p-Block Elements

Statement given in A, C and D are correct. While that in B is incorrect.

(A) $\mathrm{OF}_2$ is thermodynamically stable at 298 K because of the higher electronegativity of fluorine then oxygen.

(B) Order of stability of oxide of halogen $=\mathrm{I}>\mathrm{Cl}>\mathrm{Br}>\mathrm{F}$.

(C) Iodine pentoxide $\left[\mathrm{I}_2 \mathrm{O}_5\right]$ easily oxidises corbon monoxide to carbon dioxide at room temperature.

$ 5 \mathrm{CO}+\mathrm{I}_2 \mathrm{O}_5 \longrightarrow \mathrm{I}_2+5 \mathrm{CO} $

(D) $\mathrm{Cl}_2 \mathrm{O}$ is used as a bleaching agent for paper pulp and textiles.

Shape of $\mathrm{XeF}_6=\mathrm{bp}=6, \mathrm{lp}=1$

Shape = Distorted octahedral

Shape of $\mathrm{XeOF}_4=\mathrm{bp}=5, \mathrm{lp}=1$

Shape $=$ Square pyramidal

Among the given statements (A) and $(C)$ are correct while $(B)$ is an incorrect statement. $\mathrm{MgH}_2$ is not an interstitial hydride. Instead, it has significant covalent character and has polymeric structure.

Among the given statements C and D are incorrect.

(C) In case of aluminium, the sum of first three ionisation energies decreases considerably and hence, its value can never be positive. Its $E^{\circ}{ }_M{ }^{3+} / M(\mathrm{~V})$ value is -1.66 (i.e. in negative).

(D) In group 13, the +1 oxidation state becomes more and more stable as we move down the group i.e.

$\mathrm{Al}<\mathrm{Ga}<\mathrm{In}<\mathrm{Tl}$. This is because of the inert pair effect. Thus, in $\mathrm{Tl}+1$ oxidation state is more predominant. Whereas the +3 oxidation state is strongly oxidising in nature.

The melting point of carbon is exceedingly high. The value decreases on moving down the group because the $M-M$ bonds become weaker and thus they are easy to break. So, the correct order of melting point is

Among the given statements (c) is correct while others are incorrect.

• $\mathrm{N}_2$ is colourless, odourless, tasteless and non-toxic gas.

• Ozone is thermodynamically less stable than oxygen because its decomposition into oxygen results in the liberation of heat ( $\Delta H$ is negative) and increase in entropy ( $\Delta S$ is positive). These two effects support each other that results in large negative $\Delta G$ for its conversion into oxygen.

• $\mathrm{Cl}_2$ is greenish yellow gas with pungent and suffocating odour.

Among the given oxo-acids of sulphur, $\mathrm{H}_2 \mathrm{~S}_2 \mathrm{O}_7$ contains $\mathrm{S}-\mathrm{O}-\mathrm{S}$ bond.

All the given pairs are correctly matched.

• $\mathrm{Cl}_2$ is used in the preparation of phosgene.

$ \mathrm{CO}(\mathrm{~g})+\mathrm{Cl}_2(\mathrm{~g}) \rightleftharpoons \mathrm{COCl}_2(\mathrm{~g}) $

• Iodine pentenoxide ( $\mathrm{I}_2 \mathrm{O}_5$ ) easily oxidises carbon monoxide $(\mathrm{CO})$ to carbon dioxide at room temperature.

$ 5 \mathrm{CO}+\mathrm{I}_2 \mathrm{O}_5 \longrightarrow \mathrm{I}_2+5 \mathrm{CO}_2 $

This reaction can be used to estimation the concentration of CO in a gaseous sample.

• $\mathrm{O}_3$ is used as a germicide, a disinfectant and for sterilising water.

Kernite is the mineral of boron It's formula is $\mathrm{Na}_2 \mathrm{~B}_4 \mathrm{O}_6(\mathrm{OH})_2 \cdot 3 \mathrm{H}_2 \mathrm{O}$ While cryolite is the mineral of aluminium. It's formula is $\mathrm{Na}_3 \mathrm{AlF}_6$.

