Halogen family Questions in English

(A) $SbF_5$ is the strongest fluoride ion acceptor because it readily accepts a fluoride ion to form the $[SbF_6]^-$ complex,which is a very stable octahedral species. This reaction is driven by the high Lewis acidity of $SbF_5$.

(A) The ease of hydrolysis depends on the availability of empty $d$-orbitals and the polarity of the $M-Cl$ bond.
$CCl_4$ does not undergo hydrolysis because carbon lacks $d$-orbitals.
$SiCl_4$ undergoes hydrolysis due to the presence of empty $d$-orbitals.
$PCl_5$ is highly reactive towards water due to the high positive charge on $P$ and empty $d$-orbitals.
$AlCl_3$ is ionic in nature and undergoes rapid hydrolysis.
Thus,the increasing order of ease of hydrolysis is $CCl_4 < SiCl_4 < PCl_5 < AlCl_3$.

(B) Interhalogen compounds are formed by the combination of two different halogens. The compound $IF$ is highly unstable and undergoes disproportionation reaction as follows: $5IF \rightarrow 2I_2 + IF_5$. Therefore,$IF$ does not exist as a stable compound.

(D) The compound $I_4O_9$ is a mixed oxide of iodine where it exists in two oxidation states,$+3$ and $+5$. It is represented as $I^{3+}(IO_3^-)_3$. Upon heating or decomposition,it yields $I_2O_5$ and $I_2$. Therefore,all the given statements are correct.

(D) The chemical formula $HIO_4 \cdot 2H_2O$ or $H_5IO_6$ represents orthoperiodic acid,which is commonly referred to as para periodic acid in this context.

(D) Pseudo-halide ions are polyatomic anions that resemble halide ions in their chemical properties. Common examples include $CN^-$,$SCN^-$,$SeCN^-$,$TeCN^-$,and $N_3^-$. The ion $S_4N_2^{2-}$ does not exhibit the characteristics of a pseudo-halide ion.

(A) In Deacon's process,chlorine is produced by the oxidation of hydrogen chloride gas by atmospheric oxygen in the presence of $CuCl_2$ as a catalyst at $450 \ ^oC$.
The chemical equation is: $4HCl + O_2 \xrightarrow{CuCl_2, \ 450 \ ^oC} 2Cl_2 + 2H_2O$.

(B) Glass is primarily composed of sodium silicate $(Na_2SiO_3)$. Hydrofluoric acid $(HF)$ reacts with the silica/silicates in glass to form sodium hexafluorosilicate $(Na_2SiF_6)$ and water.
The chemical reaction is:
$Na_2SiO_3 + 6HF \to Na_2SiF_6 + 3H_2O$

(A) $ClO_3$ is a radical species with an odd number of electrons,making it paramagnetic. $Cl_2O_6$ exists as an ionic solid $[ClO_2]^+[ClO_4]^-$ in the solid state and as a dimer in the gas phase,where all electrons are paired,making it diamagnetic.

(B) The reducing property of halide ions depends on their ability to lose electrons,which is related to their standard oxidation potential. As we move down the group in the periodic table,the size of the halide ion increases,making it easier to lose an electron. Therefore,the reducing power increases in the order: $F^- < Cl^- < Br^- < I^-$. Thus,the correct increasing order is $F^- < Cl^- < Br^- < I^-$.

(B) The thermal decomposition of orthoperiodic acid $(H_5IO_6)$ occurs in steps:
$1$. At $100 \ ^\circ C$,it loses water to form metaperiodic acid: $H_5IO_6 \xrightarrow{100 \ ^\circ C} HIO_4 + 2H_2O$.
$2$. Upon strong heating at $200 \ ^\circ C$,metaperiodic acid decomposes to form iodine pentoxide: $2HIO_4 \xrightarrow{200 \ ^\circ C} I_2O_5 + H_2O + \frac{3}{2}O_2$.
Thus,the final product obtained upon strong heating is $I_2O_5$.

(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 form oxoacids like other halogens. It only forms $HOF$ (hypofluorous acid),which is often not classified as a typical oxoacid in the same series as $HClO$,$HBrO$,and $HIO$.

(A) Fluorine is the most electronegative element and acts as a strong oxidizing agent. It reacts with water to liberate $O_2$ gas.
The chemical reaction is:
$2F_2(g) + 2H_2O(l) \to 4HF(aq) + O_2(g)$

(D) The strongest oxidizing halogen is $Fluorine$ $(F_2)$ due to its high electronegativity and high hydration energy. Therefore,the pair $D$ is incorrectly matched.

