p-Block Elements

The reaction in each case is,

$ \begin{aligned} 2 \mathrm{NaBH}_4+\mathrm{I}_2 \longrightarrow & \mathrm{~B}_2 \mathrm{H}_6 +2 \mathrm{NaI}+\mathrm{H}_2 \uparrow \end{aligned} $

II. $\mathrm{B}_2 \mathrm{H}_6+3 \mathrm{O}_2 \longrightarrow \mathrm{~B}_2 \mathrm{O}_3+3 \mathrm{H}_2 \mathrm{O}$

III. $\mathrm{BF}_3+\mathrm{NaH} \longrightarrow \mathrm{B}_2 \mathrm{H}_6+6 \mathrm{NaF}$

IV. $\mathrm{B}_2 \mathrm{H}_6+6 \mathrm{H}_2 \mathrm{O} \longrightarrow 2 \mathrm{~B}(\mathrm{OH})_3$$ +6 \mathrm{H}_2 \uparrow $

Hence, only in I and IV, $\mathrm{H}_2$ is released.

Statement given in option (d) is incorrect regarding the gas evolved of reaction. The correct form is,

$\mathrm{CO}_2$ is not poisonous gas.

$ \mathrm{CaCO}_3+2 \mathrm{HCl} \rightarrow \mathrm{CaCl}_2+\mathrm{H}_2 \mathrm{O}+\mathrm{CO}_2 \uparrow $

$\mathrm{NaNO}_2$ gives more number of oxides on reacting with HCl .

Among the given statements, option (c) is incorrect. It's correct form is $\mathrm{SeO}_2$ is colourless solid at room temperature.

Step 1: Reaction with Lithium Aluminium Hydride

When boron trifluoride ($\mathrm{BF}_3$) reacts with lithium aluminium hydride ($\mathrm{LiAlH}_4$) in ether, it forms lithium fluoride ($\mathrm{LiF}$), aluminium fluoride ($\mathrm{AlF}_3$), and diborane ($\mathrm{B}_2\mathrm{H}_6$):

$ 4 \mathrm{BF}_3+3 \mathrm{LiAlH}_4 \xrightarrow{\text { Ether }} 3 \mathrm{LiF}+3 \mathrm{AlF}_3 + \mathrm{B}_2 \mathrm{H}_6 $

Step 2: Naming the Product X

So, here $X = \mathrm{B}_2\mathrm{H}_6$ (which is called diborane).

Step 3: Reaction of X with Ammonia

Diborane ($\mathrm{B}_2\mathrm{H}_6$) reacts with ammonia ($\mathrm{NH}_3$) to first form an adduct, which is a combined compound: $\left[\mathrm{B}_2\mathrm{H}_6 \cdot 2\mathrm{NH}_3\right]$.

Step 4: Heating to Get Z

When this adduct is heated, it breaks down to give borazine ($\mathrm{B}_3\mathrm{N}_3\mathrm{H}_6$), which is compound $Z$, and hydrogen gas ($\mathrm{H}_2$):

$ 2~\mathrm{B}_2 \mathrm{H}_6 + 2~\mathrm{NH}_3 \longrightarrow \left[\mathrm{B}_2 \mathrm{H}_6 \cdot 2~\mathrm{NH}_3\right] \longrightarrow 2~\mathrm{B}_3 \mathrm{N}_3 \mathrm{H}_6 (\mathrm{Z}) + \mathrm{H}_2 $

Step 5: Counting Sigma ($\sigma$) and Pi ($\pi$) Bonds in Z

Borazine ($\mathrm{B}_3\mathrm{N}_3\mathrm{H}_6$) is a ring, and its bonds are:

There are 6 sigma ($\sigma$) bonds between B and N in the ring.
There are 3 sigma ($\sigma$) bonds between B and H.
There are 3 sigma ($\sigma$) bonds between N and H.

So, the total number of $\sigma$ bonds ($x$) is: $6 + 3 + 3 = 12$.

There are 3 pi ($\pi$) bonds ($y$) in the ring.

So, $x + y = 12 + 3 = 15$.

The reactions involved are as follows.