Inert pair effect is less prominent in Sn than in Pb . Therefore, +2 oxidation state of Sn is less stable than its +4 oxidation state. Hence, $\mathrm{Sn}^{2+}$ can easily lose two electrons to form $\mathrm{Sn}^{4+}$ and thus $\mathrm{Sn}^{2+}$ acts as a reducing agent.

$ \mathrm{Sn}^{2+} \longrightarrow \mathrm{Sn}^{4+}+2 e^{-} $

In $\mathrm{Pb},+2$ oxidation state of Pb is more stable than +4 oxidation state due to inert pair effect. Therefore, $\mathrm{Pb}^{4+}$ can easily gain two electrons to form $\mathrm{Pb}^{2+}$ and hence $\mathrm{Pb}^{4+}$ acts as an oxidising agent.

$ \mathrm{Pb}^{4+}+2 e^{-} \longrightarrow \mathrm{Pb}^{2+} $

Out of $A, B, C$ and $D, C$ is the incorrect order of thermal stability. Stability of hydrides of 15th group decreases from $\mathrm{NH}_3$ to $\mathrm{SbH}_3$. This is because their bond dissociation enthalpy decreases due to increase in size of central atom from N to Sb .

∴ Correct order of thermal stability is

$ \mathrm{NH}_3>\mathrm{PH}_3>\mathrm{AsH}_3>\mathrm{SbH}_3 . $

$\mathrm{SnO}_2+4 \mathrm{HCl} \longrightarrow \mathrm{SnCl}_2+2 \mathrm{H}_2 \mathrm{O}$

$ \mathrm{SnO}_2+2 \mathrm{NaOH} \longrightarrow \mathrm{Na}_2 \mathrm{SnO}_3+\mathrm{H}_2 \mathrm{O} $

$\mathrm{SnO}_2$ is amphoteric in nature.

$ \begin{aligned} & \mathrm{SnO}+2 \mathrm{HCl} \longrightarrow \mathrm{SnCl}_2+\mathrm{H}_2 \mathrm{O} \\ & \mathrm{SnO}+2 \mathrm{NaOH} \longrightarrow \mathrm{Na}_2 \mathrm{SnO}_2+\mathrm{H}_2 \mathrm{O} \end{aligned} $

SnO is also amphoteric in nature.

$\therefore$ The dioxide and monoxide of Sn , both are amphoteric in nature.

$ \begin{aligned} & \mathrm{PbO}_2+4 \mathrm{HCl} \longrightarrow \mathrm{PbCl}_4+2 \mathrm{H}_2 \mathrm{O} \\ & \mathrm{PbO}_2+2 \mathrm{NaOH} \longrightarrow \mathrm{Na}_2 \mathrm{PbO}_3+\mathrm{H}_2 \mathrm{O} \end{aligned} $

$\mathrm{PbO}_2$ is amphoteric in nature

$ \begin{aligned} & \mathrm{PbO}+2 \mathrm{HCl} \longrightarrow \mathrm{PbCl}_2+\mathrm{H}_2 \mathrm{O} \\ & \mathrm{H}_2 \mathrm{O}+\mathrm{PbO}+\mathrm{Ca}(\mathrm{OH})_2 \longrightarrow \mathrm{Ca}\left[\mathrm{~Pb}(\mathrm{OH})_3\right]_2 \end{aligned} $

PbO is also amphoteric in nature.

$\therefore$ The dioxide and monoxide of Pb both are amphoteric in nature.

In Group 15 elements, nitrogen does not form pentahalides. This is because nitrogen has the electronic configuration of $1s^2, 2s^2, 2p^3$, indicating it only possesses $s$ and $p$ orbitals. Nitrogen lacks $d$ orbitals, which are required to expand its covalency beyond four. Consequently, it cannot form pentahalides.

Nitrogen can attain a +5 oxidation state, achieving a stable electronic configuration of $1s^2$.

Thus, although both the assertion that nitrogen does not form pentahalides and the reasoning that nitrogen can exhibit a +5 oxidation state are true, the reason does not logically explain the assertion.

In Deacon's method, hydrogen chloride ( HCl ) is passed into a blast furnace and burnt with air to produce chlorine and heat.

