p Block Elements (Group 13 & 14) JEE Advanced previous year questions with solutions

5 solved JEE Advanced questions on p Block Elements (Group 13 & 14), free to read — no sign-in needed. The full chapter has 16 questions; sign in to attempt the remaining 11 in the exam simulator.

1. Q1JEE Advanced Adv 2022 (Paper 1)
   The treatment of galena with ${HNO}_{3}$ produces a gas that is
   1. A.paramagnetic
   2. B.bent in geometry
   3. C.an acidic oxide
   4. D.colorless
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   Answer: A,D

   Concentrated nitric acid is a strong oxidising agent and attacks most metals except noble metals such as gold and platinum. The products of oxidation depend upon the concentration of the acid, temperature and the nature of the material undergoing oxidation. $PbS+dil.{HNO}_{3}\rightarrow Pb{\left({NO}_{3}\right)}_{2}+S+NO+{H}_{2}O$ $NO$ is paramagnetic due to the presence of unpaired electron. It is a neutral oxide. It is colourless. Hence, $\left(A\right)$ and $\left(D\right)$ are correct statements.
2. Q2JEE Advanced Adv 2015 (Paper 2)
   Under hydrolytic conditions, the compounds used for the preparation of linear polymer and for chain termination, respectively, are
   1. A.$C{H}_{3}SiC{l}_{3}$ and $Si{\left(C{H}_{3}\right)}_{4}$
   2. B.${\left(C{H}_{3}\right)}_{2}SiC{l}_{2}$ and ${\left(C{H}_{3}\right)}_{3}SiCl$
   3. C.${\left(C{H}_{3}\right)}_{2}SiC{l}_{2}$ and $C{H}_{3}SiC{l}_{3}$
   4. D.$SiC{l}_{4}$ and ${\left(C{H}_{3}\right)}_{3}SiCl$
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   Answer: (B)

   For linear polymer, we need ${R}_{2}SiC{l}_{2}$ and for termination, we need ${R}_{3}SiCl$ .
3. Q3JEE Advanced Adv 2015 (Paper 2)
   Three moles of ${B}_{2}{H}_{6}$ are completely reacted with methanol. The number of moles of boron containing product formed is
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   Answer: 6

   ${B}_{2}{H}_{6}+6C{H}_{3}OH⟶2B{\left(OC{H}_{3}\right)}_{3}+6{H}_{2}$ For 3 moles of ${B}_{2}{H}_{6}$ mole of B containing product formed = 6
4. Q4JEE Advanced Adv 2012 (Paper 2)
   With respect to graphite and diamond, which of the statement(s) given below is (are) correct?
   1. A.Graphite is harder than diamond.
   2. B.Graphite has higher electrical conductivity than diamond.
   3. C.Graphite has higher thermal conductivity than diamond.
   4. D.Graphite has higher C-C bond order than diamond.
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   Answer: B,D

   (a) Diamond is harder than graphite. (b) Graphite is good conductor of electricity as each carbon is attached to three C-atoms leaving one valency free, which is responsible for electrical conduction, while in diamond, all the four valencies of carbon are satisfied, hence insulator. (c) Diamond is better thermal conductor than graphite. Whereas electrical conduction is due to availability of free electrons; thermal conduction is due to transfer of thermal vibrations from atom to atom. A compact and precisely aligned crystal like diamond thus facilitates fast movement of heat. (d) In graphite, $\mathrm{C}-\mathrm{C}$ bond acquires double bond character, hence higher bond order than in diamond.
5. Q5JEE Advanced Adv 2009 (Paper 2)
   The coordination number of $\mathrm{Al}$ in the crystalline state of $\mathrm{AlCl}_3$ is
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   Answer: 6

   Coordination number of $\mathrm{Al}=6$.