On controlled hydrolysis and condensation, ${R_3}{\rm{SiCl}}$ yields

${R_3}{\rm{Si}} - {\rm{O}} - {\rm{Si}}{R_3}$

${\left( { - {R_3}{\rm{Si}} - {\rm{O}} - {\rm{Si}}{R_3} - } \right)_n}$

$\begin{array}{l} \;\;\;RR\\ \;\;\;\;\;\;\;\;\;||\\ - {\rm{Si}} - {\rm{O}} - {\rm{Si}}\\ \;\;\;\;\;\;\;\;||\\ \;\;\;\;\;\;{\rm{OO}}\\ - {\rm{Si}} - {\rm{O}} - {\rm{Si}} - \\ \;\;\;\;\;\;\;\;|| \end{array}$

The thermal stability of ${\rm{C}}{{\rm{F}}_4}$ is

More than ${\rm{Si}}{{\rm{F}}_4}\;$

Less than ${\rm{Si}}{{\rm{F}}_4}\;$

Less than ${\rm{CC}}{{\rm{l}}_4}$

Less than ${\rm{SiC}}{{\rm{l}}_4}$

Back to 14 TH GROUP ELEMENTS (P-BLOCK)

Your Progress

0 / 187 Answered

Correct

Incorrect

Accuracy

NEET • CHEMISTRY • 14 TH GROUP ELEMENTS (P-BLOCK)

14 th group elements (p-block) practice · Mixed difficulty

Choose an answer to check it, or open “View answer” to study the explanation. Sign in to save progress to your account.

Practice format and chapter quiz

The practice player uses single-correct-answer MCQs. Chapter quizzes help you review a topic; they are not full-length NEET mock papers.

Take a chapter quiz →

Showing 181-187 of 187 questions

Preparing practice and saved progress…

181 Medium

181 of 187

Remembering +4 | -1

CO forms a volatile compound with:

View answer and explanation

Correct option: A

Nickel

Option A (Nickel) is correct because carbon monoxide (CO) reacts with nickel (Ni) to form a volatile compound, nickel carbonyl (

Ni(CO)_4

182 Medium

182 of 187

Remembering +4 | -1

Which of the following is the correct statement for red lead?

View answer and explanation

Correct option: D

It decomposes into

\mathrm{PbO}\;and\;{\mathrm{O}_2}

Option D (It decomposes into

\mathrm{PbO}\;and\;{\mathrm{O}_2}

) is correct because red lead (

\mathrm{P}{\mathrm{b}_3}{\mathrm{O}_4}

) decomposes upon heating to lead(II) oxide (PbO) and oxygen.

183 Easy

183 of 187

Understanding +4 | -1

Fullerenes are the pure form of carbon because

View answer and explanation

Correct option: C

Fullerenes do not have dangling edge or surface bonds

The correct answer is C (Fullerenes do not have dangling edge or surface bonds) because the closed cage structure of fullerenes eliminates dangling bonds, making them chemically stable and relatively pure.

184 Medium

184 of 187

Understanding +4 | -1

On controlled hydrolysis and condensation,

{R_3}\mathrm{SiCl}

yields

View answer and explanation

Correct option: A

{R_3}\mathrm{Si} - \mathrm{O} - \mathrm{Si}{R_3}

Option A is correct because controlled hydrolysis and condensation of

R_3SiCl

yields

R_3Si-O-SiR_3

185 Medium

185 of 187

Remembering +4 | -1

Rose metal is an alloy of.

View answer and explanation

Correct option: A

Sn + Pb + Bi

Option A (Sn + Pb + Bi) is correct because Rose metal is an alloy composed of tin (Sn), lead (Pb), and bismuth (Bi).

186 Medium

186 of 187

Understanding +4 | -1

The thermal stability of

\mathrm{C}{\mathrm{F}_4}

View answer and explanation

Correct option: B

More than

\mathrm{Si}{\mathrm{F}_4}\;

Option B (More than

\mathrm{Si}{\mathrm{F}_4}\;

) is correct because the C-F bond is stronger than the Si-F bond due to smaller size and higher electronegativity of carbon.

187 Medium

187 of 187

Understanding +4 | -1

Choose the correct statement:

View answer and explanation

Correct option: D

Diamond is

s{p^3}

hybridized and graphite is

s{p^2}

hybridized

Option D (Diamond is

s{p^3}

hybridized and graphite is

s{p^2}

hybridized) is correct because it accurately describes the hybridization states of carbon atoms in diamond and graphite.

Revision notes and key takeaways

Review 14 TH GROUP ELEMENTS (P-BLOCK) in your textbook or class notes, then use this set to check your understanding.

Showing 181-187 of 187 questions

Your Progress

14 th group elements (p-block) practice · Mixed difficulty

Continue Your Practice

Your practice