$\mathrm{PCl}_5(B)$ in solid state exists as $\left[\mathrm{PCl}_4^{+}\right]\left[\mathrm{PCl}_6^{-}\right]$as ionic lattice. $\mathrm{PCl}_5$ on hydrolysis gives phosphoric acid.

$ \mathrm{PCl}_5+4 \mathrm{H}_2 \mathrm{O} \longrightarrow \underset{\text { (Phosphoric acid) }}{\mathrm{H}_3 \mathrm{PO}_4}+5 \mathrm{HCl} $

In $\mathrm{PCl}_5$, the all bonds are not equivalent.

Trigonal bipyramidal

Paining of electrons takes place.

Thus, option (c) is correct.

The hydrolysis of diborane $\left(\mathrm{B}_2 \mathrm{H}_6\right)$ gives boric acid $\left(\mathrm{H}_3 \mathrm{BO}_3\right)$ as the compound $X$. The reaction is

$ \mathrm{B}_2 \mathrm{H}_6+6 \mathrm{H}_2 \mathrm{O} \rightarrow 2 \mathrm{H}_3 \mathrm{BO}_3+6 \mathrm{H}_2 $

The properties of boric acid are as follows.

• Acidity : Boric acid is a weak monobasic Lewis acid. It does not donate protons but rather accepts a hydroxide ion from water, making it a monobasic acid. Therefore,

statement II is correct, and statement I (tribasic acid) is incorrect.

• Structure : In the solid state, boric acid has a layered structure, where planar $\mathrm{BO}_3$ units are linked by hydrogen bonds. This structure gives it a slippery feel. Therefore, statement III is correct.

• Solubility : Boric acid has low solubility in cold water, although its solubility increases with temperature. It is not considered highly soluble. Therefore, statement IV is incorrect.

Based on the analysis, the correct statements about compound $X$ are II and III.

Statement I : Graphite's structure is well-known to be layered, with sheets of hexagonal rings of carbon atoms stacked on top of one another. Thus, this statement is correct.

• Statement II : Buckminster fullerene ( $\mathrm{C}_{60}$ ) has a conjugated $\pi$-electron system spread over its spherical surface. This delocalisation of electrons gives it aromatic properties, making it an aromatic compound. Therefore, this statement is incorrect.

• Statement III The distance between the layers in graphite is approximately 335 pm . The value 141.5 pm represents the carbon-carbon bond length within a single layer, not the interlayer distance. Thus, this statement is incorrect.

• Statement IV In both graphite and buckminster fullerene, each carbon atom is bonded to three other carbon atoms, resulting in $s p^2$ hybridisation. The unhybridised $p$-orbitals form delocalised $\pi$-systems. Thus, this statement is correct.

The correct statements are I and IV.

Among the given orders, order given in options (b) i.e.

$ \mathrm{N}_2 \mathrm{O}<\mathrm{NO}<\mathrm{N}_2 \mathrm{O}_3<\mathrm{N}_2 \mathrm{O}_4<\mathrm{N}_2 \mathrm{O}_5 \text { (a } $

(Acidic Nature) is correct.

The correct order for remaining options are

(a) $\mathrm{H}_2 \mathrm{~S}<\mathrm{H}_2 \mathrm{Se}<\mathrm{H}_2 \mathrm{Te}<\mathrm{H}_2 \mathrm{O}$

(Boiling point)

(c) $\mathrm{HF}<\mathrm{HCl}<\mathrm{HBr}<\mathrm{HI}$

(Acidic Nature)

(d) $\mathrm{H}_2 \mathrm{Te}<\mathrm{H}_2 \mathrm{Se}<\mathrm{H}_2 \mathrm{~S}<\mathrm{H}_2 \mathrm{O}$

(Bond angle)

Helium ( $X$ ) is used as a diluent for oxygen in modern dividing apparatus and argon $(Y)$ used to provide an inert atmosphere in high temperature

metallurgical process. Thus $Y$ and $X$ respectively are $\mathrm{Ar}, \mathrm{He}$.

Orthophosphorus acid (or phosphorous acid) on heating disproportionates to orthophosphoric acid (or phosphoric acid) and phosphine. The reaction involved is as follows.

