p-Block Elements

The conjugate base of phosphorus acid is $\mathrm{H}_2 \mathrm{PO}_3^{-}$conjugate base of oleum is, $\mathrm{HS}_2 \mathrm{O}_7^{-}$

$ \begin{aligned} & \mathrm{H}_3 \mathrm{PO}_3(\mathrm{aq}) \rightleftharpoons \mathrm{H}_2 \mathrm{PO}_3^{-}+\mathrm{H}^{+} \\ & \mathrm{H}_2 \mathrm{~S}_2 \mathrm{O}_7 \rightleftharpoons \mathrm{HS}_2 \mathrm{O}_7^{-}+\mathrm{H}^{+} \end{aligned} $

Statement I:

CCl₄ does not undergo hydrolysis but SiCl₄ undergoes hydrolysis.

Explanation:

• In SiCl₄, the central atom Si has vacant 3d orbitals.

  → Hence, water (a nucleophile) can attack the Si atom forming hydrolyzed products like silicic acid (via intermediate Si(OH)₄).

• In CCl₄, the central atom C does not have vacant d orbitals (carbon is limited to 2s and 2p orbitals only).

  → Therefore, nucleophilic attack by water is not possible, so CCl₄ does not undergo hydrolysis.

✅ Statement I is correct.

Statement II:

Thermal and chemical stability of GeX₄ is more than GeX₂.

Explanation:

• Germanium belongs to Group 14.

  → As we go down the group, +2 oxidation state becomes more stable than +4 (due to the inert pair effect).

• However, for Ge, the +4 oxidation state is still more stable than the +2 state (the inert pair effect is not very pronounced yet).

So:

• GeX₄ (Ge⁴⁺) is more stable thermally and chemically than GeX₂ (Ge²⁺).

✅ Statement II is also correct.

✅ Final Answer:

Option A: Both Statement-I and Statement-II are correct.

$ \text { The disproportion reaction is, } $

$ 4 \mathrm{H}_3 \mathrm{PO}_3 \longrightarrow \underset{\text { (Phosphoric acid) }}{3 \mathrm{H}_3 \mathrm{PO}_4}+\mathrm{PH}_3 $

Statements given in $C$ and $D$ are correct, while $A$ and $B$ are incorrect.

Their correct forms are

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

(It Liberate $\mathrm{H}_2$ with NaOH )

(B) Sodium metaborate formula is $\mathrm{NaBO}_2$

The number of amphoteric oxides is 4 i.e.

$\mathrm{SnO}_2, \mathrm{PbO}_2, \mathrm{PbO}, \mathrm{SnO}$

Acidic oxide : $\mathrm{CO}_2, \mathrm{GeO}_2, \mathrm{GeO}$

Neutral oxide : CO

The complete hydrolysis reaction is,

$ \mathrm{XeF}_6+3 \mathrm{H}_2 \mathrm{O} \longrightarrow \underset{(\text { Product } X)}{\mathrm{XeO}_3+6 \mathrm{Hg}} $

$ \mathrm{XeO}_3 \text { : Xenon trioxide }=s p^3 $

The correct order of electronegativity of group 13 elements is,

$ \mathrm{B}>\mathrm{Tl}>\mathrm{In}>\mathrm{Ga}>\mathrm{Al} $

Electronegativity of 13th group generally decreases down the group with some exception due to inert pair effect.

Statement I: CO reduces the oxygen-carrying ability of blood.

• True.

  Carbon monoxide (CO) binds strongly with hemoglobin to form carboxyhemoglobin, which prevents hemoglobin from carrying oxygen effectively.

  ⇒ ✅ True

Statement II: Producer gas contains CO and N₂.

• True.

  Producer gas is produced by passing air over red-hot coke:

  $ \text{C} + \text{O}_2 \rightarrow \text{CO}_2 \quad \text{and} \quad \text{C} + \text{CO}_2 \rightarrow 2\text{CO} $

  Since air is used (not pure oxygen), nitrogen is also present.

  Components ≈ CO + N₂.

  ⇒ ✅ True

Statement III: C–O bond length in CO₂ is 115 pm.

• True.

  In CO₂, there are two equivalent C=O bonds due to resonance.

  Experimental bond length ≈ 116 pm — very close to 115 pm.

