Halogen family Questions in English

(A) The reducing character of hydrogen halides 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 dissociation enthalpy is $HF > HCl > HBr > HI$.
Since $HI$ has the weakest bond,it can easily release $H^+$ and $e^-$,making it the strongest reducing agent among the given options.

(B) Euchlorine is a mixture of chlorine gas $(Cl_2)$ and chlorine dioxide $(ClO_2)$. It is obtained by the action of hydrochloric acid on potassium chlorate $(KClO_3)$.

(D) Chlorine acts as a bleaching agent in the presence of moisture. $Cl_2 + H_2O \rightarrow 2HCl + [O]$. The nascent oxygen $[O]$ produced is responsible for the bleaching action,which is an $Oxidation$ process. Therefore,the bleaching action of chlorine is due to $Oxidation$.

(A) The reaction of chlorine gas with cold and dilute sodium hydroxide is a disproportionation reaction.
The chemical equation is: $Cl_2 + 2NaOH \rightarrow NaCl + NaClO + H_2O$.
In this reaction,$Cl_2$ is reduced to $NaCl$ and oxidized to $NaClO$ (sodium hypochlorite).
Therefore,the correct product is $NaClO$.

(B) Interhalogen compounds are formed by the reaction of two different halogens.
$1$. They are generally more reactive than halogens (except $F_2$) because the bond between two different halogens is weaker than the bond between two similar halogens.
$2$. They are covalent in nature.
$3$. They are volatile compounds with low boiling points.
$4$. Many of them are unstable and some are even explosive (e.g.,$ClF_3$ is highly reactive and potentially explosive).
Therefore,the statement that they are not explosive is incorrect.

(A) In the context of the halogen group $(Group \ 17)$,the elements are non-metals and generally act as oxidizing agents rather than bases. However,when considering the acidity of their corresponding hydrohalic acids $(HF, HCl, HBr, HI)$,the acidity increases down the group $(HF < HCl < HBr < HI)$. Conversely,the basicity of their conjugate bases $(F^-, Cl^-, Br^-, I^-)$ decreases down the group. Therefore,the fluoride ion $(F^-)$ is the strongest conjugate base among the halides. Since the question asks for the most basic element in terms of its ionic form or chemical character,$F$ is the most electronegative and its corresponding ion is the most basic.

(B) Iodine forms several oxides,but $I_2O_5$ is the only stable and true covalent oxide of iodine.
It is a white crystalline solid and is used in the estimation of carbon monoxide $(CO)$ in the air.
$I_2O_5 + 5CO \rightarrow I_2 + 5CO_2$.

(D) As we move down the group in the halogen family ($F$ to $I$),the atomic size increases due to the addition of new shells.
Because of the increase in atomic size,the effective nuclear charge experienced by the incoming electron decreases.
Consequently,the tendency to gain an electron decreases as the atomic number increases.
Therefore,halogens gain electrons less easily as the atomic number increases.

(C) When chlorine gas is passed over dry slaked lime $(Ca(OH)_2)$,it reacts to form bleaching powder $(CaOCl_2)$ and water.
The chemical equation is:
$Ca(OH)_2 + Cl_2 \rightarrow CaOCl_2 + H_2O$
Thus,the product formed is bleaching powder,which is $CaOCl_2$.

(D) Fluorine is the most electronegative element and acts as a strong oxidizing agent.
When fluorine reacts with dilute $NaOH$,it forms oxygen difluoride $(OF_2)$:
$2F_2 + 2NaOH \rightarrow OF_2 + 2NaF + H_2O$
When fluorine reacts with concentrated $NaOH$,it forms oxygen $(O_2)$:
$2F_2 + 4NaOH \rightarrow O_2 + 4NaF + 2H_2O$
Thus,the products are $OF_2$ and $O_2$ respectively. However,looking at the provided options,option $D$ is the closest match for the first product,but the correct products are $OF_2$ and $O_2$. Given the standard nature of this question,the intended answer is $OF_2$ and $O_2$. Since $O_2$ is not explicitly in the options as a pair,we select the option that correctly identifies the first product.

(A) The ability to accept a fluoride ion $(F^-)$ is a characteristic of Lewis acids.
$SbF_5$ (Antimony pentafluoride) is a very strong Lewis acid because the central $Sb$ atom is highly electron-deficient and can easily accommodate the lone pair from the fluoride ion to form the stable $[SbF_6]^-$ complex.
Among the given options,$SbF_5$ is the strongest fluoride ion acceptor,often used to create superacids like fluoroantimonic acid $(HSbF_6)$.

