Halogen family Questions in English

(D) Iodine $(I_2)$ is purified by the process of sublimation.
This is because iodine exists as a solid at room temperature and readily sublimes upon heating.
In contrast,fluorine $(F_2)$ and chlorine $(Cl_2)$ are gases,and bromine $(Br_2)$ is a liquid at room temperature,making sublimation inapplicable for their purification.

(C) To prevent iodine deficiency,which can lead to goiter and other health issues,small amounts of potassium iodide $(KI)$ or sodium iodide $(NaI)$ are added to table salt.
This is known as iodized salt.
Therefore,the correct option is $C$.

(D) The correct statement is that $SF_4$ is polar and reactive,while $SF_6$ is non-polar and chemically inert.
$SF_6$ is chemically inert because the six $F$ atoms sterically protect the central $S$ atom from attack by reagents,preventing reactions like hydrolysis.
$SF_4$ is polar due to its see-saw geometry and the presence of a lone pair on the $S$ atom.
$SF_4$ is typically prepared by the reaction: $3SCl_2 + 4NaF \rightarrow SF_4 + S_2Cl_2 + 4NaCl$.

(A) The acidic strength of hydrohalic acids $(HX)$ depends on the bond dissociation enthalpy of the $H-X$ bond.
As we move down the group from $F$ to $I$,the atomic size of the halogen increases,which leads to an increase in the bond length and a decrease in the bond dissociation enthalpy.
Therefore,the $H-I$ bond is the weakest and breaks most easily to release $H^+$ ions,making $HI$ the strongest acid.
The correct order of increasing acidity is $HF < HCl < HBr < HI$.

(A) The bond strength of hydrogen halides depends on the bond dissociation enthalpy.
As the size of the halogen atom increases from $F$ to $I$,the bond length increases,which leads to a decrease in bond strength.
Therefore,the correct order of bond strength is $HF > HCl > HBr > HI$.

(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 shortest bond length and the highest bond dissociation enthalpy,it is the weakest acid among the given options.

(B) 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 of chlorine,the oxidation states are:
$HClO$ $(+1)$
$HClO_2$ $(+3)$
$HClO_3$ $(+5)$
$HClO_4$ $(+7)$
Since the oxidation state of $Cl$ is highest in $HClO_4$ $(+7)$,it is the strongest acid.
The order of acidic strength is: $HClO < HClO_2 < HClO_3 < HClO_4$.

(B) $SO_2$ bleaches the colour of flowers by reduction,whereas $Cl_2$ bleaches the colour of flowers by oxidation.

(A) $MnO_2$ acts as an oxidizing agent in acidic medium.
It can oxidize $Cl^-$,$Br^-$,and $I^-$ to their respective halogens ($Cl_2$,$Br_2$,$I_2$).
However,$F^-$ has a very high standard oxidation potential and is the strongest reducing agent among halides,while $F_2$ is the strongest oxidizing agent.
Therefore,$MnO_2$ cannot oxidize $F^-$ to $F_2$.

(A) The correct answer is $(A)$.
Fluorine $(F_2)$ is the most powerful oxidizing agent among the halogens.
This is because it has the highest standard reduction potential,$E^o = +2.87 \ V$,which indicates a very strong tendency to gain electrons.

(D) $HOCl$ (or $HClO$) is the strongest oxidising agent among the given oxyacids of chlorine.
The oxidising power depends on the ease with which the chlorine atom can be reduced.
In $HOCl$,the oxidation state of chlorine is $+1$,which can easily be reduced to $Cl^-$ (oxidation state $-1$).
The correct order of oxidising power is $HOCl > HClO_2 > HClO_3 > HClO_4$.

(D) The reducing power of halide ions depends on their ability to lose electrons,which is inversely proportional to their electronegativity and directly proportional to their ionic size.
As we move down the group from $F^{-}$ to $I^{-}$,the ionic size increases,making it easier to lose an electron.
Therefore,the reducing strength order is $F^{-} < Cl^{-} < Br^{-} < I^{-}$.
Thus,$I^{-}$ is the strongest reducing agent among the given options.

(D) $Ag$ has the electronic configuration $[Kr] \, 4d^{10} \, 5s^1$,making $+1$ its stable oxidation state.
$Fe^{3+}$ is more stable than $Fe^{2+}$ due to the half-filled $3d^5$ electronic configuration.
$Sn^{2+}$ is less stable than $Sn^{4+}$ because the $+4$ oxidation state is more stable for tin.
In $p$-block elements,the inert pair effect makes the lower oxidation state more stable as we move down the group. Thus,$Pb^{2+}$ is more stable than $Pb^{4+}$ due to the inert pair effect.