$ 4 \mathrm{HCl}+\mathrm{O}_2 \xrightarrow[723 \mathrm{~K}]{\mathrm{CuCl}_2} 2 \mathrm{H}_2 \mathrm{O}+\mathrm{Cl}_2 $

The nitrate of potassium metal of heating liberates $\mathrm{KNO}_2$ and $\mathrm{O}_2$, the reaction involved is as follows:

$ 2 \mathrm{KNO}_3(s) \xrightarrow{\Delta} 2 \mathrm{KNO}_2(s)+\mathrm{O}_2(g) $

So, nitrate of potassium does not librate $\mathrm{NO}_2$ gas on heating.

Let's analyze each statement one by one:

$\textbf{(I)}\quad \text{"The stability of +1 oxidation state follows the order } \mathrm{Tl} > \mathrm{In} > \mathrm{Ga} \text{."}$

In group 13, the inert pair effect increases as we go down the group. This makes the +1 oxidation state particularly stable in thallium (Tl), moderately stable in indium (In), and least stable in gallium (Ga).

Thus, statement (I) is correct.

$\textbf{(II)}\quad \text{"Boron has the lowest melting point and boiling point as it is a non‐metal."}$

Even though boron is a non‐metal (or more correctly a metalloid), it is unique in having very high melting and boiling points. For example, boron has a melting point over 2000°C, contrary to the idea that non‐metals have low melting and boiling points.

Therefore, statement (II) is incorrect.

$\textbf{(III)}\quad \text{"Boron shows a maximum covalency of 4 in its compounds."}$

Being a second-period element, boron does not have d-orbitals to expand its octet. Hence, it can accommodate at most four electron pairs around it. This is seen in species like the tetrahydroborate ion, $\mathrm{BH_4^-},$ where boron forms four bonds.

So, statement (III) is correct.

$\textbf{(IV)}\quad \text{"The order of atomic radius is } \mathrm{Ga} > \mathrm{Al} > \mathrm{In} \text{."}$

Generally, atomic radii increase down a group. For group 13, the expected order is:

$\mathrm{Al} < \mathrm{Ga} < \mathrm{In}.$

The given order is not consistent with the trend (in fact, indium should be larger than gallium), so statement (IV) is incorrect.

$\textbf{(V)}\quad \text{"Aluminium is passive to concentrated nitric acid."}$

Aluminium forms a thin, adherent oxide layer that protects it from reacting with concentrated nitric acid.

Thus, statement (V) is correct.

Summing up, the correct statements are (I), (III), and (V).

The answer is Option A: I, III and V only.

First, let us identify what the species $Q$ is in the given reaction:

$ \mathrm{P}_4 + 3\,\mathrm{NaOH} + 3\,\mathrm{H}_2\mathrm{O} \;\longrightarrow\; Q \;+\; 3\,\mathrm{NaH}_2\mathrm{PO}_2. $

This is a classic reaction in which white phosphorus $\mathrm{P}_4$ reacts with a base (e.g., $\mathrm{NaOH}$) and water to form phosphine ($\mathrm{PH}_3$) and sodium hypophosphite ($\mathrm{NaH}_2\mathrm{PO}_2$):

$ \mathrm{P}_4 + 3\,\mathrm{NaOH} + 3\,\mathrm{H}_2\mathrm{O} \;\longrightarrow\; \mathrm{PH}_3 \;+\; 3\,\mathrm{NaH}_2\mathrm{PO}_2. $

Hence, $Q = \mathrm{PH}_3$ (phosphine).

Checking Each Option

We want to find the reaction that does NOT produce phosphine $(\mathrm{PH}_3)$.

Option A: $\mathrm{Ca}_3 \mathrm{P}_2 + \mathrm{H}_2 \mathrm{O}$

Calcium phosphide and water react to form phosphine ($\mathrm{PH}_3$) and calcium hydroxide:

$ \mathrm{Ca}_3\mathrm{P}_2 + 6\,\mathrm{H}_2\mathrm{O} \;\longrightarrow\; 3\,\mathrm{Ca}(\mathrm{OH})_2 + 2\,\mathrm{PH}_3. $

Phosphine is produced. So $Q$ is formed here.

Option B: $\mathrm{H}_3 \mathrm{PO}_3 \xrightarrow{\Delta}$

On heating phosphorous acid, it disproportionates to form phosphoric acid ($\mathrm{H}_3\mathrm{PO}_4$) and phosphine ($\mathrm{PH}_3$):

$ 4\,\mathrm{H}_3\mathrm{PO}_3 \;\xrightarrow{\Delta}\; 3\,\mathrm{H}_3\mathrm{PO}_4 + \mathrm{PH}_3. $

Phosphine is produced. So $Q$ is formed here too.