(D) The iodine oxides $I_2O_4$ and $I_4O_9$ are ionic in nature.
$I_2O_4$ is formulated as $IO^+ \cdot IO_3^-$,which contains iodine in $+1$ and $+5$ oxidation states.
$I_4O_9$ is formulated as $I^{3+} \cdot (IO_3^-)_3$,which contains iodine in $+3$ and $+5$ oxidation states.
Upon heating,$I_4O_9$ decomposes to form $I_2O_5$ and $I_2$ ($2I_4O_9 \rightarrow 2I_2O_5 + 3O_2$ is not the standard path,but it is known to yield $I_2O_5$ upon thermal decomposition).
Therefore,all the given statements are correct.

(D) Periodic acid exists in different forms depending on the degree of hydration.
$HIO_4$ is known as metaperiodic acid.
When $HIO_4$ reacts with water,it forms $H_5IO_6$,which is $HIO_4 \cdot 2H_2O$.
This form $(H_5IO_6)$ is commonly known as orthoperiodic acid.

(A) The industrial preparation of chlorine by the oxidation of hydrogen chloride gas with atmospheric oxygen in the presence of $CuCl_2$ as a catalyst at $723 \ K$ is known as Deacon's process.
The chemical reaction is: $4HCl + O_2 \xrightarrow{CuCl_2} 2Cl_2 + 2H_2O$.

(C) Hydrofluoric acid $(HF)$ reacts with silica $(SiO_2)$ present in glass to form silicon tetrafluoride $(SiF_4)$.
The chemical reaction is: $SiO_2(s) + 4HF(aq) \rightarrow SiF_4(g) + 2H_2O(l)$.
Further,$SiF_4$ can react with excess $HF$ to form hydrofluorosilicic acid $(H_2SiF_6)$: $SiF_4(g) + 2HF(aq) \rightarrow H_2SiF_6(aq)$.
Therefore,$HF$ is stored in wax-coated bottles or plastic containers instead of glass.

(A) $ClO_3$ is an odd-electron molecule with an unpaired electron,making it paramagnetic.
$Cl_2O_6$ exists as a dimer in the solid state as $[ClO_2]^+[ClO_4]^-$,which contains only paired electrons,making it diamagnetic.
Therefore,$ClO_3$ is paramagnetic and $Cl_2O_6$ is diamagnetic.

(B) The reducing power of halide ions depends on their ability to lose electrons,which is related to their oxidation potential.
As the size of the halide ion increases down the group $(F^- < Cl^- < Br^- < I^-)$,the hold of the nucleus on the valence electrons decreases.
Therefore,it becomes easier to remove an electron from the larger ion,making it a stronger reducing agent.
Thus,the increasing order of reducing power is $F^- < Cl^- < Br^- < I^-$.

(A) The oxidizing power of oxoacids of chlorine depends on the ease with which they can release oxygen or accept electrons.
In $HOCl$,the oxidation state of $Cl$ is $+1$.
In $HClO_2$,the oxidation state of $Cl$ is $+3$.
In $HClO_3$,the oxidation state of $Cl$ is $+5$.
In $HClO_4$,the oxidation state of $Cl$ is $+7$.
$HOCl$ is the least stable and most reactive among these,as it can easily release oxygen to act as a strong oxidizing agent.
Therefore,the oxidizing power decreases in the order: $HOCl > HClO_2 > HClO_3 > HClO_4$.

(D) The reducing power of halide ions depends on their ability to lose electrons,which is related to their 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,$I^{-}$ is the strongest reducing agent among the given options.

(A) Iodized salt is prepared by adding small amounts of iodine compounds to common salt $(NaCl)$.
$KIO_3$ (potassium iodate) and $KI$ (potassium iodide) are the standard compounds used for this purpose because they are stable and provide the necessary iodine content.

(B) Iodised salt is common table salt $(NaCl)$ that has been fortified with a small amount of iodine-containing salts.
These salts are typically $KI$ (potassium iodide),$KIO_3$ (potassium iodate),$NaI$ (sodium iodide),or $NaIO_3$ (sodium iodate).
Among the given options,$NaIO_3$ is the correct chemical form present in iodised salt.

(B) When chlorine water $(Cl_2)$ is added to a solution of $KBr$,the following reaction occurs: $2KBr + Cl_2 \to 2KCl + Br_2$.
Initially,the solution turns brown due to the formation of $Br_2$.
However,in the presence of excess chlorine water,$Br_2$ reacts further with $Cl_2$ to form $BrCl$ (bromine monochloride),which is orange-red in color: $Br_2 + Cl_2 \to 2BrCl$.

