > Br > I)$,the acidic strength follows the order $HClO_4 > HBrO_4 > HIO_4$.

(D) $SO_2$ acts as a bleaching agent by the process of reduction,whereas $Cl_2$ acts as a bleaching agent by the process of oxidation.
$SO_2$ reacts with water to produce nascent hydrogen,which reduces the coloured matter to a colourless substance:
$SO_2 + 2H_2O \rightarrow H_2SO_4 + 2[H]$
$Cl_2$ reacts with water to produce nascent oxygen,which oxidizes the coloured matter to a colourless substance:
$Cl_2 + H_2O \rightarrow 2HCl + [O]$
Therefore,the correct option is $D$.

(D) $1$. Hydrolysis of $F_2$: $2F_2 + 2H_2O \rightarrow 4HF + O_2$.
$2$. Hydrolysis of $XeF_4$: $6XeF_4 + 12H_2O \rightarrow 4Xe + 2XeO_3 + 24HF + 3O_2$.
$3$. Hydrolysis of $XeF_2$: $2XeF_2 + 2H_2O \rightarrow 2Xe + 4HF + O_2$.
Since $O_2$ is produced in all these reactions,the correct option is $D$.

(C) In photography,$Na_2S_2O_3 \cdot 5H_2O$ (sodium thiosulfate),commonly known as hypo,is used as a fixing agent.
It reacts with the undecomposed silver bromide $(AgBr)$ on the photographic film to form a water-soluble complex,sodium dithiosulfatoargentate$(I)$,which can be washed away.
The chemical reaction is: $AgBr(s) + 2Na_2S_2O_3(aq) \rightarrow Na_3[Ag(S_2O_3)_2](aq) + NaBr(aq)$.

(D) mixed anhydride is an oxide that reacts with water to produce two different acids.
$N_2O_4$ reacts with water to form a mixture of nitrous acid $(HNO_2)$ and nitric acid $(HNO_3)$: $N_2O_4 + H_2O \to HNO_2 + HNO_3$.
$Cl_2O_6$ reacts with water to form a mixture of chloric acid $(HClO_3)$ and perchloric acid $(HClO_4)$: $Cl_2O_6 + H_2O \to HClO_3 + HClO_4$.
Therefore,both $N_2O_4$ and $Cl_2O_6$ are mixed anhydrides.

(A) The acidic strength of oxoacids of the same halogen increases with an increase in the oxidation state of the central atom.
For the given oxoacids,the oxidation states of chlorine are:
$HOCl$: $+1$
$HClO_2$: $+3$
$HClO_3$: $+5$
$HClO_4$: $+7$
Since the acidic strength is directly proportional to the oxidation state of the central atom,the order is $HClO_4 > HClO_3 > HClO_2 > HOCl$.

(D) $1$. Electrolysis of aqueous $NaCl$ (brine) produces $Cl_2$ gas at the anode: $2Cl^- \rightarrow Cl_2 + 2e^-$.
$2$. Oxidation of conc. $HCl$ by $KMnO_4$ produces $Cl_2$ gas: $2KMnO_4 + 16HCl \rightarrow 2KCl + 2MnCl_2 + 8H_2O + 5Cl_2$.
$3$. Oxidation of conc. $HCl$ by $MnO_2$ produces $Cl_2$ gas: $MnO_2 + 4HCl \rightarrow MnCl_2 + 2H_2O + Cl_2$.
$4$. The reaction between $NaCl$ and conc. $H_2SO_4$ produces $HCl$ gas,not $Cl_2$ gas: $NaCl + H_2SO_4 \rightarrow NaHSO_4 + HCl$.

(B) The boiling point of $HF$ is the highest among hydrogen halides due to strong intermolecular $H$-bonding.
For the remaining hydrogen halides ($HCl$,$HBr$,$HI$),the boiling point increases with increasing molecular mass due to stronger van der Waals forces in the order $HCl < HBr < HI$.
Since volatility is inversely proportional to the boiling point,the compound with the lowest boiling point is the most volatile.
Therefore,$HCl$ is the most volatile hydrogen halide.

