p-Block Elements

(a) IBr is an interhalogen compound with the chemical symbol IBr. It is a black solid that melts near room temperature. It is used for iodometry.

(b) ClF3 is an inter halogen compound. This is colourless, poisonous, corrosive and extremely reactive gas. It condenses to a pale-greenish/yellow liquid, the form in which it is most often sold.

(c) BrF$_3$ is also an interhalogen compound. It is yellow-green liquid with a pungent odor. It is soluble in sulphuric acid but reacts violently with water.

(d) ICl$_3$ is an interhalogen compound of iodine and chlorine. It is bright yellow-orange solid but on exposure to light it turns red.

Helium is used as a cryogenic agent due to its very low melting point.

BaF₂ is not an anti-fluorite structure. It is of AB₂ type i.e. fluorite (CaF₂) structure. While Rb₂S, K₂O and Li₂O are anti-fluorite structures.

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.

Buckminister fullerene is a type of fullerene with formula $\mathrm{C}_{60}$. It has a cage like fused sings that resembles a soccer ball.

Buckminister fullerene contain even number of carbon atoms with 12-(twelve) five membered rings and 20 six membered rings, i.e.

$ (X)=20 \Rightarrow(Y)=12 $

Hence, option (c) is the correct answer.

Orthophosphoric acid is $\mathrm{H}_3 \mathrm{PO}_4$, It dissociates as follows :

(i) $\mathrm{H}_3 \mathrm{PO}_4 \rightleftharpoons \mathrm{H}_2 \mathrm{PO}_4^{-}+\mathrm{H}^{+}$

(ii) $\mathrm{H}_2 \mathrm{PO}_4^{-} \rightleftharpoons \mathrm{HPO}_4^{2-}+\mathrm{H}^{+}$

(iii) $\mathrm{HPO}_4^{2-} \longrightarrow \mathrm{PO}_4^{3-}+\mathrm{H}^{+}$

Hence, has 3 dissociable protons. Therefore, option (a) is the correct answer.

I. Structural formula of borax $\left(\mathrm{Na}_2 \mathrm{~B}_4 \mathrm{O}_7 \cdot 10 \mathrm{H}_2 \mathrm{O}\right)$ is $\mathrm{Na}_2\left[\mathrm{~B}_4 \mathrm{O}_5(\mathrm{OH})_4\right] \cdot 8 \mathrm{H}_2 \mathrm{O}$ ⇒ correct statement.

II. Aqueous solution of borax is basic in nature, because borax in water dissociates into $\mathrm{H}_3 \mathrm{BO}_3$ (weak acid) and NaOH (strong base).

⇒ Incorrect statement.

III. Cobalt (Co) gives blue colour of cobalt metaborate $\left[\mathrm{Co}\left(\mathrm{BO}_2\right)_2\right]$ in borax bead $\left(\mathrm{NaBO}_2+\mathrm{B}_2 \mathrm{O}_3\right)$ test ⇒ Correct statement.

So, the correct statements are (I) and (III).

Among group-14 elements, Pb does not show catenation property. The order of catenation is,

$ \mathrm{C} \gg \mathrm{Si}>\mathrm{Ge} \simeq \mathrm{Sn} $

When white phosphorus $\left(\mathrm{P}_4\right)$ is boiled with conc. NaOH solution, we get phosphine $\left(\mathrm{PH}_3\right)$ gas in laboratory.

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

Thermodynamically most stable allotrope of sulphur is rhombic sulphur $\left(\mathrm{S}_\alpha\right)$ which is a molecular crystal made of $\mathrm{S}_8$ units.

The solution of borax is basic in nature, because it has formula $\mathrm{Na}_2\left[\mathrm{~B}_4 \mathrm{O}_5(\mathrm{OH})_4\right] \cdot 8 \mathrm{H}_2 \mathrm{O}$.

It is a white powdery substance and widly used as house hold cleaner and booster for laundry detergent.

The molecular formula of metaphosphoric acid is $\mathrm{HPO}_3$. It's structure can be given as :

It has general formula $\left(\mathrm{HPO}_3\right)_n$.

where, $n$ denotes number of phosphoric acid units present in the ring, with $n=3$ or more.

In metaphosphoric acid oxidation state of phosphorus is +5 with less number of H -atoms.

Hence, option (b) is the correct answer.

The final acid product obtained during the synthesis of $\mathrm{H}_2 \mathrm{SO}_4$ by contact process is $\mathrm{H}_2 \mathrm{~S}_2 \mathrm{O}_7$, i.e. oleum.

The following reaction(s) are involved in the formation of oleum.

(i) $2 \mathrm{SO}_2+\mathrm{O}_2 \rightleftharpoons 2 \mathrm{SO}_3$

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

Hence, option (d) is the correct answer.

For the most, group 14 elements do not react with water. One interesting consequence of this is that $\operatorname{tin}(\mathrm{Sn})$ is often sprayed as a protective layer on irons cans to prevent the can from corroding. It is stable to water under ambient conditions but on heating with stream, tin reacts with water to form tin dioxide $\mathrm{SnO}_2$ and hydrogen.

When copper metal is treated with cold and dilute nitric acid. It yields copper nitrate, water and nitric oxide.$ 3 \mathrm{Cu}+8 \mathrm{HNO}_3 \underset{(\text { Cold })}{(\text { dil. })} \longrightarrow \begin{aligned} & 3 \mathrm{Cu}\left(\mathrm{NO}_3\right)_2 +4 \mathrm{H}_2 \mathrm{O}+2 \mathrm{NO} \end{aligned}$

$\mathrm{NO}_2$ (Nitrogen dioxide) is not a colourless compound. It is a brown gas, highly reactive and paramagnetic. Oxidation state of N in $\mathrm{NO}_2$ is +4 .

The reaction can be completed as follows

Na₂B₄O₇ + conc. H₂SO₄ + 5H₂O $\to$ Na₂SO₄ + $\mathop {4{H_3}B{O_3}}\limits_{(A)} $

H₃BO₃ + 3C₂H₅OH $\to$ $\mathop {B(O{C_2}{H_5}) + 3{H_2}O \to 3}\limits_{Volatile\,Burn\,with\,green\,edged\,flame} $

The inorganic halide (A) is SO₂Cl₂.

$\mathop {S{O_2}C{l_2}}\limits_{(A)} + 2{H_2}O\buildrel {} \over \longrightarrow \mathop {{H_2}S{O_4}}\limits_{(B)} + \mathop {2HCl}\limits_{(C)} $

$\mathop {S{O_2}C{l_2}}\limits_{(A)} + 4NaOH\buildrel {} \over \longrightarrow \mathop {N{a_2}S{O_4}}\limits_{(D)} + \mathop {2NaCl}\limits_{(E)} + 2{H_2}O$

$\mathop {N{a_2}S{O_4}}\limits_{(D)} + BaC{l_2}\buildrel {} \over \longrightarrow \mathop {BaS{O_4}}\limits_{White\,ppt.} + 2NaCl$

$\mathop {NaCl}\limits_{(E)} + AgN{O_3}\buildrel {} \over \longrightarrow \mathop {AgCl}\limits_{White\,ppt.} + NaN{O_3}$