$ 4 \mathrm{H}_3 \mathrm{PO}_3 \rightarrow \underset{(A)}{3 \mathrm{H}_3} \mathrm{PO}_4+\underset{(B)}{\mathrm{PH}_3} $

Thus, $A$ and $B$ are $\mathrm{H}_3 \mathrm{PO}_4$ and $\mathrm{PH}_3$ respectively.

Among the given options, HCl and $\mathrm{H}_2 \mathrm{~S}$ are electron rich hydrides. Electron rich hydrides are those compounds where the central atom has more electron than required to form single covalent bonds with its surrounding atoms.

Among the given statements, statement given in option (c) is incorrect about compounds of group 13 elements. It's correct form is, Diborane $\left(\mathrm{B}_2 \mathrm{H}_6\right)$ is an example of electron deficient hydride.

Among the given statements, statement given in option (c) is incorrect about oxidation state of group 14 elements, it's correct form is lead in +2 state is not a good reducing agent. This is because it readily accepts two more electrons to achieve the stable +4 oxidation state.

Statment given in I, II and IV are correct, while III is incorrect. Its correct form is given in statement II.

The reaction involved are as follows.

Option A: Chlorine oxidises ferrous salts to ferric salts in acidic medium.

• Ferrous (Fe²⁺) has an oxidation state of +2. Ferric (Fe³⁺) has an oxidation state of +3.

• Chlorine (Cl₂) is a strong oxidizing agent.

• In acidic medium, chlorine readily oxidizes Fe²⁺ to Fe³⁺:

  2Fe²⁺ + Cl₂ → 2Fe³⁺ + 2Cl⁻

• This statement is correct.

Option B: Chlorine oxidises iodine to periodic acid in water.

• Iodine (I₂) has an oxidation state of 0.

• Periodic acid (HIO₄) has iodine in a +7 oxidation state.

• When chlorine reacts with iodine in the presence of water, it typically oxidizes iodine to iodic acid (HIO₃), where iodine is in a +5 oxidation state:

  I₂ + 5Cl₂ + 6H₂O → 2HIO₃ + 10HCl

• To obtain periodic acid (HIO₄), a much stronger oxidizing agent than chlorine (like hot concentrated nitric acid, or by electrolysis, or certain peroxo compounds) is generally required, or specific conditions like strongly alkaline medium with hypochlorite. Simple "chlorine in water" does not usually produce periodic acid from iodine.

• Therefore, this statement is incorrect.

Option C: Chlorine acts as a bleaching agent due to oxidation.

• When chlorine dissolves in water, it forms hypochlorous acid (HOCl):

  Cl₂ + H₂O ⇌ HCl + HOCl

• Hypochlorous acid is a powerful oxidizing agent. It decomposes to release nascent oxygen ([O]), which oxidizes colored substances into colorless ones, thereby achieving bleaching.

  HOCl → HCl + [O]

  Colored substance + [O] → Colorless substance

• Thus, the bleaching action of chlorine is indeed due to oxidation.

• This statement is correct.

Option D: Chlorine is manufactured by Deacon's process.

• Deacon's process is an industrial method for the production of chlorine. It involves the catalytic oxidation of hydrogen chloride gas with atmospheric oxygen, usually using copper(II) chloride as a catalyst at high temperatures:

  4HCl(g) + O₂(g) ⇌ 2Cl₂(g) + 2H₂O(g)

• While electrolysis of brine is the predominant modern method, Deacon's process is a known and historically significant method for chlorine manufacturing.

• This statement is correct.

Based on the analysis, Option B is the incorrect statement.

The final answer is $\boxed{B}$

Both the Assertion and the Reason are correct statements, but the Reason does not explain the Assertion properly.

Phosphorus can make both phosphorus(III) chloride and phosphorus(V) chloride because phosphorus has empty $d$-orbitals in its outer shell. This lets phosphorus use these orbitals to form more than four bonds, so it can form compounds like $\mathrm{PCl}_5$ (phosphorus(V) chloride).

Nitrogen cannot form nitrogen(V) chloride ($\mathrm{NCl}_5$) because nitrogen does not have $d$-orbitals in its valence shell. It can only make up to four bonds, so it cannot expand its octet to form five bonds like phosphorus can.

Therefore, the main reason why phosphorus can form $\mathrm{PCl}_5$ but nitrogen cannot form $\mathrm{NCl}_5$ is the availability of $d$-orbitals in phosphorus, not the difference in electronegativity.