Option C: $\mathrm{PH}_4 \mathrm{I} + \mathrm{KOH}$

Phosphonium iodide reacts with a strong base (KOH) to release phosphine ($\mathrm{PH}_3$) plus $\mathrm{KI}$ and $\mathrm{H}_2\mathrm{O}$:

$ \mathrm{PH}_4\mathrm{I} + \mathrm{KOH} \;\longrightarrow\; \mathrm{PH}_3 + \mathrm{KI} + \mathrm{H}_2\mathrm{O}. $

Phosphine is produced. So $Q$ is formed in this reaction as well.

Option D: $\mathrm{PCl}_3 + \mathrm{H}_2\mathrm{O}$

Phosphorus trichloride hydrolyzes (usually vigorously) to form phosphorous acid ($\mathrm{H}_3\mathrm{PO}_3$) and hydrochloric acid ($\mathrm{HCl}$):

$ \mathrm{PCl}_3 + 3\,\mathrm{H}_2\mathrm{O} \;\longrightarrow\; \mathrm{H}_3\mathrm{PO}_3 + 3\,\mathrm{HCl}. $

Phosphine ($\mathrm{PH}_3$) is not produced in this reaction.

Conclusion

Since we are looking for the reaction in which phosphine $(Q)$ is not a product, the correct choice is:

$ \boxed{\text{Option D: } \mathrm{PCl}_3 + \mathrm{H}_2 \mathrm{O} \longrightarrow \mathrm{H}_3\mathrm{PO}_3 + 3\,\mathrm{HCl} \ (\text{no } \mathrm{PH}_3).} $

The formula of peroxydisulphuric acid is $\mathrm{H}_2 \mathrm{~S}_2 \mathrm{O}_8$. It's structure is given by

From the structure, we can clearly see that, the oxidation state for both the sulphur atom is +6 and it has two $\mathrm{S}-\mathrm{OH}$ bonds.

The outer electronic configuration is $n s^2 n p^1$, i.e. the element belongs to group 13 (boron family). As the element in crystalline form is unreactive towards oxygen, it might be boron which does not react with air in crystalline form. In amorphous form it reacts to form $\mathrm{B}_2 \mathrm{O}_3$. Boron trioxide which is acidic in nature. Down the group the nature of oxide changes from acidic to amphoteric ( $\mathrm{Al}_2 \mathrm{O}_3$ and gallium oxide) to basic (indium and thallium oxide). Thus, given element is boron.

The element that does not show catenation is Pb (lead). Catenation is the property/tendency of element to link with one another through covalent bond forming long chains and rings. Down the group, this tendency decreases in order $\mathrm{C} \gg \mathrm{Si}>\mathrm{Ge} \approx \mathrm{Sn}$.

$ \text { ∴ Only } \mathrm{H}_2 \mathrm{~S}_2 \mathrm{O}_8 \text { has peroxy linkage. } $

The balanced reaction is

$ 3 \mathrm{Cl}_2+8 \mathrm{NH}_3 \longrightarrow 6 \mathrm{NH}_4 \mathrm{Cl}+\mathrm{N}_2 $

If 3 moles of chlorine is reacting with 8 moles of ammonia.

$\therefore 1$ mole of chlorine will react with $8 / 3$ moles of ammonia.

Enthalpy of vaporisation $\propto$ van der Waals' forces and van der Waals' forces $\propto$ Molar mass

↑ is molar mass, $\uparrow$ will be enthalpy of vaporisation.

∴ The correct order of enthalpy of vaporisation of noble gases is

(A) is false because $\left[\mathrm{B}\left(\mathrm{H}_2 \mathrm{O}\right)_6\right]^{3+}$ does not exist, while $\left[\mathrm{B}(\mathrm{OH})_4\right]$ is $s p^3$ hybridised and have tetrahedral complex.

The boron atom is electron deficient. It has 6 electrons instead of 8 electrons in its valence shell. Hence, it acts as a Lewis acid by readily accepting the lone pair from oxygen from a $\mathrm{H}_2 \mathrm{O}$ molecule or $\mathrm{OH}^{-}$ion. Hence, ( R ) is true.

