p-Block Elements

(1) Fluorspar is $\mathrm{CaF}_2$

Statement-I: Oxygen can have oxidation state from -2 to +2 , so statement I is incorrect

Statement- II: On moving down the group stability of +4 oxidation state increases whereas stability of +6 oxidation state decreases down the group, according to inert pair effect.

So both statements are wrong.

Fluorine is the element among the options that does not exhibit a variable oxidation state, unlike its fellow halogens. This is because fluorine is the most electronegative element, and it always forms compounds in the oxidation state of -1 as it has a strong tendency to gain one electron to achieve a stable electronic configuration.

Solid Boron has very strong crystalline lattice so its melting point unusually high in group 13 elements

When $ \text{PCl}_5 $ is fluorinated in a polar organic solvent, it forms ionic isomers. The formation of these ionic species can be understood as follows:

1. $ \text{PCl}_5 $ in a polar organic solvent can dissociate into ionic forms, such as $ \left[\text{PCl}_4\right]^+ $ and $ \left[\text{PCl}_6\right]^- $.


2. Upon fluorination, $ \text{PCl}_5 $ can be converted into $ \left[\text{PF}_6\right]^- $ and $ \left[\text{PCl}_4\right]^+ $.


3. Further fluorination can lead to the formation of mixed ionic compounds where $ \text{PCl}_5 $ can form isomers like $ \left[\text{PCl}_4\right]^+ \left[\text{PCl}_4 \text{F}_2\right]^- $.

The two ionic isomers formed are:

1. $ \left[\text{PCl}_4\right]^+ \left[\text{PCl}_4 \text{F}_2\right]^- $, which are colorless crystals.


2. $ \left[\text{PCl}_4\right]^+ \left[\text{PF}_6\right]^- $, which are white crystals.

Hence, the ionic isomers formed when $ \text{PCl}_5 $ is fluorinated in a polar organic solvent are :

To determine the products of the disproportionation of nitrous acid ($ \mathrm{HNO}_2 $) at room temperature, let's analyze the chemical behavior of $ \mathrm{HNO}_2 $ in aqueous solution.

Disproportionation is a redox reaction in which a single substance is simultaneously oxidized and reduced, yielding two different products. In the case of nitrous acid ($ \mathrm{HNO}_2 $), the reaction can be represented as follows:

$ 3 \mathrm{HNO}_2 \rightarrow \mathrm{HNO}_3 + \mathrm{H_2O} + 2 \mathrm{NO} $

Breaking this down, we see that nitrous acid disproportionates into nitric acid ($ \mathrm{HNO}_3 $), water ($ \mathrm{H_2O} $), and nitrogen monoxide ($ \mathrm{NO} $). Additionally, in an aqueous solution, nitric acid dissociates to form $ \mathrm{H_3O}^+ $ (hydronium ion) and $ \mathrm{NO_3}^- $ (nitrate ion).

Therefore, the species formed from this disproportionation reaction are $ \mathrm{H_3O}^+ $, $ \mathrm{NO_3}^- $, and $ \mathrm{NO} $.

From the given options, the correct answer is:

Option A

$ \mathrm{H_3O}^+ $, $ \mathrm{NO_3}^- $, and $ \mathrm{NO} $

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} $

The correct pair is (ii) and (iii) only.

• Boron fibres are used in high-stiffness, high-strength composites (e.g. aerospace), not in bullet-proof vests.

• Metal borides (e.g. TiB₂, ZrB₂) are extremely hard and find use in protective shields/armor.

• Borax (Na₂B₄O₇·10H₂O) provides B₂O₃ in glass formulations and is a key ingredient in glass wool.

Answer: Option D (ii, iii only).

$\mathrm{XeF}_4$ violently reacts with water and gives following.

$ \begin{array}{r} 6 \mathrm{XeF}_4+12 \mathrm{H}_2 \mathrm{O} \longrightarrow 4 \mathrm{Xe}+2 \mathrm{XeO}_3 \\ \,\,\,\,\text { Produc } x \\ +24 \mathrm{HF}+30 \end{array} $

Statement I is correct, while Statement II is not correct.

For clarification, $\mathrm{GeCl}_4$ is more stable than $\mathrm{GeCl}_2$. This is because, in $\mathrm{GeCl}_4$, germanium is in the +4 oxidation state, which is more stable than the +2 state, primarily due to the inert pair effect.