(A) Chlorine is a stronger oxidizing agent than $Br_2$ and $I_2$,so it can displace them from their salt solutions.
However,fluorine is more electronegative and a stronger oxidizing agent than chlorine.
Therefore,chlorine cannot displace fluorine from $NaF$.

(B) Due to the absence of $d$-orbitals,fluorine cannot exhibit higher oxidation states.

(D) The halogen group $(Group \ 17)$ consists of Fluorine,Chlorine,Bromine,Iodine,and Astatine. Therefore,Astatine is the correct answer.

(C) The oxidizing power of a halogen in aqueous solution depends on the overall energy change involved in the process: $\frac{1}{2}X_2(g) + e^- \rightarrow X^-(aq)$. This process involves three steps: $(1)$ Enthalpy of dissociation $(\frac{1}{2} \Delta H_{diss})$, $(2)$ Electron affinity $(\Delta H_{eg})$, and $(3)$ Enthalpy of hydration $(\Delta H_{hyd})$. Ionization potential is related to the removal of an electron, which is not involved in the reduction process of halogens to halide ions. Therefore, it is not a factor determining the oxidizing strength in aqueous solution.

(A) Fluorine is the most electronegative element in the periodic table. Due to its extremely high electronegativity and the absence of $d$-orbitals in its valence shell,it cannot exhibit positive oxidation states. It only shows an oxidation state of $-1$.

(D) As we move down the group,the metallic character increases. $I_2$ (Iodine) exhibits some metallic character due to its larger size and lower ionization energy compared to other halogens.

(D) The bond strength is directly proportional to the bond dissociation energy.
Lower bond dissociation energy indicates a weaker bond.
Comparing the given values: $F_2 (155 \ kJ \ mol^{-1})$,$Cl_2 (244 \ kJ \ mol^{-1})$,$Br_2 (193 \ kJ \ mol^{-1})$,and $I_2 (151 \ kJ \ mol^{-1})$.
Since $I_2$ has the lowest bond dissociation energy of $151 \ kJ \ mol^{-1}$,it has the weakest bond.

(C) The strength of oxyacids of halogens increases with the increase in the oxidation state of the central halogen atom.
In $HOCl$,the oxidation state of $Cl$ is $+1$.
In $HOClO$,the oxidation state of $Cl$ is $+3$.
In $HOClO_2$,the oxidation state of $Cl$ is $+5$.
In $HOClO_3$,the oxidation state of $Cl$ is $+7$.
Since $HOClO_3$ has the highest oxidation state $(+7)$,it is the strongest acid.

(C) Caliche,also known as Chile saltpeter,is $NaNO_3$.
It contains approximately $0.3\%$ of iodine in the form of sodium iodate $(NaIO_3)$.

(D) The reaction of fluorine with water is unique because it produces ozonised oxygen $(O_3)$ along with oxygen $(O_2)$.
$2F_2(g) + 2H_2O(l) \to 4HF(aq) + O_2(g)$
$3F_2(g) + 3H_2O(l) \to 6HF(aq) + O_3(g)$
Thus,$F_2$ is the correct answer.

(B) Iodide ions $(I^-)$ can be oxidized to iodine $(I_2)$ by strong oxidizing agents like nitric acid $(HNO_3)$.
The reaction is: $6I^- + 8HNO_3 \to 3I_2 + 8NO + 4H_2O + 4H^+$.
Furthermore,iodine $(I_2)$ can be further oxidized to iodic acid $(HIO_3)$ by concentrated $HNO_3$:
$I_2 + 10HNO_3 \to 2HIO_3 + 10NO_2 + 4H_2O$.

(B) The greenish-yellow gas is chlorine $(Cl_2)$.
When $Cl_2$ reacts with hot and concentrated potassium hydroxide $(KOH)$,it forms potassium chlorate $(KClO_3)$:
$3Cl_2 + 6KOH \xrightarrow{\Delta} KClO_3 + 5KCl + 3H_2O$
$KClO_3$ is a strong oxidizing agent used in the manufacture of safety matches and fireworks.

(A) Iodine is used as an antiseptic in the form of tincture of iodine,which is a $2-3 \%$ solution of $I_2$ in alcohol-water mixture.

(A) The reaction between $KMnO_4$ and $HCl$ is a well-known laboratory method for the preparation of chlorine gas. The reaction is as follows:
$2KMnO_4 + 16HCl \rightarrow 2KCl + 2MnCl_2 + 8H_2O + 5Cl_2$
This reaction proceeds rapidly at room temperature or with mild heating to produce $Cl_2$ gas.