$\mathrm{N}_2+\mathrm{O}_2 \rightleftharpoons 2 \mathrm{NO}$

Temperature range for this reaction is $1600-2200 \mathrm{~K}$.

Reaction is endothermic when temperature is high more and more heat will be absorbed and reaction will shift in forward reaction.

So, the correct option is (b), i.e. 2000 K .

(A) is true because both rhombic and monoclinic sulphur have $\mathrm{S}_8$ molecules.

Rhombic $S$ is octahedral in form and monoclinic S is crystalline in nature. As rhombic is melted in a dish and then cooled, monoclinic S is obtained. Therefore, $(R)$ is false.

Aluminium produces sodium tetrahydroxoaluminate $(P)$ and hydrogen $(Q)$ as it reacts with excess of sodium hydroxide solution.

Carbon $(C)$ shows catenation to maximum extent because of its small size and tendency to form strong covalent bonds. The least abundant element on earth is germanium ( Ge ). $\therefore X$ is carbon $(\mathrm{C})$ and $Y$ is germanium (Ge).

Group 16 elements are also called chalcogen. They are so called because most of the copper ores have copper in the form of oxides and sulphides.

$ \text { For reaction, } $

$ \text { Hence, shape of } \mathrm{BrF}_5 \text { is square pyramidal. } $

Hydrolysis of $\mathrm{XeF}_4$ will give $\mathrm{XeO}_3$ and Xe along with HF and $\mathrm{O}_2$. It is a type of disproportionation reaction in which $\mathrm{XeF}_4$ undergoes oxidation as well as reduction.

$ \begin{aligned} & 6 \mathrm{XeF}_4+12 \mathrm{H}_2 \mathrm{O} \longrightarrow 4 \mathrm{Xe}+2 \mathrm{XeO}_3+24 \mathrm{HF}+3 \mathrm{O}_2 \\ & \therefore \quad a=6 \\ &\,\,\,\,\,\,\,\,\,\,\,\,\,\, b=12 \\ &\,\,\,\,\,\,\,\,\,\,\,\,\,\, p=4 \\ &\,\,\,\,\,\,\,\,\,\,\,\,\,\, q=2 \\ & \,\,\,\,\,\,\,\,\,\,\,\,\,\,r=24 \\ & \text { and } \quad s=3 \end{aligned} $

Phosphorus acid gives two $\mathrm{H}^{+}$ ions thus, it is dibasic.

$ \mathrm{H}_3 \mathrm{PO}_3 \longrightarrow 2 \mathrm{H}^{+}+\mathrm{HPO}_3^{2-} $

Phosphoric acid gives three $\mathrm{H}^{+}$ions. Thus, it is tribasic.

$ \mathrm{H}_3 \mathrm{PO}_4 \longrightarrow 3 \mathrm{H}^{+}+\mathrm{PO}_4^{3-} $

The order of back-bonding in boron halides is

$ \mathrm{BF}_3>\mathrm{BCl}_3>\mathrm{BBr}_3>\mathrm{BI}_3 $

due to the strength of overlapping which is in order

$ 2 p-2 p>2 p-3 p>2 p-4 p>2 p-5 p $

Hence, stronger is the back-bonding, lesser the availability of empty $p$-orbital and thus, weaker the tendency to act as a Lewis acid.

So, the correct order of Lewis acidic character of boron trihalides is

$ \mathrm{BI}_3>\mathrm{BBr}_3>\mathrm{BCl}_3>\mathrm{BF}_3 . $

Oxides of Sn (i.e. SnO and $\mathrm{SnO}_2$ ) and Pb , (i.e. $\mathrm{PbO}_2$ ) are amphoteric in nature However, oxide of silicon ( $\mathrm{SiO}_2$ ) is acidic in nature.

The test for nitrate ion is called "brown ring test". It can be performed by adding $\mathrm{FeSO}_4$ to a solution of nitrate followed by careful addition of conc. $\mathrm{H}_2 \mathrm{SO}_4$ along the sides of test tube. In this reaction, NO (nitric oxide) is formed in situ which then reacts with the iron (II) complex to give brown ring along the sides of test tube.