The explanation provides information on the uses of certain carbon allotropes:

Graphite - Crucibles:

This statement is correct. Graphite's high melting point and stability make it ideal for use in crucibles, which are containers used to heat substances to very high temperatures.

Activated Charcoal - Water filters:

This statement is also correct. Activated charcoal is commonly used in water filters due to its ability to absorb impurities and contaminants.

Carbon Black - Fuel:

This statement is incorrect. Carbon black is not used as a fuel. Instead, it is obtained as soot from the incomplete combustion of carbon-rich fuels. It is primarily used to strengthen rubber in tires, as well as in various pigments and coatings.

Therefore, the correct matches for the uses of carbon allotropes are in statements (i) and (ii).

The reaction between sodium nitrite ($\mathrm{NaNO}_2$) and hydrochloric acid ($\mathrm{HCl}$) proceeds as follows:

First, sodium nitrite reacts with hydrochloric acid to form nitrous acid ($\mathrm{HNO}_2$) and sodium chloride ($\mathrm{NaCl}$):

$ \mathrm{NaNO}_2 + \mathrm{HCl} \longrightarrow \mathrm{HNO}_2 + \mathrm{NaCl} $

However, $\mathrm{HNO}_2$ is unstable and decomposes further:

$ \mathrm{HNO}_2 \longrightarrow \mathrm{NO} \uparrow + \mathrm{NO}_2 \uparrow + \mathrm{H}_2\mathrm{O} $

Overall, the balanced reaction is:

$ 2 \mathrm{NaNO}_2 + 2 \mathrm{HCl} \longrightarrow 2 \mathrm{NaCl} + \mathrm{H}_2\mathrm{O} + \mathrm{NO(g)} \uparrow + \mathrm{NO}_2(g) \uparrow $

In this reaction, two gaseous products, nitric oxide ($\mathrm{NO}$) and nitrogen dioxide ($\mathrm{NO}_2$), are formed.

Hydrated sodium aluminium silicate is known as zeolite.

Chemical Formula:

$ \mathrm{Na}_2 \mathrm{O} \cdot \mathrm{Al}_2 \mathrm{O}_3 \cdot 2 \mathrm{SiO}_2 \cdot 4 \mathrm{H}_2 \mathrm{O} $

Zeolites are microporous, aluminosilicate minerals commonly used as commercial adsorbents and catalysts.

Buckminster fullerene is an allotrope of carbon known for its aromatic nature.

$ \begin{aligned} &\text { The complete reaction is as follows, }\\ &\mathrm{P}_2 \mathrm{O}_3+3 \mathrm{H}_2 \mathrm{O} \longrightarrow 2 \mathrm{H}_3 \mathrm{PO}_3+2 \mathrm{H}_2 \uparrow \end{aligned} $

$ \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\text { Product } X $

$\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, \text { (Phosphorous acid) } $

$ \text { Red } \mathrm{P}_4+\text { Alkali } \longrightarrow \mathrm{H}_4 \mathrm{P}_2 \mathrm{O}_6 $

$ \text { The complete reaction is a follows, } $

Let’s examine each statement:

Statement i: "Basic structural unit of silicates is $-R_2 SiO-$"

The basic building block of silicates is actually the $\mathrm{SiO_4}^{4-}$ tetrahedron. The unit $-R_2SiO-$ is representative of silicones (or polysiloxanes), where $R$ is an organic group. In silicates, there is no organic $R$ group.

Therefore, statement i is incorrect.

Statement ii: "Silicones are biocompatible"

Silicones are known for their excellent biocompatibility and are widely used in medical implants, devices, and other applications where interacting with biological tissues is required.

This statement is correct.

Statement iii: "Producer gas contains CO and $\mathrm{N}_2$"

Producer gas is produced by the partial combustion of carbonaceous materials in air. Its primary components include carbon monoxide ($CO$), hydrogen ($H_2$), carbon dioxide ($CO_2$), and nitrogen ($\mathrm{N}_2$).

This statement is also correct.

Since only statements ii and iii are correct, the correct option is:

Option B: ii, iii only.

$\mathrm{SiCl}_4$ on hydrolysis gives HCl and $\mathrm{Si}(\mathrm{OH})_4$ which is solid. The complete reaction is

$ \mathrm{SiCl}_4+4 \mathrm{H}_2 \mathrm{O} \longrightarrow 4 \mathrm{HCl}+\mathrm{Si}(\mathrm{OH})_4 $

$ (X) \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,(Solid)$

Reaction of calcium phosphide with water gives phosphine.