  ⇒ ✅ True

✅ All three statements (I, II, III) are correct.

✅ Correct option: Option B — I, II and III

$ \begin{aligned} &\text { } \mathrm{P}_4+\mathrm{SOCl}_2=\mathrm{PCl}_3 \text { and }\\ &\begin{aligned} & \mathrm{P}_4+\mathrm{SO}_2 \mathrm{Cl}_2 \longrightarrow \mathrm{PCl}_3 \\ & \mathrm{PCl}_3=\text { central atom }=\mathrm{P} \end{aligned} \end{aligned} $

P has 5 valence electron 3

Statement (d) is incorrect regarding silica. The correct form is, While graphite has a two dimension layered structure, $\mathrm{SiO}_2$ has three dimensional network structure.

$ \text{The oxoacid made when red } P_4 \text{ reacts with alkali is :} $

Name of the acid :

It is called Hypophosphoric acid, with the formula $\mathrm{H}_4\mathrm{P}_2\mathrm{O}_6$.

Counting bonds :

There are two P=O (double) bonds in this acid.

There is one P–P (single) bond in this acid.

Among the given options, statement given in option (b) is incorrect regarding group 13 elements. Its the correct form is, orthoboric acid is a weak monobasic acid.

Statement I and II are true, while III is incorrect.

The correct form of statement III is, $\mathrm{GeCl}_2$ is less stable the $\mathrm{GeCl}_4$.

When sodium nitrate react with HCl , two nitrogen oxide (nitrogen dioxide $\left(\mathrm{NO}_2\right)$ and nitric oxide $\left.(\mathrm{NO})\right]$ are formed. $\mathrm{NO}_2$ is acid and NO is neutral.

The complete reaction is,

$ \begin{array}{r} 2 \mathrm{NaNO}_2+2 \mathrm{HCl} \longrightarrow 2 \mathrm{NaCl}+\underset{X}{\mathrm{NO}_2(g)} +\underset{Y}{\mathrm{NO}(g)}+\mathrm{H}_2 \mathrm{O} \end{array} $

The correct reaction to produce diborane on industrial scale is, Reaction of $\mathrm{BF}_3$ with NaH at 450 K The reaction involved is,

$ 2 \mathrm{BF}_3+6 \mathrm{NaH} \xrightarrow{450 \mathrm{~K}} \mathrm{~B}_2 \mathrm{H}_6+6 \mathrm{NaF} $

The reducing property is incorrectly ordered.

The correct order is, $\mathrm{H}_2 \mathrm{~S}<\mathrm{H}_2 \mathrm{Se}<\mathrm{H}_2 \mathrm{Te}<\mathrm{H}_2 \mathrm{Po}$ Reducing strength increases down the group in chalcogen due to decreasing bond dissociation enthalpy.

Statement I:

$\mathrm{Al_2O_3}$ is amphoteric in nature.

• Explanation:

  Aluminum oxide reacts both with acids and with bases, forming salts in each case.

  $ \mathrm{Al_2O_3 + 6HCl → 2AlCl_3 + 3H_2O} $

  $ \mathrm{Al_2O_3 + 2NaOH + 3H_2O → 2Na[Al(OH)_4]} $

  Hence, aluminum oxide is amphoteric. ✅

So, Statement I is correct.

Statement II:

$\mathrm{Tl_2O_3}$ is more basic than $\mathrm{Ga_2O_3}$.

We are comparing Group 13 element oxides:

$ B_2O_3 < Al_2O_3 < Ga_2O_3 < In_2O_3 < Tl_2O_3 $

→ Trend: From B to Tl, metallic character increases, so basic character of oxides increases (acidic → amphoteric → basic).

Thus, Tl₂O₃ (Thallium(III) oxide) is more basic than Ga₂O₃ (Gallium oxide). ✅

So, Statement II is also correct.

✅ Final Answer:

Option A: Both Statement I and Statement II are correct.

When white phosphorus ($ P_4 $) is heated with concentrated NaOH solution in an inert atmosphere of CO₂, we are told it gives a salt 'X' and a gas 'Y'.