(C) Fluorine is the most electronegative element and has the least tendency to form $p\pi-d\pi$ double bonds. Due to its small size and high electronegativity,it forms only one oxyacid,which is $HOF$ (hypofluorous acid). In contrast,other halogens like $Cl$,$Br$,and $I$ form multiple oxyacids due to the availability of vacant $d$-orbitals.

(C) Chlorine water is yellow in colour due to the presence of dissolved $Cl_2$.
On standing,$Cl_2$ reacts with water to form hydrochloric acid and hypochlorous acid.
The reaction is: $Cl_2 + H_2O \to HCl + HOCl$.
Since $HOCl$ is unstable,it further decomposes to give $HCl$ and nascent oxygen,which causes the bleaching action and loss of colour.

(C) The reaction of $CaF_2$ and $SiO_2$ with concentrated $H_2SO_4$ produces silicon tetrafluoride gas $(SiF_4)$:
$2CaF_2 + SiO_2 + 2H_2SO_4 \longrightarrow SiF_4 + 2CaSO_4 + 2H_2O$
When $SiF_4$ is hydrolyzed,it forms hydrofluosilicic acid and a white gelatinous precipitate of silicic acid $(H_2SiO_3)$:
$3SiF_4 + 3H_2O \longrightarrow 2H_2SiF_6 + H_2SiO_3$ (white gelatinous precipitate of silicic acid).

(B) The molecular geometry of $BrF_5$ is square pyramidal,not trigonal bipyramidal.
$Br$ in $BrF_5$ has $7$ valence electrons,forms $5$ bonds with $F$ atoms,and has $1$ lone pair,resulting in $sp^3d^2$ hybridization and square pyramidal geometry.

(A) The reaction of chlorine with hot and concentrated sodium hydroxide is a disproportionation reaction.
$3Cl_2 + 6NaOH \to 5NaCl + NaClO_3 + 3H_2O$
In this reaction,chlorine is both oxidized and reduced.
The products formed are sodium chloride $(NaCl)$ and sodium chlorate $(NaClO_3)$.
Thus,the ionic species present are $Cl^{-}$ and $ClO_3^{-}$.

(D) $SO_2$ bleaches the colour of flowers by reduction,whereas $Cl_2$ bleaches by oxidation.
$Cl_2$ reacts with water to produce nascent oxygen,which acts as an oxidizing agent to bleach the colour:
$Cl_2 + H_2O \rightarrow 2HCl + [O]$
$SO_2$ reacts with water to produce nascent hydrogen,which acts as a reducing agent to remove oxygen from the coloured substance,making it colourless:
$SO_2 + 2H_2O \rightarrow H_2SO_4 + 2[H]$

(A) When $KBr$ reacts with concentrated $H_2SO_4$,$HBr$ is initially formed: $KBr + H_2SO_4 \to KHSO_4 + HBr$.
However,concentrated $H_2SO_4$ is a strong oxidizing agent.
It further oxidizes the produced $HBr$ to $Br_2$ gas: $2HBr + H_2SO_4 \to Br_2 + SO_2 + 2H_2O$.
Therefore,it cannot be used to prepare pure $HBr$.

(C) The structure of interhalogen compounds depends on the hybridization of the central atom and the number of lone pairs.
$BrF_5$ and $IF_5$ are $AX_5E$ type molecules,which exhibit $sp^3d^2$ hybridization and have a square pyramidal geometry.
$IF_3$ is an $AX_3E_2$ type molecule,which exhibits $sp^3d$ hybridization and has a $T$-shaped geometry.
Therefore,$IF_3$ is not square pyramidal.

(D) Artificial smoke screens are produced using $TiCl_4$ (Titanium tetrachloride).
When $TiCl_4$ is released into the atmosphere,it reacts with moisture in the air to form $TiO_2$ (Titanium oxide) and $HCl$ (Hydrogen chloride) gas.
The reaction is: $TiCl_4 + 2H_2O \rightarrow TiO_2 + 4HCl$.
The resulting $TiO_2$ particles form a dense white smoke that acts as a screen.