(C) The halogens belong to Group $17$ of the periodic table.
They have $7$ electrons in their outermost valence shell.
The general electronic configuration for the valence shell of halogens is $ns^2 np^5$.

(A) The elements of group $17$,known as the halogens,are: $F, Cl, Br, I$,and $At$ (Astatine).
Therefore,Astatine is a halogen.

(B) The given atomic numbers are $9 (F)$,$17 (Cl)$,$35 (Br)$,$53 (I)$,and $85 (At)$.
These elements belong to group $17$ of the periodic table.
The general valence shell electronic configuration of these elements is $ns^2 np^5$.
Therefore,these elements are known as halogens.

(C) Halogens have the highest electronegativity and the smallest size in their respective periods,allowing them to gain an electron easily to achieve a stable noble gas configuration. Therefore,they form anions most readily.

(C) The correct answer is $(C)$.
As we move down the group in halogens,the atomic size increases due to the addition of new shells.
As the atomic size increases,the distance between the nucleus and the valence electrons increases,leading to a decrease in the effective nuclear attraction.
Consequently,the energy required to remove an electron,known as the ionization potential,decreases.

(A) The reactivity of halogens decreases down the group in the periodic table.
Fluorine $(F)$ is the most reactive halogen due to its high electronegativity and small atomic size.
As we move down the group from $F$ to $I$,the atomic size increases and electronegativity decreases,leading to a decrease in the tendency to gain electrons.
Therefore,the correct order of reactivity is $F > Cl > Br > I$.

(D) The correct answer is $D$. The standard electrode potential of a halogen depends on three factors: enthalpy of dissociation,ionization enthalpy,and hydration enthalpy.
Fluorine has a very low $F-F$ bond dissociation energy due to the high inter-electronic repulsion between the lone pairs of the small $F$ atoms.
This low bond dissociation energy,combined with high hydration enthalpy,makes fluorine the strongest oxidizing agent among the halogens.

(D) The oxidizing power of halogens depends on their ability to gain electrons,which is related to their standard reduction potential $(E^\circ)$.
As we move down the group from $F_2$ to $I_2$,the standard reduction potential decreases.
Therefore,the oxidizing power decreases in the order $F_2 > Cl_2 > Br_2 > I_2$.
Thus,the correct increasing order is $I_2 < Br_2 < Cl_2 < F_2$.

(C) The acidic strength of hydrohalic acids $(HX)$ increases down the group as the bond dissociation enthalpy decreases.
The order of acidic strength is $HF < HCl < HBr < HI$.
Therefore,$HF$ is the least acidic among the given halogen acids due to the strong $H-F$ bond resulting from the small size of the fluorine atom.

(A) The reduction potential of halogens decreases down the group $(F_2 > Cl_2 > Br_2 > I_2)$.
Since $F_2$ has the highest standard reduction potential,it acts as the strongest oxidizing agent and is most easily reduced.

(B) The reactivity of halogens decreases down the group in the periodic table as the atomic size increases and the electronegativity decreases.
Therefore,the correct order of reactivity is $F_2 > Cl_2 > Br_2 > I_2$.

(D) The strongest reducing agent among the halide ions is $I^{-}$.
Reducing power is directly related to the tendency to lose electrons.
Due to its large size,the iodide ion $(I^{-})$ has the maximum tendency to lose electrons,making it the strongest reducing agent.

(D) The oxidising power of an element is directly related to its tendency to gain electrons (reduction potential).
Among halogens,the standard reduction potential decreases down the group from $F$ to $I$.
Therefore,the order of oxidising strength is $I_2 < Br_2 < Cl_2 < F_2$.
Thus,the correct order is $I < Br < Cl < F$.

(C) The acidity of hydrogen halides in aqueous solution depends on the bond dissociation enthalpy of the $H-X$ bond.
As the size of the halogen atom increases from $F$ to $I$,the $H-X$ bond length increases and the bond strength decreases.
Consequently,the ease of releasing $H^+$ ions increases,leading to an increase in acidity in the order: $HF < HCl < HBr < HI$.
Therefore,$HI$ is the most acidic among the given options.