Statement given in I and II are correct regarding boron, while statement III and IV are incorrect.

Their correct forms are,

III. in diborane oxidation state of hydrogen is -1 .

IV. boric acid is not a tribasic acid, it is monobasic acid.

The complete reaction is as follows,

$ \underset{\substack{(X) \\ \text { Hypophosphorus acid }}}{\mathrm{H}_3 \mathrm{PO}_2(a q)}+2 \mathrm{AgNO}_3(a q) \xrightarrow{\mathrm{H}_2 \mathrm{O}} \underset{(Y)}{\mathrm{H}_3 \mathrm{PO}_4}+2 \mathrm{Ag}+2 \mathrm{HNO}_3 $

The reaction is,

$ \mathrm{Zn}+\underset{\text { (conc.) }}{4 \mathrm{HNO}_3} \longrightarrow \mathrm{Zn}\left(\mathrm{NO}_3\right)_2+\underset{(A)}{2 \mathrm{NO}_2}+2 \mathrm{H}_2 \mathrm{O} $

$ 4 \mathrm{Zn}+\underset{\text { (dil.) }}{10 \mathrm{HNO}_3} \longrightarrow 4 \mathrm{Zn}\left(\mathrm{NO}_3\right)_2+\underset{(B)}{\mathrm{N}_2 \mathrm{O}}+5 \mathrm{H}_2 \mathrm{O} $

Oxidation state of nitrogen,

(A) $\mathrm{NO}_2=+4$

(B) $\mathrm{N}_2 \mathrm{O}=+1$

Deacon's process is a chemical method use to produce chlorine gas from hydrochloric acid.

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

The complete reaction is as follows,

Complete hydrolysis, Thus, $X=\mathrm{PCl}_5, Y=\mathrm{H}_3 \mathrm{PO}_4$.

$ \text {The reaction involved is, } $

$ \mathrm{XeF}_6+\mathrm{H}_2 \mathrm{O} \longrightarrow \underset{(\text { Product } X)}{\mathrm{XeOF}_4}+2 \mathrm{HF} $

Thus the shape of $(X) \mathrm{XeF}_4$ is square pyramidal.

$\mathrm{H}_3 \mathrm{PO}_3$ and $\mathrm{H}_2 \mathrm{SO}_4$ both have basicity of $2 . \mathrm{H}_3 \mathrm{PO}_3$ has two $\mathrm{P}-\mathrm{OH}$ bond meaning it can donate two $\left[\mathrm{OH}^{-}\right]$ions when in solution.

$\mathrm{H}_2 \mathrm{SO}_4$ also have two ionisation H -atom attached to oxygen, giving it a basicity of 2 .

The correct order of acidic strength is,

$ \mathrm{Al}_{2} \mathrm{O}_{3} < \mathrm{SiO}_{2} < \mathrm{P}_{2} \mathrm{O}_{3} < \mathrm{SO}_{2} $

As electronegativity difference between element and oxygen decreases the acidic character of oxides increases. The electronegativity also increases with increasing oxidation number. In general, as non-metallic character increases across the period, the acidic character of their oxides increases.

Borax, scientifically referred to as sodium tetraborate decahydrate, is commonly used as a household cleaner and a booster for laundry detergent.

Chemical Formula

The chemical formula for borax is:

$ \mathrm{Na}_{2}\left[\mathrm{B}_{4} \mathrm{O}_{5}(\mathrm{OH})_{4}\right] \cdot 8 \mathrm{H}_{2} \mathrm{O} $

Here, the values are:

$ x = 4 $ (the number of hydroxyl groups)

$ y = 8 $ (the number of water molecules)

Adding these values gives:

$ x + y = 4 + 8 = 12 $

$ \mathrm{H}_{4} \mathrm{P}_{2} \mathrm{O}_{5}, \mathrm{H}_{4} \mathrm{P}_{2} \mathrm{O}_{7} $ and $ \left(\mathrm{HPO}_{3}\right)_{3} $

oxoacids of phosphorous contain $ \mathrm{P}-\mathrm{O}-\mathrm{P} $ bonds.