Step 1: Recall the known reaction

White phosphorus reacts with hot concentrated NaOH (and no oxygen) to produce phosphine gas (PH₃) and sodium hypophosphite (NaH₂PO₂) as the salt:

$ P_4 + 3NaOH + 3H_2O \longrightarrow PH_3 + 3NaH_2PO_2 $

Step 2: Identify 'X' and 'Y'

• Salt $ X = \text{NaH}_2\text{PO}_2 $ (sodium hypophosphite)

• Gas $ Y = \text{PH}_3 $ (phosphine)

Step 3: Find oxidation states of phosphorus in each

For $ X = \text{NaH}_2\text{PO}_2 $

Let oxidation state of P = $ x $.

$ (+1) + 2(+1) + x + 2(-2) = 0 $

$ 1 + 2 + x - 4 = 0 $

$ x = +1 $

So in NaH₂PO₂, P has oxidation state +1.

For $ Y = \text{PH}_3 $

Let oxidation state of P = $ x $.

$ x + 3(+1) = 0 $

$ x = -3 $

Thus, in PH₃, P has oxidation state −3.

✅ Answer:

Oxidation state of central atom in $ X $ and $ Y $ are respectively:

$ +1 ~\text{and}~ -3 $

✅ Correct option: (B) $ +1, -3 $

Statement given in option (a) is incorrect and it's correct form is, when aluminium react with conc. $\mathrm{HNO}_3$, it forms a protective oxide layer on surface.

In buckminster fullerene $\left(C_{60}\right)$, there are 20 six member rings ( $X$ ) and 12 five member rings $(Y)$ are present. $X+Y=20+12=32$

Among given, statement (a) is incorrect. It's correct form is, On heating potassium permanganate, it produces.

Potassium manganate, maganese oxide and oxygen gas

$ \begin{aligned} 2 \mathrm{KMnO}_4 \xrightarrow{\Delta} & \mathrm{~K}_2 \mathrm{MnO}_4+\mathrm{MnO}_2 + \mathrm{O}_2 \uparrow \end{aligned} $

$ \text { Diborane : } \mathrm{B}_2 \mathrm{H}_6 $

There are four terminal $\mathrm{B}-\mathrm{H}$ bonds. These are 2-center-2-electron bonds Thus $X=4$.

There are two bridging $\mathrm{B}-\mathrm{H}-\mathrm{B}$ bonds. These are 3-centre-2-electron bonds. $Y=2 \Rightarrow X+Y=4+2=6$

The complete reaction is as follows

(i) $\mathrm{P}_4 \mathrm{O}_{10}+4 \mathrm{HNO}_3 \longrightarrow 4 \mathrm{HPO}_3 +2 \mathrm{~N}_2 \mathrm{O}_5$

$ \begin{aligned}(ii) \mathrm{P}_4+2 \mathrm{OHNO}_3 \rightarrow & 4 \mathrm{H}_3 \mathrm{PO}_4 +2 \mathrm{ONO}_2+4 \mathrm{H}_2 \mathrm{O} \end{aligned} $

Hence, oxides are $\mathrm{N}_2 \mathrm{O}_5$ and $\mathrm{NO}_2$.

Both statement I and statement II are incorrect. Their correct form are The order of electronegativity is

$ \mathrm{B}>\mathrm{Tl}>\mathrm{In}>\mathrm{Al} $

Boric acid is not a protonic acid, because it does not donate $\left(\mathrm{H}^{+}\right)$proton, it accept a lone pair of electrons from water.

$\left[\mathrm{SiCl}_6\right]^{2-}$ does not exist.

This is due to the larger size of Cl atom, which make it difficult for six chloride ions to be accomodated around the relatively small silicon ions leading to steric hindrance and instability.

The complete reaction is as follows,

Deacon's process

$ \begin{aligned} 4 \mathrm{HCl}+\mathrm{O}_2 & \longrightarrow 2 \mathrm{H}_2 \mathrm{O}+2 \mathrm{Cl}_2(\mathrm{~g}) \uparrow 5 \mathrm{Cl}_2+6 \mathrm{H}_2 \mathrm{O}+\mathrm{I}_2 \longrightarrow 10 \mathrm{HCl}+2 \mathrm{HIO}_3 \end{aligned} $

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} $

Here’s the reaction we need to identify P and Q for:

$P + Q \longrightarrow \left[ B(\mathrm{OH})_4 \right]^{-} + \mathrm{H}_3 \mathrm{O}^{+}$

Explanation:

The balanced reaction is:

$\mathrm{B}(\mathrm{OH})_3 + 2 \mathrm{H}_2 \mathrm{O} \longrightarrow \left[ \mathrm{B}(\mathrm{OH})_4 \right]^{-} + \mathrm{H}_3 \mathrm{O}^{+}$

This means that $P = \mathrm{H}_3 \mathrm{BO}_3$ and $Q = 2 \mathrm{H}_2 \mathrm{O}$. In this reaction, $\mathrm{H}_3 \mathrm{BO}_3$ acts as a Lewis acid.