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

Phosphine

Heating white phosphorous with conc. NaOH solution in inert atmosphere also gives phosphine.

$ \begin{aligned} & \mathrm{P}_4+3 \mathrm{NaOH}+3 \xrightarrow{3 \mathrm{H}_2 \mathrm{O}} \mathrm{PH}_3+3 \mathrm{NaH}_2 \mathrm{PO}_2 \\ & \text { } \end{aligned} $

The complete reaction is as follows,

$ \mathrm{P}_4+3 \mathrm{OH}^{-}+3 \mathrm{H}_2 \mathrm{O} \longrightarrow \mathrm{PH}_3+3 \mathrm{H}_2 \mathrm{PO}_4^{-} $

Here, $x$ is $\mathrm{H}_2 \mathrm{PO}_2^{-}$. The conjugate acid of $x$ is hypophosphorus acid.

$ \mathrm{H}_3 \mathrm{PO}_2 \rightleftharpoons \mathrm{H}^{+}+\mathrm{H}_2 \mathrm{PO}_2^{-} $

Statement given in Eqs. (i) and (ii) are correct, while (iii) is incorrect. It's correct form is

Boric acid is monobasic acid in water.

Disproportionation is a type of redox reaction in which an element undergoes both oxidation and reduction to form two different compounds.

The disproportionation reaction for orthophosphorus acid is:

$ 4 \mathrm{H}_3 \mathrm{PO}_3 \xrightarrow{\Delta} \mathrm{H}_3 \mathrm{PO}_4 + \mathrm{PH}_3 $

In this reaction:

Orthophosphorus acid ($\mathrm{H}_3 \mathrm{PO}_3$) is initially at a +3 oxidation state.

It gets oxidized to phosphoric acid ($\mathrm{H}_3 \mathrm{PO}_4$), where phosphorus is at a +5 oxidation state.

Simultaneously, it is reduced to phosphine ($\mathrm{PH}_3$), where phosphorus is at a -3 oxidation state.

Let's look at the reaction step by step.

Aluminium carbide, typically represented as $\mathrm{Al_4C_3},$ reacts with water to produce aluminium hydroxide and methane. The balanced reaction with H₂O is:

$\mathrm{Al_4C_3 + 12\,H_2O \rightarrow 4\,Al(OH)_3 + 3\,CH_4}.$

When deuterium oxide (D₂O) is used instead of water, every hydrogen atom is replaced by deuterium. Hence, aluminium hydroxide becomes aluminium deuteroxide, $\mathrm{Al(OD)_3},$ and methane becomes fully deuterated methane, $\mathrm{CD_4}.$

The modified reaction becomes:

$\mathrm{Al_4C_3 + 12\,D_2O \rightarrow 4\,Al(OD)_3 + 3\,CD_4}.$

So, the product $X'$ in the reaction is $\mathrm{CD_4}.$

Thus, the correct answer is Option D: $\mathrm{CD_4}.$

When chlorine is reacted with hot concentrated NaOH solution, the products are sodium chloride $(\mathrm{NaCl})$, sodium chlorate $\left(\mathrm{NaClO}_3\right)$ and water $\left(\mathrm{H}_2 \mathrm{O}\right)$.

$ \begin{aligned} 3 \mathrm{Cl}_2+6 \mathrm{NaOH} \longrightarrow & 5 \mathrm{NaCl}+\mathrm{NaClO}_3 + 3 \mathrm{H}_2 \mathrm{O} \end{aligned} $

During the water-gas shift reaction, carbon monoxide (CO) + H₂ reacts with steam (H₂O) to produce carbon dioxide (CO₂) and hydrogen gas (H₂). The chemical equation for the reaction is :

Therefore, option B is correct, which states that carbon monoxide is oxidized to carbon dioxide. This reaction is typically catalyzed by metal catalysts such as iron, nickel, or copper, which promote the reaction rate by providing a surface for the reactants to adsorb and react.

Option A is not correct as water is not evaporated during the reaction, but rather reacts with carbon monoxide to produce hydrogen gas and carbon dioxide.

Option C is not correct as carbon is not directly involved in the reaction.

Option D is also not correct as carbon dioxide is not reduced to carbon monoxide during the water-gas shift reaction.