Reactions involved

$ \begin{aligned} & 2 \mathrm{NO}_3^{-}+6 \mathrm{H}^{+}+6 \mathrm{Fe}^{2+} \longrightarrow 6 \mathrm{Fe}^{3+}+2 \mathrm{NO}+4 \mathrm{H}_2 \mathrm{O} \end{aligned} $

The colour of $\mathrm{F}_2, \mathrm{Cl}_2, \mathrm{Br}_2$ and $\mathrm{I}_2$ molecules is yellow, greenish yellow, red and violet respectively. They absorb different quanta of radiations in visible region which results in the excitation of outer electrons to higher energy levels.

∴ The correct match is- A-III, B-IV, C-I, D-II

The oxidation state of central atom, Xe increases from +2 to +4 to +6 in $\mathrm{XeF}_2, \mathrm{XeF}_4$ and $\mathrm{XeF}_6$ respectively. Thus, its tendency to accept electrons also increases and thus, its oxidising power increases.

Also, $\mathrm{XeF}_6$ can give more fluorine than $\mathrm{XeF}_4$ and $\mathrm{XeF}_2$. So, the fluorinating agent order is $\mathrm{XeF}_6>\mathrm{XeF}_4>\mathrm{XeF}_2$. Hence, the correct decreasing order of the given ' $\mathrm{Xe}^{\prime}$ compound to act as both fluorinating and oxidising agent is $\mathrm{XeF}_6>\mathrm{XeF}_4>\mathrm{XeF}_2$.

When aluminium chloride is dissolved in excess of water, hydrated aluminium ion and chloride ions are formed.

$ \mathrm{AlCl}_3+\underset{\text { (Excess) }}{6 \mathrm{H}_2 \mathrm{O}} \longrightarrow\left[\mathrm{Al}\left(\mathrm{H}_2 \mathrm{O}\right)_6\right]^{3+} $

$ (a q)+3 \mathrm{Cl}^{-}(a q) $

(A) The $\mathrm{N}_2 \mathrm{O}$ produced by heating $\mathrm{NH}_4 \mathrm{NO}_3$ is a gas.

$ \mathrm{NH}_4 \mathrm{NO}_3 \xrightarrow{\text { Heat }} \mathrm{N}_2 \mathrm{O}+2 \mathrm{H}_2 \mathrm{O} $

It is diamagnetic as it does not have unpaired electron.

(B) No gaseous product is formed $2 \mathrm{NaNO}_2+\mathrm{H}_2 \mathrm{SO}_4 \longrightarrow 2 \mathrm{HNO}_2+\mathrm{Na}_2 \mathrm{SO}_4$

$ \begin{aligned} &\text { }\\ &\begin{aligned}(C)\,\, 2 \mathrm{~Pb}\left(\mathrm{NO}_3\right)_2 \xrightarrow{673 \mathrm{~K}} & 4 \mathrm{NO}_2 +2 \mathrm{PbO}+\mathrm{O}_2 \end{aligned} \end{aligned} $

Nitric oxide and oxygen are gases produced in this reaction.

$\mathrm{O}_2$ is paramagnetic because it has two unpaired electrons.

$\mathrm{NO}_2$ is paramagnetic as it has unpaired electron at nitrogen.

$2 \mathrm{Se}_2 \mathrm{Cl}_2 \longrightarrow \mathrm{SeCl}_4+3 \mathrm{Se}$

In a disproportionation reaction, the element is oxidised and reduced simultaneously.

The reactions in which a more reactive element displaces another element from its compounds are called displacement reactions.

$ \begin{aligned} & 2 \mathrm{Al}(s)+2 \mathrm{NaOH}(l)+6 \mathrm{H}_2 \mathrm{O}(l) \longrightarrow 2 \mathrm{Na}\left[\mathrm{Al}(\mathrm{OH})_4\right](s)+3 \mathrm{H}_2(g) \end{aligned} $

The reaction of aluminium with sodium hydroxide gives a salt and hydrogen gas. In this reaction, aluminium displaces sodium from its compound, therefore, it is a displacement reaction.

The electronegativity decreases down the group. $\mathrm{C}, \mathrm{Ge}$ and Pb all belong to 14th group. So, the order should be C $>\mathrm{Ge}>\mathrm{Pb}$. But the actual order is $\mathrm{C}> \mathrm{Pb}>\mathrm{Ge}$. This can be explained on the basis of shielding effect.