$\mathrm{P}_4$ reacts with NaOH. On heating, it gives $\mathrm{NaH}_2 \mathrm{PO}_2$ and $\mathrm{PH}_3$.

$\mathrm{P}_4+3 \mathrm{NaOH}+3 \mathrm{H}_2 \mathrm{O} \xrightarrow{\Delta} 3 \mathrm{NaH}_2 \mathrm{PO}_2+\mathrm{PH}_3$

This reaction is a disproportionation reaction as oxidation state of P changes from 0 to +1 and $-$3 . Which implies that, it undergoes oxidation and reduction simultaneously.

$\mathrm{S}_8$ reacts with NaOH on heating to given $\mathrm{Na}_2 \mathrm{~S}$ and $\mathrm{Na}_2 \mathrm{~S}_2 \mathrm{O}_3$.

This reaction is a disproportionation reaction as oxidation state of $S$ changes from 0 to $-$2 and +2 . Which implies that, it undergoes oxidation and reduction simultaneously. But, $\mathrm{N}_2$ is an inert gas. So, it does not react with NaOH . Thus, only $\mathrm{P}_4$ and $\mathrm{S}_8$ undergo disproportionation when heated with NaOH solution.

(I) Size of boron atom is very small, so it cannot accommodate 4 large halide atoms and thus, does not form dimer. Also, the electron deficiency of boron is removed by $p \pi-p \pi$ back-bonding. Hence, the boron trihalides do not form dimers.

(II) In boron family, as we move down the group, two oxidation states are shown by them, i.e. +1 and +3 . However, boron shows only +3 oxidation state because of the absence of inert pair effect. Boron does not show +1 oxidation state.

(III) Boron shows a maximum covalency of four because of absent of $d$-orbitals.

(IV) Boron does not form $\mathrm{BF}_6^{6-}$ ion because of absence of empty $d$-orbitals.

Hence, only statements III and IV are correct.

When $\mathrm{SO}_3$ gas is passed through $\mathrm{H}_2 \mathrm{SO}_4$ solution, it get absorbed in concentrated $\mathrm{H}_2 \mathrm{SO}_4$. The final product formed here is oleum or fuming sulphuric acid.

$\mathrm{SO}_3(g)+\mathrm{H}_2 \mathrm{SO}_4 \longrightarrow \underset{\text { Oleum }}{\mathrm{H}_2 \mathrm{S}_2 \mathrm{O}_7}$

I. Boron has a high melting point of 2352 K . So, statement I is correct.

II. Boron has a low density of $2.37 \mathrm{~gcm}^{-3}$. So, statement II is incorrect.

III. Boron has low electrical conductivity at low temperature. So, statement III is incorrect.

IV. B-10 isotope is a neutron absorber due to the high neutron cross-section of isotope ${ }^{10}$ B. So, statement IV is also correct. Hence, statements I and IV are correct.

HCl gas is dried by passing through the drying agent i.e. concentrated sulphuric acid. So, assertion is correct.

HCl reacts with $\mathrm{NH}_3$ to give white fumes of ammonium chloride.

$\mathrm{HCl}+\mathrm{NH}_3 \longrightarrow \underset{\text { White fumes }}{\mathrm{NH}_4 \mathrm{Cl}}$

So, reason is also correct. But reason is not a correct explanation of assertion.

Boric acid $\left(\mathrm{H}_3 \mathrm{BO}_3\right)$ is an acid because its molecule accepts $\mathrm{OH}^{-}$ from water releasing proton.

$\mathrm{B}(\mathrm{OH})_3+\mathrm{H}_2 \mathrm{O} \rightleftharpoons\left[\mathrm{B}(\mathrm{OH})_4\right]^{-}+\mathrm{H}^{+}$