(A) The acid strength of hydrohalic acids depends on the bond dissociation enthalpy of the $H-X$ bond.
As we move down the group from $F$ to $I$,the atomic size of the halogen increases,which leads to an increase in the bond length and a decrease in the bond dissociation enthalpy.
Consequently,the release of the $H^+$ ion becomes easier.
Therefore,the acid strength increases in the order: $HF < HCl < HBr < HI$.

(D) Fluorine $(F)$ is the most electronegative element in the periodic table.
Due to its high electronegativity and the absence of $d$-orbitals in its valence shell,it cannot expand its octet or share electrons in a way that results in a positive oxidation state.
Therefore,fluorine always exhibits an oxidation state of $-1$ in its compounds.
In contrast,other halogens like chlorine $(Cl)$,bromine $(Br)$,and iodine $(I)$ possess vacant $d$-orbitals and lower electronegativity,allowing them to exhibit positive oxidation states such as $+1, +3, +5$,and $+7$.

(B) The basic character of an element is inversely proportional to its ionization potential $(I.P.)$.
Among the halogens,the ionization potential decreases down the group $(F > Cl > Br > I)$.
Therefore,Iodine $(I)$ has the lowest $I.P.$ value,making it the most basic element among the given options.
The correct option is $(B)$.

(A) The affinity of halogens for hydrogen depends on the bond dissociation energy and the electronegativity of the halogen atom.
Fluorine $(F_2)$ is the most electronegative element and has the highest tendency to form a bond with hydrogen to form $HF$.
The decreasing order of reactivity of halogens towards hydrogen is $F_2 > Cl_2 > Br_2 > I_2$.
Therefore,$F_2$ has the maximum affinity for hydrogen.

(C) When potassium dichromate $(K_2Cr_2O_7)$ is heated with concentrated hydrochloric acid $(HCl)$,it acts as an oxidizing agent and oxidizes $HCl$ to chlorine gas $(Cl_2)$.
The chemical reaction is as follows:
$K_2Cr_2O_7 + 14HCl \rightarrow 2KCl + 2CrCl_3 + 7H_2O + 3Cl_2 \uparrow$
Therefore,$Cl_2$ gas is evolved.

(C) The correct answer is $C$. The bleaching action of $Cl_2$ in moist conditions is permanent due to the formation of nascent oxygen.
$Cl_2 + H_2O \to HCl + HClO$
$HClO \to HCl + [O]$
Overall reaction: $Cl_2 + H_2O \to 2HCl + [O]$
Coloured matter + $[O] \to$ Colourless matter.
Since the bleaching is caused by oxidation,it is permanent.

(D) The correct statement analysis is as follows:
$1$. Zinc is amphoteric and reacts with $NaOH$ to form sodium zincate: $Zn + 2NaOH \rightarrow Na_2ZnO_2 + H_2$. This statement is correct.
$2$. Carbon monoxide acts as a reducing agent in the blast furnace: $Fe_2O_3 + 3CO \rightarrow 2Fe + 3CO_2$. This statement is correct.
$3$. Mercury $(II)$ iodide dissolves in excess $KI$ to form a soluble complex: $HgI_2 + 2KI \rightarrow K_2[HgI_4]$. This statement is correct.
$4$. Tin $(IV)$ chloride $(SnCl_4)$ is prepared by the action of dry chlorine gas on tin metal $(Sn + 2Cl_2 \rightarrow SnCl_4)$. Dissolving tin in concentrated $HCl$ produces tin $(II)$ chloride $(SnCl_2)$,not tin $(IV)$ chloride. Therefore,this statement is incorrect.

(B) Glass,which is a mixture of sodium and calcium silicates,reacts with hydrofluoric acid $(HF)$ to form sodium and calcium fluorosilicates.
$Na_2SiO_3 + 6HF \to Na_2SiF_6 + 3H_2O$
$CaSiO_3 + 6HF \to CaSiF_6 + 3H_2O$
The etching of glass is based on these reactions.

(C) The head of a safety matchstick contains a mixture of $K_2Cr_2O_7$,sulfur $(S)$,and red phosphorus $(P)$.

(A) Aqua-regia is a freshly prepared mixture of concentrated $HNO_3$ and concentrated $HCl$ in the ratio of $1 : 3$ by volume.
It is used for dissolving noble metals like gold and platinum.