(C) $1$. Bond dissociation energy: Due to the small size of the $F$ atom,the lone pairs on the $F$ atoms in $F_2$ experience strong inter-electronic repulsion,making the $F-F$ bond weaker than the $Cl-Cl$ bond. Thus,the order is $Cl-Cl > F-F$. Statement $A$ is incorrect.
$2$. Thermal stability: Thermal stability decreases down the group as the bond length increases and bond dissociation energy decreases. The correct order is $HF > HCl > HBr > HI$. Statement $B$ is incorrect.
$3$. Reducing power: Reducing power increases as the bond dissociation energy decreases down the group. The correct order is $HI > HBr > HCl > HF$. Statement $C$ is correct.

(B) The oxyacids of chlorine are $HOCl$,$HOClO$ $(HClO_2)$,$HOClO_2$ $(HClO_3)$,and $HOClO_3$ $(HClO_4)$.
In all these oxyacids,the central chlorine atom is $sp^3$ hybridised.
Also,all these oxyacids are monobasic because they contain only one $OH$ group attached to the chlorine atom.
Therefore,they show similarity in their hybridisation state and basicity.

(D) Hydrolysis of halides typically occurs via the coordination of the lone pair of electrons from a water molecule to the central atom's vacant $d$-orbitals.
In $NF_3$,the nitrogen atom does not have vacant $d$-orbitals,and the fluorine atom is highly electronegative,which prevents the attack of water molecules.
Therefore,$NF_3$ is inert towards hydrolysis at room temperature.

(D) In $SF_6$,the $S$ atom is sterically protected by six fluorine atoms,preventing the attack of $H_2O$ molecules.
$NF_3$ cannot be hydrolysed because $N$ lacks vacant $d$-orbitals and $F$ is highly electronegative,making the $N-F$ bond very stable.
$TeF_6$ is susceptible to hydrolysis due to the large size of the $Te$ atom,which allows for the coordination of water molecules.
$NCl_3$ is readily hydrolysed to form $NH_3$ and $HOCl$.
Therefore,$SF_6$ and $NF_3$ cannot be hydrolysed at room temperature.

(D) The prefix $hypo-$ is typically used for acids with a lower oxidation state than the $-ous$ acid.
For $HClO_3$ (chloric acid),the corresponding $-ous$ acid is $HClO_2$ (chlorous acid) and the $hypo-$ acid is $HClO$ (hypochlorous acid).
However,the question asks for the parent acid that does not have a corresponding $hypo-$ acid in the standard nomenclature series.
$HClO_3$ is the parent acid for chlorous and hypochlorous acid,but strictly speaking,$HClO_3$ itself is not a $hypo-$ acid.
Among the given options,$HClO_3$ is the correct choice as it represents the chloric acid series.

(C) Concentrated nitric acid acts as a strong oxidizing agent and oxidizes iodine $(I_2)$ to iodic acid $(HIO_3)$.
The balanced chemical equation for the reaction is:
$I_2 + 10HNO_3 \to 2HIO_3 + 10NO_2 + 4H_2O$

(A) The thermal stability of oxyacids of chlorine increases with the increase in the oxidation state of the central chlorine atom.
The oxidation states of $Cl$ in the given compounds are:
$HOClO_3$ $(+7)$,$HOClO_2$ $(+5)$,$HOClO$ $(+3)$,and $HOCl$ $(+1)$.
Therefore,the decreasing order of thermal stability is: $HOClO_3 > HOClO_2 > HOClO > HOCl$.
Thus,$HOClO_3$ is the most stable compound.

(B) The estimation of carbon monoxide $(CO)$ is carried out by using iodine pentoxide $(I_2O_5)$.
The chemical reaction is: $5CO + I_2O_5 \to 5CO_2 + I_2$
The liberated iodine $(I_2)$ is then titrated against a standard sodium thiosulphate $(Na_2S_2O_3)$ solution:
$I_2 + 2Na_2S_2O_3 \to 2NaI + Na_2S_4O_6$
Thus,$I_2O_5$ is the correct reagent.