Chemistry M.C.Q (1 Marks)

In diamond, each carbon atom undergoes $sp ^3$ hybridisation and linked to four other carbon atoms by using hybridised orbitals in tetrahedral fashion.

Buckminsterfullerene contains six membered and five membered rings and hence is a cage like molecule.

Graphite is very soft and slippery. Hence, it is used as a dry lubricant in machines running at high temperature.

due to inert pair effect $Tl^{+}$is more stable than $Tl^{+3}$.

Non- planar

In graphite each carbon is bonded with three other carbon atoms. So hybridisation of carbon atom is $sp ^{2}$.

Boron belongs to $2^{\text {nd }}$ period and it does not have vacant $d-$orbital.

As a result, $Pb(II)$ is more stable than $Pb(IV)$

$\mathrm{Sn}(\mathrm{IV})$ is more stable than $\mathrm{Sn}(\mathrm{II})$

$\mathrm{Pb}(\mathrm{IV})$ is easily reduced to $\mathrm{Pb}(\mathrm{II})$

$\mathrm{Pb}(\mathrm{IV})$ is oxidizing agent

$\mathrm{Sn}(\mathrm{II})$ is easily oxidized to $\mathrm{Sn}(\mathrm{IV})$

$\mathrm{Sn}(\mathrm{II})$ is reducing agent

$\mathrm{B}(\mathrm{OH})_{3}+\mathrm{H}_{2} \mathrm{O} \longrightarrow\left[\mathrm{B}(\mathrm{OH})_{4}\right]^{(-)}(\mathrm{aq})+\mathrm{H}^{(-)}(\mathrm{aq})$

maximum $C.N.$ of $A l^{+3}$ is six so it forms $AlF_{6}^{3-}$

$\begin{array}{*{20}{c}}
  {Me} \\ 
  | \\ 
  {Me - Si - OH} \\ 
  | \\ 
  {Me} 
\end{array}$ $+$ $\begin{array}{*{20}{c}}
  {Me} \\ 
  | \\ 
  {HO - Si - Me} \\ 
  | \\ 
  {Me} 
\end{array}$ $\xrightarrow[{ - {H_2}O}]{}$ $\mathop {\begin{array}{*{20}{c}}
  {\,\,Me\,\,\,\,\,\,\,\,\,\,\,\,\,Me} \\ 
  {|\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,|} \\ 
  {Me - Si - O - Si - Me} \\ 
  {|\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,|} \\ 
  {\,\,Me\,\,\,\,\,\,\,\,\,\,\,\,\,Me} 
\end{array}}\limits_{Final\,product\,(\dim er)} $

These retain metallic conductivity. These resemble the parent metal in chemical properties (reactivity) but differ in physical properties like hardness, melting point, etc.

Adduct form $^{\mathrm{n}}$ order $\mathrm{BI}_{3}>\mathrm{BBr}_{3}>\mathrm{BCl}_{3}>\mathrm{BF}_{3}$

As in $\mathrm{BF}_{3}$ strong back bonding occur. So adduct form $^{\mathrm{n}}$ decreases. Back bonding $\alpha \frac{1}{\text { adduct }}$ form $^{\text {n }}$

(ii) Size of halides increase down the group, hence the ability of back bonding decreases making the molecule more acidic.

Hence, correct order of Lewis acid for boron-halides is $\mathrm{Bl}_{3}>\mathrm{BBr}_{3}>\mathrm{BCl}_{3}>\mathrm{BF}_{3}$

(ii) Size of halides increase down the group, hence the ability of back bonding decreases making the molecule more acidic.

Hence, correct order of Lewis acid for boron-halides is $\mathrm{Bl}_{3}>\mathrm{BBr}_{3}>\mathrm{BCl}_{3}>\mathrm{BF}_{3}$

$\mathrm{Si}+4 \mathrm{HF} \rightarrow \mathrm{SiF}_{4}+4 \mathrm{H}^{+}$

$\begin{array}{l}\downarrow \mathrm{HF} \\\mathrm{H}_{2} \mathrm{SiF}_{6}\end{array}$

The reaction is as follows:

$\mathrm{SiCl}_{4}+4 \mathrm{H}_{2} \mathrm{O} \rightarrow \mathrm{Si}(\mathrm{OH})_{4}+4 \mathrm{HCl}$

Hence, silicic acid is formed when $\mathrm{SiCl}_{4}$ undergoes hydrolysis.

Hence, option $\mathrm{C}$ is the correct answer.

$\mathrm{BF}_{3}+3 \mathrm{H}_{2} \mathrm{O} \rightarrow \mathrm{B}(\mathrm{OH})_{3}+3 \mathrm{HF}$

$3 \mathrm{HF}+3 \mathrm{BF}_{3} \rightarrow 3 \mathrm{H}\left[\mathrm{BF}_{4}\right]$

$\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_$

$4 \mathrm{BF}_{3}+3 \mathrm{H}_{2} \mathrm{O} \rightarrow \mathrm{B}(\mathrm{OH})_{3}+3 \mathrm{H}\left[\mathrm{BF}_{4}\right]$

Anhydrous aluminium chloride is hydrolysed partially with the moisture in the atmosphere to give HCl gas.

This HCI combines with the moisture in the air and appears white in colour.

$\mathrm{AlCl}_{3}+3 \mathrm{H}_{2} \mathrm{O} \longrightarrow \mathrm{Al}(\mathrm{OH})_{3}+3 \mathrm{HCl}(\text { white fumes })$

Hence, the correct answer is option D.

Explanation:

Borax on strong heating loses its water of crystallization and then shrinks forming a transparent glassy bead of sodium metaborate and boric anhydride. Boric anhydride, being non-volatile displaces more volatile acidic oxides and combine with basic oxides present to form metaborates which are identified through their characteristic colours

But diborane, $B _2 H _6$ contains two electrons each, three centred bonds. Each Boron atom is in a link with four hydrogen atoms. This makes tetrahedral geometry.

Hence, each Boron atom is $sp ^3$ - hybridized.