The p-Block Elements's Question 18 | Chemistry XI - Part 2 | NCERT Exemplar Questions

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MCQ (Multiple Correct Answer)

$\mathrm{M}_3 \mathrm{SiO}$ is used during polymerisation of organo silicones because

chain length of organo silicone polymers can be controlled by adding $\mathrm{Me}_3 \mathrm{SiCl}$

$\mathrm{Me}_3 \mathrm{SiCl}$ blocks the end terminal of silicone polymer

$\mathrm{Me}_3 \mathrm{SiCl}$ improves the quality and yield of the polymer

$\mathrm{Me}_3 \mathrm{SiCl}$ acts as a catalyst during polymerisation

MCQ (Multiple Correct Answer)

Which of the following statements are correct ?

Fullerenes have dangling bonds

Fullerenes are cage-like molecules

Graphite is thermodynamically most stable allotrope of carbon

Graphite is slippery and hard and therefore used as a dry lubricant in machines.

MCQ (Multiple Correct Answer)

Which of the following statements are correct? Answer on the basis of given figure.

The two bridged hydrogen atoms and the two boron atoms lie in one plane

Out of six $\mathrm{B}-\mathrm{H}$ bonds two bonds can be described in terms of 3 centre 2 electron bonds

Out of six $\mathrm{B}-\mathrm{H}$ bonds four $\mathrm{B}-\mathrm{H}$ bonds can be described in terms of 3 centre 2 electron bonds

The four terminal $\mathrm{B}-\mathrm{H}$ bonds are two centre-two electron regular bonds

MCQ (Multiple Correct Answer)

Identify the correct resonance structures of carbon dioxide from the ones given below

$\mathrm{O}-\mathrm{C} \equiv \mathrm{O}$

$\mathrm{O}=\mathrm{C}=\mathrm{O}$

${ }^{-} \mathrm{O} \equiv \mathrm{C}-\mathrm{O}^{+}$

${ }^{-} \mathrm{O}-\mathrm{C} \equiv \mathrm{O}^{+}$

Subjective

Draw the structures of $\mathrm{BC}_3 \cdot \mathrm{NH}_3$ and $\mathrm{Ald}_3$ (dimer).

Explanation

In the $\mathrm{BCl}_3$, due to small size of $\mathrm{B}^{3+}, \mathrm{BCl}_3$ is covalent. Still, the octet of B remains incomplete. In $\mathrm{NH}_3, \mathrm{~N}$ has a lone pair of electrons. Hence, N shares lone pair of electron with B to complete the octet. Hence, $\mathrm{BCl}_3$ acts as Lewis acid and $\mathrm{NH}_3$ as Lewis bases.

$$\mathrm{H}_3 \mathrm{~N}:+\mathrm{BCl}_3 \longrightarrow \mathrm{H}_3 \mathrm{~N} \longrightarrow \mathrm{BCl}_3 \text { or } \mathrm{BCl}_3 \cdot \mathrm{NH}_3$$

$\mathrm{AlCl}_3$ forms dimer by completing octet of Al involving $p$-orbitals to accept electron pair from chlorine.

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