(C) Fluorine $(F_2)$ is a very strong oxidizing agent. It reacts with water to liberate oxygen gas according to the following reaction:
$2H_2O + 2F_2 \to 4HF + O_2$
Therefore,the correct option is $C$.

(D) Among the given elements,$P$,$Na$,and $S$ react directly with oxygen to form their respective oxides.
$P_4 + 5O_2 \longrightarrow P_4O_{10}$
$S + O_2 \longrightarrow SO_2$
$4Na + O_2 \longrightarrow 2Na_2O$
Chlorine $(Cl)$ does not react directly with oxygen under normal conditions. Therefore,the correct option is $D$.

(D) Hypo (sodium thiosulfate,$Na_2S_2O_3$) is used in photography as a fixing agent. It dissolves the unexposed silver bromide from the photographic film or paper,thereby making the image permanent.

(B) 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 strength decreases.
Therefore,the order of bond dissociation energy is $HF > HCl > HBr > HI$.
Consequently,the correct order of thermal stability is $HF > HCl > HBr > HI$.

(D) $Iodine$ has the least affinity for water and is only slightly soluble in it.
However,it dissolves in $10\%$ aqueous solution of $KI$ due to the formation of a complex ion,$I_3^-$.
The reaction is:
$I_2 + KI \rightleftharpoons KI_3$ or $I_2 + I^- \rightleftharpoons I_3^-$ (complex ion).

(A) The reducing power of hydrohalic acids $(HX)$ increases down the group as the bond dissociation energy decreases.
$HI$ has the lowest bond dissociation energy among $HF$,$HCl$,$HBr$,and $HI$ due to the large size of the $I$ atom.
Therefore,$HI$ is the strongest reducing agent.

(A) In $1774$,the scientist $Carl \ Wilhem \ Scheele$ discovered chlorine gas by the action of $HCl$ on $MnO_2$.
However,at that time,he believed it was a compound containing oxygen.
It was $Humphry \ Davy$ who later identified chlorine as an element in $1810$ and named it chlorine due to its greenish-yellow color.

(B) Carnallite is $KCl \cdot MgCl_2 \cdot 6H_2O$. The mother liquor left after the crystallisation of $KCl$ from carnallite contains about $0.25\%$ of bromine as $MgBr_2$ and $KBr$.

(D) . Chlorine reacts with sodium hydroxide $(NaOH)$ under different conditions as follows:
$1$. With cold and dilute $NaOH$:
$2NaOH + Cl_2 \xrightarrow{\text{Cold}} NaCl + NaClO + H_2O$
(Here,$NaClO$ is sodium hypochlorite).
$2$. With hot and concentrated $NaOH$:
$6NaOH + 3Cl_2 \xrightarrow{\text{Heat}} 5NaCl + NaClO_3 + 3H_2O$
(Here,$NaClO_3$ is sodium chlorate).
Since all the products mentioned in the options are formed under different conditions,the correct answer is $D$.

(A) The bromine $(Br_2)$ acts as an oxidizing agent and displaces iodine $(I_2)$ from the potassium iodide $(KI)$ present in the starch iodide paper.
$Br_2 + 2KI \rightarrow 2KBr + I_2$
The liberated iodine $(I_2)$ then reacts with the starch to form a blue-black complex,which is commonly referred to as a blue color.

(D) Deacon's process is used for the industrial manufacture of chlorine.
In this process,hydrogen chloride $(HCl)$ is oxidized by atmospheric oxygen $(O_2)$ in the presence of a catalyst,copper$(II)$ chloride $(CuCl_2)$,at a temperature of about $723 \ K$.
The chemical equation for the reaction is:
$4HCl + O_2 \xrightarrow{CuCl_2} 2Cl_2 + 2H_2O$

(A) The acidic strength of hydrohalic acids increases down the group: $HF < HCl < HBr < HI$.
This is because the bond dissociation enthalpy decreases as the size of the halogen atom increases,making it easier to release the $H^{+}$ ion.
$HF$ is the weakest acid because of the strong $H-F$ bond and the presence of strong intermolecular $H$-bonding,which restricts the release of $H^{+}$ ions.
Therefore,the correct option is $A$.

(B)

Hydride	$HF, HCl, HBr, HI$
Boiling point (in $K$)	$293, 189, 206, 238$

Volatility is inversely proportional to the boiling point. Since $HCl$ has the lowest boiling point $(189 \ K)$ among the given hydrogen halides,it is the most volatile compound.