(C) $HCl$ has the lowest boiling point among the given hydrogen halides due to the absence of strong intermolecular hydrogen bonding compared to $HF$ and weaker van der Waals forces compared to $HBr$ and $HI$. Therefore,it is the most volatile substance.

(B) The thermal stability of hydrogen halides $(H - X)$ decreases down the group as the bond dissociation enthalpy decreases.
As the size of the halogen atom increases from $F$ to $I$,the bond length increases and the bond strength decreases.
Therefore,the correct order of thermal stability is $HF > HCl > HBr > HI$.

(A) Among the given options,$HCl$ is the most stable towards heat due to its high bond dissociation energy.
$HOCl$ is an oxyacid and is highly unstable towards heat,decomposing as follows:
$2HOCl \xrightarrow{\Delta} 2HCl + O_2$

(D) The reducing character of hydrogen halides increases down the group $(HF < HCl < HBr < HI)$.
$HF$ is a very weak reducing agent and cannot reduce $H_2SO_4$,$KMnO_4$,or $K_2Cr_2O_7$.

(B) Chlorine acts as a bleaching agent only in the presence of moisture.
$H_2O + Cl_2 \to 2HCl + [O]$
The nascent oxygen $[O]$ produced is responsible for the bleaching action.

(B) $N_2$ contains a triple bond $(N \equiv N)$,which results in the highest bond dissociation enthalpy.
$O_2$ has a double bond $(O = O)$,giving it a higher bond dissociation enthalpy than the halogens.
Due to the small size of the $F$ atom,there is significant inter-electronic repulsion between the lone pairs in $F_2$,making its bond weaker than that of $Cl_2$.
Therefore,the correct order of bond dissociation enthalpy is $F_2 < Cl_2 < O_2 < N_2$.

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

(D) When $Cl_2$ is in excess,it reacts with $NH_3$ to form $NCl_3$ (nitrogen trichloride) and $HCl$. The reaction where $N_2$ and $NH_4Cl$ are formed occurs when $NH_3$ is in excess.
The reaction for excess $Cl_2$ is: $NH_3 + 3Cl_2 \to NCl_3 + 3HCl$.
The reaction for excess $NH_3$ is: $8NH_3 + 3Cl_2 \to 6NH_4Cl + N_2$.
Therefore,the statement in option $D$ is incorrect because $Cl_2$ in excess produces $NCl_3$ and $HCl$.

(C) The thermal stability of hydrogen halides $(HX)$ 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 order of bond strength is $HF > HCl > HBr > HI$.
Consequently,the order of thermal stability is $HF > HCl > HBr > HI$.

(B) The acidic strength of oxoacids increases as the oxidation state of the central atom increases.
In the given series,the oxidation states of $Cl$ are:
$HClO: +1$
$HClO_2: +3$
$HClO_3: +5$
$HClO_4: +7$
Therefore,the correct order of acidic strength is $HClO < HClO_2 < HClO_3 < HClO_4$.

(C) The bond dissociation energy of $F_2$ is lower than that of $Cl_2$ due to the small size of the fluorine atom,which leads to significant inter-electronic repulsion between the lone pairs of the two atoms.
Therefore,the correct order of bond dissociation energy is $Cl_2 > Br_2 > F_2 > I_2$.
Thus,the order given in option $(C)$ is incorrect.

(C) The oxidizing power of halogens follows the order $F_2 > Cl_2 > Br_2 > I_2$. Thus,option $(A)$ is correct.
The electron gain enthalpy of halogens follows the order $Cl_2 > F_2 > Br_2 > I_2$. The value for $F_2$ is less negative than $Cl_2$ due to its small size and inter-electronic repulsions. Thus,option $(B)$ is incorrect.
The bond dissociation energy of halogens follows the order $Cl_2 > Br_2 > F_2 > I_2$. The $F-F$ bond is weaker due to high inter-electronic repulsion between lone pairs. Thus,option $(C)$ is incorrect.
The electronegativity of halogens follows the order $F_2 > Cl_2 > Br_2 > I_2$. Thus,option $(D)$ is correct.
Since the question asks for the arrangement that does not represent the correct trend,both $(B)$ and $(C)$ are incorrect. However,in standard multiple-choice questions of this type,if only one must be chosen,$(C)$ is often cited as the most significant anomaly.