(A) The reaction of chlorine with cold and dilute sodium hydroxide is a disproportionation reaction.
The balanced chemical equation is:
$2NaOH + Cl_2 \to NaCl + NaOCl + H_2O$
In this reaction,chlorine is both oxidized and reduced.
The ionic products formed are chloride $(Cl^{-})$ and hypochlorite $(OCl^{-})$.

(C) The ability of halogens to act as oxidizing agents decreases down the group $(F_2 > Cl_2 > Br_2 > I_2)$.
$Br_2$ can only oxidize the iodide ion $(I^-)$ to iodine $(I_2)$ because $I^-$ is a stronger reducing agent than $Br^-$,$Cl^-$,and $F^-$.
$Br_2$ cannot oxidize $Cl^-$ or $F^-$ because $Br_2$ is a weaker oxidizing agent than $Cl_2$ and $F_2$.
Therefore,the reaction is: $2NaI + Br_2 \to 2NaBr + I_2$.

(C) The first reaction represents the oxidation of water by fluorine,where $F_2$ acts as a strong oxidizing agent: $2F_{2(g)} + 2H_2O_{(l)} \to 4H^{+}_{(aq)} + 4F^{-}_{(aq)} + O_{2(g)}$. Thus,$X = F$.
The second reaction represents the disproportionation of halogens (like chlorine) in water: $Cl_{2(g)} + H_2O_{(l)} \to HCl_{(aq)} + HOCl_{(aq)}$. Thus,$Y = Cl$.

(A) Fluorine is the most electronegative element and has a small atomic size.
It has the electronic configuration $1s^2 2s^2 2p^5$.
Due to the absence of vacant $d-$orbitals in the valence shell,it cannot expand its octet or show positive oxidation states.
Therefore,it exhibits only a $-1$ oxidation state.
The reason provided ($2s^2 2p^5$ configuration) is the fundamental cause for its inability to show other oxidation states,making it the correct explanation for the assertion.

(D) Fluorine $(F_2)$ is the most reactive halogen due to its extremely low bond dissociation energy,which makes the $F-F$ bond very easy to break.
Therefore,the Assertion that fluorine has lower reactivity is incorrect,while the Reason that the $F-F$ bond has low bond dissociation energy is correct.

(A) All halogens are coloured because they absorb light in the visible region of the electromagnetic spectrum.
This absorption of visible light causes the excitation of electrons to higher energy levels,which results in the characteristic colours of the halogens.
Therefore,the Assertion is correct,and the Reason is the correct explanation for the Assertion.

(A) In $SF_6$,the sulphur atom is sterically protected by $six$ fluorine atoms,which prevents the attack of $H_2O$ molecules.
Additionally,the sulphur atom in $SF_6$ is coordinatively saturated.
In contrast,$SF_4$ has a lone pair on the sulphur atom and is not sterically hindered to the same extent,allowing $H_2O$ to attack the sulphur atom and undergo hydrolysis.
Thus,both the assertion and the reason are correct,and the reason correctly explains the assertion.

(A) $SnI_4$ is an orange-colored solid.
The colour arises due to the charge transfer phenomenon.
In $SnI_4$,an electron is transferred from the iodide ion $(I^-)$ to the tin center $(Sn^{4+})$ upon absorption of light,which is known as ligand-to-metal charge transfer $(LMCT)$.
This absorption of light in the blue region of the visible spectrum results in the compound appearing orange.

(A) Tribromooctaoxide has the chemical formula $Br_3O_8$.
In this structure,the central bromine atom is bonded to two other bromine atoms.
Each of the terminal bromine atoms is bonded to three oxygen atoms via double bonds and to the central bromine atom.
The central bromine atom is bonded to two oxygen atoms via double bonds.
This results in a neutral molecule where all bromine atoms achieve a stable oxidation state.
The other options represent anionic species,which do not correspond to the neutral $Br_3O_8$ molecule.

(A) The statement '$All$ form monobasic oxyacids' is incorrect because $fluorine$ does not form a series of oxyacids like other halogens due to its high electronegativity and small size. It only forms $HOF$ ($Fluoric(I)$ acid).
All halogens act as oxidizing agents.
$Chlorine$ has the highest electron-gain enthalpy among all halogens.
Except for $fluorine$,all halogens show positive oxidation states $(+1, +3, +5, +7)$ in their oxyacids.