Diborane undergoes cleavage reaction with Lewis base (CO) to give borane adduct, $ \mathrm{BH}_{3} \cdot \mathrm{CO} $ which is compound $ X $. When alkyl metal hyaride

( NaH ) react with diborane in presence of ether, a tetrahedral compound $ \left(\mathrm{NaBH}_{4}\right) $ is formed.

The complete reaction is as follow

$ \begin{aligned} & 2 \mathrm{CO}+\mathrm{B}_2 \mathrm{H}_6 \longrightarrow \underset{\substack{\text { Compound } X}}{2 \mathrm{BH}_3 \mathrm{CO}} \\\\ & 2 \mathrm{NaH}+\mathrm{B}_2 \mathrm{H}_6 \xrightarrow{\text { Ether }} \underset{\substack{\text { Compound } Y}}{2 \mathrm{NaBH}_4} \end{aligned} $

(ii) Diamond has directional covalent bonds.

Diamonds are structured with each carbon atom forming strong covalent bonds with four other carbon atoms, creating a tetrahedral lattice. This arrangement results in directional covalent bonding, contributing to diamond's extreme hardness and high melting point.

(iv) Glass is a man-made silicate.

Glass is an amorphous solid primarily composed of silica ($\text{SiO}_2$) and is synthesized through the high-temperature melting and cooling of sand and other materials. It fits into the category of silicates because it includes silicon and oxygen.

Option B: ii, iv only

The complete reaction is as follows

$ \begin{array}{c} \mathrm{P}_{2} \mathrm{O}_{3}+3 \mathrm{H}_{2} \mathrm{O} \longrightarrow \underset{\text { (Product } X)}{2 \mathrm{H}_{3} \mathrm{PO}_{3}} \\\\ \mathrm{P}_{4} \mathrm{O}_{10}+6 \mathrm{H}_{2} \mathrm{O} \longrightarrow \underset{\text { (Product } Y)}{4 \mathrm{H}_{3} \mathrm{PO}_{4}} \end{array} $

Product $ X $ and $ Y $ both consist of one $ \mathrm{P}=\mathrm{O} $ bond each.

Carbon reaction with hot and conc. $ \mathrm{H}_{2} \mathrm{SO}_{4} $ gives,

$ \mathrm{C}+2 \mathrm{H}_{2} \mathrm{SO}_{4} \longrightarrow \mathrm{CO}_{2}+2 \mathrm{SO}_{2}+2 \mathrm{H}_{2} \mathrm{O} $

Both the oxides, $ \mathrm{CO}_{2} $ and $ \mathrm{SO}_{2} $ are acidic in nature.

Among the given species there are total 4 Lewis acids, these are $ \mathrm{AlCl}_{3} $, $ \mathrm{SnCl}_{4}, \mathrm{CO}_{2}, \mathrm{Ag}^{+} $.

$ \mathrm{HSO}_{4}^{-} $ is amphoteric compound.

$ \mathrm{NH}_{3} $ is Lewis base.

When an organic compound containing phosphorus is oxidised with $ \mathrm{Na}_{2} \mathrm{O}_{2} $ it gives $ \mathrm{Na}_{3} \mathrm{PO}_{4} $.

$ \mathrm{Na}_{3} \mathrm{PO}_{4} $ when treated with $ \mathrm{HNO}_{3} $ will form phosphoric acid $ \left(\mathrm{H}_{3} \mathrm{PO}_{4}\right) $. When $ \mathrm{H}_{3} \mathrm{PO}_{4} $ is treated with ammonium molybdate in presence of nitric acid, it forms a yellow ppt. The reaction is as follow

$\begin{gathered}\underset{(C \text { Compound } X)}{\mathrm{Na}_3 \mathrm{PO}_4}+3 \mathrm{HNO}_3 \longrightarrow \mathrm{H}_3 \mathrm{PO}_4+3 \mathrm{NaNO}_2 \\\\ \mathrm{H}_3 \mathrm{PO}_4+12\left(\mathrm{NH}_4\right)_2 \mathrm{MoO}_4+21 \mathrm{HNO}_3 \longrightarrow \\\\ 21 \mathrm{NH}_4 \mathrm{NO}_3+\left(\mathrm{NH}_4\right)_3 \mathrm{PO}_4 \cdot 12 \mathrm{MoO}_3 \downarrow \\\\ \text { Yellow ppt } \\\\ \text { (Product } Y \text { ) }\end{gathered}$

$\mathrm{XeF}_6$ on complete hydrolysis gives

$\mathrm{XeO}_3$.