The valence shell electronic configuration of Ge is $[\mathrm{Ar}] 3 d^{10} 4 s^2 4 p^2$ and that of Pb is $[\mathrm{Xe}] 6 s^2 4 f^{14} 5 d^{10} 6 p^2$. Due to the poor shielding effect of intervening ' $f$ ' electrons in Pb , the $Z_{\text {eff }}$ gets highly increased in Pb . Therefore, the electronegativity of Pb is higher than Ge .

Reaction of ammonia with nitrous acid results in production of $\mathrm{N}_2$ gas.

Thermal decomposition of ammonium dichromate results into $\mathrm{N}_2$ production.

$ \left(\mathrm{NH}_4\right)_2 \mathrm{Cr}_2 \mathrm{O}_7 \longrightarrow \mathrm{Cr}_2 \mathrm{O}_3+4 \mathrm{H}_2 \mathrm{O}+\mathrm{N}_2 $

Treating aqueous solution of ammonium salt with nitrite results into $\mathrm{N}_2$ production.

$ \mathrm{NH}_4^{+}+\mathrm{NO}_2^{-} \longrightarrow \mathrm{N}_2+2 \mathrm{H}_2 \mathrm{O} $

Reaction of ammonia with sodiumhypochlorite is a oxidation reduction reaction in which NaOCl acts as oxidising agent and $\mathrm{NH}_3$ acts as reducing agent. It does not results into $\mathrm{N}_2$ production.

$ 2 \mathrm{NH}_3+\mathrm{NaOCl} \longrightarrow \mathrm{NaCl}+\mathrm{N}_2 \mathrm{H}_4+\mathrm{H}_2 \mathrm{O} $

Gradual increase in boiling points down the group happens as the relative molecular mass of the molecules increases. But the water has anomalously highest boiling point due to strong intermolecular hydrogen bonding in water molecules. Therefore, the order is $\mathrm{H}_2 \mathrm{O}>\mathrm{H}_2 \mathrm{Te}>\mathrm{H}_2 \mathrm{Se}>\mathrm{H}_2 \mathrm{~S}$.

Carnallite is not the mineral of fluorine, whereas all others are mineral of fluorine. Carnallite (also carnalite) is an evaporite mineral, a hydrated potassium magnesium chloride with formula $\mathrm{KMgCl}_3 \cdot 6\left(\mathrm{H}_2 \mathrm{O}\right)$.

The element that can even diffuse through silica glass is He due to very small size.

In the given reaction :

$ \begin{aligned} & (X)=\mathrm{HCl} \\ & (Y)=\mathrm{Mg} \end{aligned} $

The reaction occurs as follows :

$ \begin{aligned} \text { Borax } \xrightarrow[\mathrm{H}_2 \mathrm{O}]{\mathrm{HCl}(X)} \mathrm{H}_3 \mathrm{BO}_3 \xrightarrow{\Delta} & \mathrm{BrO}_3 \xrightarrow[\Delta]{\mathrm{Mg}(Y)} \text { (B) Product } \end{aligned} $

$ \left\{\mathrm{Na}_2\left[\mathrm{~B}_4 \mathrm{O}_5(\mathrm{HO})_4\right] \times 8 \mathrm{H}_2 \mathrm{O}\right\} $

$ \begin{aligned}(i)\,\, & \mathrm{Na}_2 \mathrm{~B}_4 \mathrm{O}_7 \cdot 10 \mathrm{H}_2 \mathrm{O}+2 \mathrm{HCl} \\ & \quad \text { Borax } \\ & \longrightarrow 4 \mathrm{~B}(\mathrm{OH})_3+2 \mathrm{NaCl}+5 \mathrm{H}_2 \mathrm{O} \end{aligned} $

$ \begin{aligned}(ii) \mathrm{H}_3 \mathrm{BO}_3+\mathrm{Mg} \xrightarrow{\Delta} & \mathrm{Mg}\left(\mathrm{BO}_2\right)_2 +2 \mathrm{H}_2 \mathrm{O}+\mathrm{H}_2 \end{aligned} $

Hence, option (a) is the correct answer.