(A) For the reaction with hot and concentrated $NaOH$:
$6 NaOH + 3 Cl_{2} \rightarrow NaClO_{3} + 5 NaCl + 3 H_{2}O$
Here,$(A)$ is $NaClO_{3}$ (sodium chlorate).
For the reaction with dry $Ca(OH)_{2}$:
$2 Ca(OH)_{2} + 2 Cl_{2} \rightarrow Ca(OCl)_{2} + CaCl_{2} + 2 H_{2}O$
Here,$(B)$ is $Ca(OCl)_{2}$ (calcium hypochlorite).
Thus,the products $(A)$ and $(B)$ are $NaClO_{3}$ and $Ca(OCl)_{2}$ respectively.

(A) The reaction of chlorine with hot and concentrated $NaOH$ is:
$3 Cl_2 + 6 NaOH \rightarrow 5 NaCl + NaClO_3 + 3 H_2O$
Here,$(X)$ is $NaCl$ and $(Y)$ is $NaClO_3$.
$NaCl$ reacts with $AgNO_3$ to give a white precipitate of $AgCl$:
$NaCl + AgNO_3 \rightarrow AgCl \downarrow + NaNO_3$
In the chlorate ion $(ClO_3^-)$,the central $Cl$ atom is bonded to three oxygen atoms. There are $3$ resonance structures.
The total number of bonds between $Cl$ and $O$ atoms across the resonance structures is $5$ (one double bond and two single bonds in each resonance contributor).
Average bond order $= \frac{\text{Total number of bonds}}{\text{Number of resonating positions}} = \frac{5}{3} \approx 1.67$.

(N/A) $F_{2}$ can oxidize $Cl^{-}$ to $Cl_{2}$,$Br^{-}$ to $Br_{2}$,and $I^{-}$ to $I_{2}$ as:
$F_{2(aq)} + 2Cl^{-}_{(aq)} \longrightarrow 2F^{-}_{(aq)} + Cl_{2(g)}$
$F_{2(aq)} + 2Br^{-}_{(aq)} \longrightarrow 2F^{-}_{(aq)} + Br_{2(l)}$
$F_{2(aq)} + 2I^{-}_{(aq)} \longrightarrow 2F^{-}_{(aq)} + I_{2(s)}$
On the other hand,$Cl_{2}$,$Br_{2}$,and $I_{2}$ cannot oxidize $F^{-}$ to $F_{2}$. The oxidizing power of halogens increases in the order: $I_{2} < Br_{2} < Cl_{2} < F_{2}$. Hence,fluorine is the best oxidant among halogens.
$HI$ and $HBr$ can reduce $H_{2}SO_{4}$ to $SO_{2}$,but $HCl$ and $HF$ cannot. Therefore,$HI$ and $HBr$ are stronger reductants than $HCl$ and $HF$.
$2HI + H_{2}SO_{4} \longrightarrow I_{2} + SO_{2} + 2H_{2}O$
$2HBr + H_{2}SO_{4} \longrightarrow Br_{2} + SO_{2} + 2H_{2}O$
Again,$I^{-}$ can reduce $Cu^{2+}$ to $Cu^{+}$,but $Br^{-}$ cannot.
$4I^{-}_{(aq)} + 2Cu^{2+}_{(aq)} \longrightarrow Cu_{2}I_{2(s)} + I_{2(aq)}$
Hence,hydroiodic acid $(HI)$ is the best reductant among hydrohalic compounds. Thus,the reducing power of hydrohalic acids increases in the order: $HF < HCl < HBr < HI$.

(N/A) The stronger oxidizing power of fluorine compared to chlorine is due to the following factors:
$(i)$ Low enthalpy of dissociation of the $F-F$ bond.
$(ii)$ High hydration enthalpy of the $F^-$ ion.
These factors collectively make the standard electrode potential $(E^{\Theta})$ of fluorine more positive, thereby making it a stronger oxidizing agent.

Property	$F$
Atomic number	$9$
Atomic mass $/g \ mol^{-1}$	$19.00$
Electronic configuration	$[He] 2s^2 2p^5$
Covalent radius $/pm$	$64$
Ionic radius $X^- / pm$	$133$
Ionisation enthalpy $/kJ \ mol^{-1}$	$1680$
Electron gain enthalpy $/kJ \ mol^{-1}$	$-333$
Electronegativity	$4.0$
$\Delta_{Hyd} H(X^-) / kJ \ mol^{-1}$	$515$
Bond dissociation enthalpy $/kJ \ mol^{-1}$	$158.8$
$E^{\Theta} / V$	$2.87$

(N/A) 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 absence of $d$-orbitals,it cannot exhibit any positive oxidation state.
Other halogens (like $Cl, Br, I$) possess vacant $d$-orbitals in their valence shell. This allows them to expand their octets and promote electrons to higher energy levels,enabling them to exhibit $+1, +3, +5$ and $+7$ oxidation states in addition to $-1$.