$ \begin{aligned} \mathrm{XeF}_6+\mathrm{H}_2 \mathrm{O} & \longrightarrow \mathrm{XeOF}_4+2 \mathrm{HF} \\\\ \mathrm{XeOF}_4+\mathrm{H}_2 \mathrm{O} & \longrightarrow \mathrm{XeO}_2 \mathrm{~F}_2+2 \mathrm{HF} \\\\ \mathrm{XeO}_2 \mathrm{~F}_2+\mathrm{H}_2 \mathrm{O} & \longrightarrow \mathrm{XeO}_3+2 \mathrm{HF} \end{aligned} $

So, the compound formed on complete hydrolysis of xenon is $\mathrm{XeO}_3$.

$\mathrm{XeO}_3$ has three double bond and each double bond has $1 \sigma$ and $1 \pi$ bond.

Total $\sigma+\pi=3+3=6$

The complete balanced reaction is as follows,

$ \begin{aligned} & 2 \mathrm{BF}_3+6 \mathrm{NaH} \xrightarrow{450 \mathrm{~K}} \underset{\substack{\text { Diborane } \\\\ \text { (Product } X \text { ) }}}{\mathrm{B}_2 \mathrm{H}_6}+6 \mathrm{NaF} \\\\ & \mathrm{~B}_2 \mathrm{H}_6+6 \mathrm{H}_2 \mathrm{O} \longrightarrow \underset{\substack{\text { Boric acid } \\\\ \text { (Product } Y \text { ) }}}{2 \mathrm{H}_3 \mathrm{BO}_3}+6 \mathrm{H}_2 \uparrow \end{aligned} $

Hence, $X$ is electron deficient molecule while $Y$ is an Lewis acid.

$\mathrm{In}\left[\mathrm{SiCl}_6\right]^{2-}$, the chlorine atoms are much larger which does not allow the six chlorine to surround the silicon due to repulsion between chlorine atom. Hence, due to interelectron repulsion between chlorine atom the compound $\left[\mathrm{SiCl}_6\right]^{2-}$ is unstable and does not exists.

The complete balance reaction is as follows,

$ \underset{\substack{\text { Molecule } \\\\ X}}{2 \mathrm{Cu}_2 \mathrm{~S}}+3 \mathrm{O}_2 \longrightarrow 2 \mathrm{Cu}_2 \mathrm{O}+\mathrm{SO}_2 $

$ 2 \mathrm{Cu}_2 \mathrm{O}+\underset{\left(\mathrm{M}_{\text {olecule } X)}\right.}{\mathrm{Cu}_2 \mathrm{~S}} \xrightarrow{\Delta} 6 \mathrm{Cu}+\underset{\text { (Product } Y)}{\mathrm{SO}_2(g)} $

The product $Y$ is $\mathrm{SO}_2$ which has bent or angular geometry.

In contact process of manufacture of $\mathrm{H}_2 \mathrm{SO}_4$, the arsenic purifier used in industrial plant consist of $\mathrm{Fe}_2 \mathrm{O}_3 \cdot x \mathrm{H}_2 \mathrm{O}$. It is used to remove arsenic impurities formed in contact process.

Among the given oxides, only three oxides namely $\mathrm{B}_2 \mathrm{O}_3, \mathrm{SiO}_2$ and $\mathrm{CO}_3$ are acidic oxides.

$ \begin{aligned} & \mathrm{B}_2 \mathrm{O}_3+2 \mathrm{NaOH} \longrightarrow \mathrm{NaBO}_3+\mathrm{H}_2 \mathrm{O} \\ & \mathrm{SiO}_2+2 \mathrm{NaOH} \longrightarrow \mathrm{Na}_2 \mathrm{SiO}_3+\mathrm{H}_2 \mathrm{O} \\ & \mathrm{CO}_2+2 \mathrm{NaOH} \longrightarrow \mathrm{Na}_2 \mathrm{CO}_3+\mathrm{H}_2 \mathrm{O} \end{aligned} $

while $\mathrm{Tl}_2 \mathrm{O}_3$ is basic oxide.

