p-Block Elements

In trivalent state most of the compounds being covalent are hydrolysed in water. Trichlorides on hydrolysis in water form tetrahedral $\left[\mathrm{M}(\mathrm{OH})_4\right]^{-}$ species, the hybridisation state of element $\mathrm{M}$ is $\mathrm{sp}^3$.

In case of aluminium, acidified aqueous solution forms octahedral $\left[\mathrm{Al}\left(\mathrm{H}_2 \mathrm{O}\right)_6\right]^{3+}$ ion.

$\mathrm{S}_1: \mathrm{S}_8+12 \mathrm{OH}^{\ominus} \rightarrow 4 \mathrm{~S}^{2-}+2 \mathrm{~S}_2 \mathrm{O}_3^{2-}+6 \mathrm{H}_2 \mathrm{O}$

$\mathrm{S}_2: \mathrm{ClO}_4^{\ominus}$ cannot undergo disproportionation reaction as chlorine is present in it's highest oxidation state.

(A) All group 16 elements form oxides of the $\mathrm{EO}_2$ and $\mathrm{EO}_3$ type where $\mathrm{E}=\mathrm{S}, \mathrm{Se}, \mathrm{Te}$ or $\mathrm{Po}$.

(B) $\mathrm{SO}_2$ is reducing while $\mathrm{TeO}_2$ is an oxidising agent.

(C) The reducing property increases from $\mathrm{H}_2 \mathrm{S}$ to $\mathrm{H}_2$ Te down the group.

(D)

SnO$_2$ and PbO$_2$ are amphoteric.

Option A is the correct answer. Both Statement I and Statement II are false.

Noble gases actually have low boiling points, not high. This is because noble gases are indeed monoatomic gases that exist as single atoms. They have a completely filled valence shell and do not easily form chemical bonds with other atoms. As a result, the intermolecular forces present in noble gases are only temporary dipole-induced dipole attractions, also known as dispersion forces (or London forces). However, these dispersion forces are very weak compared to other types of intermolecular forces like hydrogen bonding or dipole-dipole interactions that occur in other substances.

The dispersion forces in noble gases are weak because these gases are non-polar and since they are monoatomic, the electron cloud around the atom is symmetrical, meaning that there are no permanent dipoles within the molecules. As the atomic size of noble gases increases down the group, the dispersion forces do increase in strength due to the larger electron cloud that can be more easily distorted, resulting in higher boiling points for the heavier noble gases. Still, compared to other substances, the overall strength of the dispersion forces in noble gases is low, which means they can be liquefied at low temperatures and hence their boiling points are low, not high.

Therefore, the correct statement would be: "Noble gases are monoatomic gases. They are held together by weak dispersion forces. Because of this they are liquefied at very low temperatures. Hence, they have very low boiling points."

$\mathrm{Cr{O_2} + 4NaOH \to \mathop {N{a_2}Cr{O_4}}\limits_{(A)} + 2NaCl + 2{H_2}O}$

$\mathrm{N{a_2}Cr{O_4} + 2{H_2}{O_2} + 2HCl \to \mathop {Cr{O_5}}\limits_{(B)} + \mathop {2NaCl}\limits_{(Mis\sin g\,from\,balanced\,equation)} + 3{H_2}O}$

On moving down the group, bond strength of $\mathrm{M}-\mathrm{H}$ bond decreases, which reduces the thermal stability but increases reducing nature of hydrides, hence A and B are correct statements.

If nitrogen or sulphur is also present in the compound, the sodium fusion extract is first boiled with concentrated nitric acid to decompose cyanide or sulphide of sodium during Lassaigne's test

Nessler's Reagent Reaction :

$\underset{\text { (Nessler's Reagent) }}{2 \mathrm{~K}_2 \mathrm{HgI}_4}+\mathrm{NH}_3+3 \mathrm{KOH} \rightarrow \underset{\substack{\text { (lodine of Millon's base } \\ \text { (Brown precipitate }}}{\mathrm{HgO} . \mathrm{Hg}\left(\mathrm{NH}_2\right) \mathrm{I}}+7 \mathrm{KI}+2 \mathrm{H}_2 \mathrm{O}$

(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} $