(N/A) Fluorine is a much stronger oxidizing agent than chlorine. The oxidizing power depends on three factors:
$1.$ Bond dissociation energy
$2.$ Electron gain enthalpy
$3.$ Hydration enthalpy
The electron gain enthalpy of chlorine is more negative than that of fluorine. However,the bond dissociation energy of fluorine is much lower than that of chlorine. Also,because of its small size,the hydration energy of fluorine is much higher than that of chlorine. Therefore,the latter two factors more than compensate for the less negative electron gain enthalpy of fluorine. Thus,$F_{2}$ is a much stronger oxidizing agent than $Cl_{2}$.

(N/A) The anomalous behaviour of fluorine is due to its small size,high electronegativity,low $F-F$ bond dissociation enthalpy,and absence of $d$-orbitals in its valence shell.
$i$. Fluorine forms only one oxoacid $(HOF)$,whereas other halogens form a number of oxoacids (e.g.,$HOCl, HOClO, HOClO_2, HOClO_3$).
$ii$. The ionisation enthalpy,electronegativity,and electrode potential of fluorine are much higher than expected compared to other halogens.

(N/A) Sea water contains chlorides,bromides,and iodides of $Na$,$K$,$Mg$,and $Ca$.
It primarily contains $NaCl$.
The deposits of dried-up sea beds contain sodium chloride and carnallite,$KCl \cdot MgCl_2 \cdot 6H_2O$.
Marine life also contains iodine in their systems.
For example,sea weeds contain up to $0.5\%$ iodine as sodium iodide.
Thus,the sea is the greatest source of halogens.

(B) The bleaching action of $Cl_{2}$ is due to the process of oxidation. $Cl_{2}$ reacts with water to produce nascent oxygen,which is a powerful oxidizing agent.
$Cl_{2} + H_{2}O \longrightarrow 2HCl + [O]$
This nascent oxygen $[O]$ reacts with coloured organic substances and oxidizes them into colourless substances.
$\text{Coloured substance} + [O] \longrightarrow \text{Colourless substance}$

(N/A) Two poisonous gases that can be prepared from chlorine gas are:
$(i)$ Phosgene $(COCl_2)$
$(ii)$ Mustard gas $(ClCH_2CH_2SCH_2CH_2Cl)$

(C) The reactivity of interhalogen compounds is generally higher than that of halogens because the $I-Cl$ bond is a polar covalent bond,whereas the $I-I$ bond is a non-polar covalent bond.
Additionally,the $I-Cl$ bond is weaker than the $I-I$ bond due to the difference in electronegativity and size between iodine and chlorine atoms.
Therefore,both factors contribute to the higher reactivity of $ICl$ compared to $I_2$.

(N/A) The general electronic configuration of halogens is $ns^2 np^5$.
They require only one electron to complete their octet and attain the stable noble gas configuration.
Halogens possess high electronegativity,low dissociation energies,and high negative electron gain enthalpies.
Due to these factors,they have a strong tendency to gain an electron,which makes them strong oxidizing agents.

(N/A) Fluorine is the most electronegative element in the periodic table and has a very small atomic size. Due to these properties,it cannot exhibit positive oxidation states or form multiple oxoacids like other halogens. Therefore,it forms only one oxoacid,$HOF$ (hypofluorous acid).

(C) $i.$ It is used for the purification of water.
$ii.$ It is used as a bleaching agent for paper pulp and textiles.

(B) Almost all halogens are coloured.
This is because halogens absorb radiations in the visible region.
This results in the excitation of valence electrons to a higher energy level.
Since the amount of energy required for excitation differs for each halogen,each halogen displays a different colour.

(N/A) $(i) \ Cl_{2(g)} + H_2O_{(l)} \to HCl_{(aq)} + HOCl_{(aq)}$
$(ii) \ 2F_{2(g)} + 2H_2O_{(l)} \to 4H^+_{(aq)} + 4F^-_{(aq)} + O_{2(g)}$

(N/A) $(i)$ $Cl_2$ can be prepared from $HCl$ by Deacon's process:
$4HCl + O_2 \xrightarrow{CuCl_2} 2Cl_2 + 2H_2O$
$(ii)$ $HCl$ can be prepared from $Cl_2$ by reaction with water:
$Cl_2 + H_2O \to HCl + HOCl$

(N/A) $(i)$ $4NaCl + MnO_{2} + 4H_{2}SO_{4} \longrightarrow MnCl_{2} + 4NaHSO_{4} + 2H_{2}O + Cl_{2}$
$(ii)$ $Cl_{2} + 2NaI \longrightarrow 2NaCl + I_{2}$