(B) The oxidizing power of halogens decreases down the group: $F_2 > Cl_2 > Br_2 > I_2$.
$A$ halogen with higher oxidizing power can displace a halogen with lower oxidizing power from its salt solution.
$Br_2$ is a stronger oxidizing agent than $I_2$,so it can displace $I^-$ ions from $NaI$ to form $I_2$.
$2NaI + Br_2 \rightarrow 2NaBr + I_2$
However,$Br_2$ is a weaker oxidizing agent than $F_2$ and $Cl_2$,so it cannot displace $F^-$ from $NaF$ or $Cl^-$ from $NaCl$.
Therefore,only $I_2$ is liberated.

(B) The correct statement is $B$.
$1$. Interhalogen compounds are generally more reactive than halogens because the bond between two different halogens is weaker than the bond between two identical halogens (except $F_2$).
$2$. The oxidising power of oxoacids of chlorine increases with the number of oxygen atoms attached to the central chlorine atom,thus $HClO_4 > HClO_3 > HClO_2 > HClO$ is correct.
$3$. Chlorine shows bleaching action by oxidation,not reduction. It reacts with water to form $HCl$ and $HClO$. $HClO$ releases nascent oxygen $[O]$,which oxidizes coloured substances to colourless ones.
$4$. The stability order of halogen oxides is $I > Cl > Br$ due to the difference in electronegativity and bond polarity,but this is a complex trend; however,option $B$ is definitively correct.

(A) The complete hydrolysis of $IF_7$ is given by the reaction:
$IF_7 + 4H_2O \longrightarrow HIO_4 + 7HF$
In the product $HIO_4$ (periodic acid),the oxidation state of iodine is $+7$.
Since the oxidation state is $+7$,it is named as periodic acid,which ends with the $-ic$ suffix.
Therefore,the product is an oxyacid with an $-ic$ suffix.

(A) The balanced chemical equation for the complete hydrolysis of $ClF_5$ is:
$ClF_5 + 3H_2O \longrightarrow HClO_3 + 5HF$
In the product $HClO_3$ (chloric acid),the oxidation state of chlorine is $+5$.
Since the suffix is $-ic$ (chloric acid),the correct description is that the product is an oxyacid with an $-ic$ suffix.

(A) The complete hydrolysis of $BrF_5$ at room temperature $(R.T.)$ is given by the reaction:
$BrF_5 + 3H_2O \longrightarrow HBrO_3 + 5HF$
In the product $HBrO_3$ (bromic acid),the oxidation state of $Br$ is $+5$.
Since the suffix is $-ic$ (bromic acid),the product is an oxy acid with an $-ic$ suffix.
Therefore,the correct option is $A$.

(A) The complete hydrolysis of iodine pentafluoride $(IF_5)$ at room temperature $(R.T.)$ is represented by the balanced chemical equation:
$IF_5 + 3H_2O \longrightarrow HIO_3 + 5HF$
In this reaction,the iodine atom in $IF_5$ (oxidation state $+5$) is converted into iodic acid $(HIO_3)$.
Since the suffix for $HIO_3$ is $-ic$ (iodic acid),the correct description is that the product is an oxy acid with an $-ic$ suffix.

(A) The reaction for the hydrolysis of $I_2O_5$ is:
$I_2O_5 + H_2O \longrightarrow 2HIO_3$
In $HIO_3$ (iodic acid),the oxidation state of iodine is $+5$.
Since the suffix used for the acid with the higher oxidation state is $-ic$,$HIO_3$ is named iodic acid.
Therefore,the product is an oxy acid with an $-ic$ suffix.

(B) The complete hydrolysis of $ClF_3$ at room temperature $(R.T)$ is given by the reaction:
$ClF_3 + 2H_2O \longrightarrow HClO_2 + 3HF$
Here,$HClO_2$ is chlorous acid.
The suffix used for $HClO_2$ is $-ous$ (since the oxidation state of $Cl$ is $+3$).
Therefore,the product is an oxy acid with an $-ous$ suffix.

(D) $SF_6$ is an extremely stable molecule due to steric hindrance and the absence of vacant $d$-orbitals accessible for bonding with water molecules.
It does not undergo hydrolysis at room temperature $(R.T.)$.
Since there is no reaction,no oxy acid is formed.
Therefore,the product is not an oxy acid,and it does not possess an $-ic$ or $-ous$ suffix.

(A) The reaction for the complete hydrolysis of $IOF_5$ is:
$IOF_5 + 2H_2O \longrightarrow HIO_4 + 5HF$
In the product $HIO_4$ (periodic acid),the oxidation state of iodine is $+7$.
Since the oxidation state is $+7$,the acid is named periodic acid,which ends with the $-ic$ suffix.
Therefore,the correct option is $A$.