CO is neutral oxide and $\mathrm{Al}_2 \mathrm{O}_3$ and $\mathrm{PbO}_2$ are amphoteric oxides.

Ammonium dichromate and barium azide when subjected to thermal decomposition will liberate dinitrogen.

$ \begin{gathered} \left(\mathrm{NH}_4\right)_2 \mathrm{Cr}_2 \mathrm{O}_7 \xrightarrow{\Delta} \mathrm{Cr}_2 \mathrm{O}_3+\mathrm{N}_2 \uparrow+4 \mathrm{H}_2 \mathrm{O} \uparrow \\ \mathrm{Ba}\left(\mathrm{~N}_3\right)_2 \longrightarrow \mathrm{Ba}+3 \mathrm{~N}_2 \uparrow \end{gathered} $

Zeolite is silicate containing two metal ions $\mathrm{Na}^{+}$and $\mathrm{Al}^{3+}$. Therefore, $X$ and $Y$, respectively are $\mathrm{Na}^{+}$and $\mathrm{Al}^{3+}$.

The order of melting point is

$ \mathrm{Si}>\mathrm{Ge}>\mathrm{Pb}>\mathrm{Sn} $

The lead $(\mathrm{Pb})$ has high melting point than tin ( Sn ) because Sn forms a distorted 12 -coordinated structure. Also, Sn atoms are bonded covalently, which are weak in nature. So, Sn has a low melting point.

In hydrides of group 16 elements, boiling point increases down the group because of increasing molecular mass but water ( $\mathrm{H}_2 \mathrm{O}$ ) has exceptionally high boiling point due to intermolecular hydrogen bonding and $\mathrm{H}_2 \mathrm{~S}$ has lowest boiling point due to least molecular mass. Therefore, $X$ is $\mathrm{H}_2 \mathrm{~S}$ and $Y$ is $\mathrm{H}_2 \mathrm{O}$.

Sodium nitrite with hydrochloric acid given two oxides of nitrogen i.e. NO and $\mathrm{NO}_2$ alongwith water and sodium chloride.

The reaction involved is as follows.

Iodine monochloride ( ICl ) reacts with pure water to from HCl and iodic acid ( HOI ).

The reaction involved is a follows

$ \mathrm{ICl}+\mathrm{H}_2 \mathrm{O} \longrightarrow \mathrm{HOI}+\mathrm{HCl} $

Therefore, statement (c) is incorrect.

Aluminium reacts with dilute hydrochloric acid at room temperature. The metal dissolve in HCl , yielding aluminium chloride and colourless hydrogen gas.

$ 2 \mathrm{Al}(s)+6 \mathrm{HCl}(a q) \longrightarrow 2 \mathrm{AlCl}_3(a q)+3 \mathrm{H}_2(g) $

When aluminium reacts with aqueous alkali it produces hydrogen gas.

$ \begin{aligned} \mathrm{Al}(s)+\mathrm{NaOH}(a q) & +\mathrm{H}_2 \mathrm{O}(a q) \longrightarrow \mathrm{NaAlO}_2(a q)+\mathrm{H}_2(g) \end{aligned} $

Statement given in option (c) is incorrect.

Due to inert pair effect, the lower elements in group 14 are stablised in $(+2)$ oxidation state than $(+4)$.

Hence, $\mathrm{Pb}^{2+}$ is more stable than $\mathrm{Pb}^{4+}$.

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

Hence, $\mathrm{Pb}^{4+}$ is good oxidising agent. Since, it is unstable.

Statement given in option (b) is incorrect, because 4 is hypophosphorus acid.

$\mathrm{H}_3 \mathrm{PO}_2$ is called hypophosphorus acid, conical has only one ionisable $(\mathrm{H})$ ions, hence its basicity is 1 and is a reducing agent.

$ \begin{aligned} \mathrm{H}_3 \mathrm{PO}_2 & \\ & 3 \times 1+x+2(-2)=0 \\ \Rightarrow & 3+x-4=0 \end{aligned} $

or $[x=1] \rightarrow$ oxidation state