p-Block Elements (Groups 13-14)

5 NEET-UG practice questions on p-Block Elements (Groups 13-14), part of Chemistry. The first 5 are below; the first 5 with answers and explanations (tap Show answer after you have chosen), the rest in Learn with a free account.

1. Question 1 · Medium

   Diborane reacts readily with water. Which products form?

   1. AB and H₂O₂
   2. BB(OH)₃ and H₂
   3. CBH₃ and O₂
   4. DB₂O₃ and H₂

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   Answer: B · B(OH)₃ and H₂

   The reaction is B₂H₆ + 6H₂O → 2B(OH)₃ + 6H₂, so boric acid and hydrogen form. Each hydrogen of B₂H₆ is released as H₂ with one hydrogen from water.

   • A: Boron is not set free and H₂O₂ does not form.
   • C: Oxygen is not released, and BH₃ is not a stable product.
   • D: B₂O₃ forms on burning in oxygen, not with water.

   Remember this

   Diborane gives boric acid and H₂ with water, but B₂O₃ and water when burnt in oxygen.

2. Question 2 · Medium

   Arrange the following Group 13 elements in increasing order of atomic radius. A. Al B. B C. Ga D. In

   1. AB, C, A, D
   2. BC, B, A, D
   3. CB, A, C, D
   4. DD, A, C, B

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   Answer: A · B, C, A, D

   Radii are about B 85, Ga 135, Al 143 and In 167 pm. Gallium is smaller than aluminium because its 3d electrons shield the nuclear charge poorly.

   • B: This places Ga below B, far too small for a fourth-period element.
   • C: This follows a simple group trend and misses the Ga and Al exception.
   • D: This is the decreasing order.

   Remember this

   Atomic radius rises down Group 13, except that Ga is smaller than Al because of poor 3d shielding.

3. Question 3 · Medium

   Given below are two statements: Statement I: In graphite, each carbon atom is sp² hybridised and the layers are held together by van der Waals forces. Statement II: Buckminsterfullerene, C₆₀, contains twenty six-membered rings and twelve five-membered rings. Choose the correct answer.

   1. ABoth Statement I and Statement II are incorrect
   2. BBoth Statement I and Statement II are correct
   3. CStatement I is incorrect but Statement II is correct
   4. DStatement I is correct but Statement II is incorrect

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   Answer: B · Both Statement I and Statement II are correct

   Graphite has flat sheets of sp² carbon with weak forces between layers, which is why it is soft and slippery. C₆₀ has a soccer-ball shape with 20 hexagons and 12 pentagons. Both statements are true.

   • A: Both statements match the known structures.
   • C: Statement I is also correct, as graphite layers are weakly held.
   • D: Statement II is also correct, as C₆₀ has both ring sizes.

   Remember this

   Weak forces between graphite layers make it a lubricant; delocalised electrons make it a conductor.

4. Question 4 · Hard

   SiCl₄ reacts with water, whereas CCl₄ is not hydrolysed. Which option gives the correct product and reason?

   1. ACCl₄ reacts faster than SiCl₄, because the small carbon atom attracts a lone pair from water
   2. BCCl₄ gives C(OH)₄ and HCl, because vacant 2d orbitals of carbon accept a lone pair from water
   3. CSiCl₄ gives SiO₂ and Cl₂, because the Si-Cl bond breaks homolytically on contact with water
   4. DSiCl₄ gives Si(OH)₄ and HCl, because vacant 3d orbitals of silicon accept a lone pair from water

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   Answer: D · SiCl₄ gives Si(OH)₄ and HCl, because vacant 3d orbitals of silicon accept a lone pair from water

   The oxygen of water donates a lone pair into a vacant 3d orbital of silicon, forming an intermediate, and HCl is then lost stepwise to give Si(OH)₄. Carbon has no d orbitals in its valence shell, and the small carbon is crowded by four chlorine atoms, so CCl₄ does not react.

   • A: This reverses the observed behaviour, as CCl₄ does not react.
   • B: The second shell has no d subshell, and CCl₄ is not hydrolysed.
   • C: Hydrolysis gives HCl, not Cl₂, and no homolysis occurs.

   Remember this

   Second-period carbon lacks valence d orbitals; this explains many differences from silicon, including the maximum covalency of four.

5. Question 5 · Easy

   How many three-centre two-electron bonds are present in a molecule of diborane, B₂H₆?

   1. A1
   2. B6
   3. C4
   4. D2

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   Answer: D · 2

   Two hydrogen atoms bridge the boron atoms, and each B-H-B unit is a three-centre two-electron bond. The other four hydrogens are terminal, with normal two-centre bonds.

   • A: This counts only one of the two bridges.
   • B: This counts all the B-H links as three-centre bonds.
   • C: This counts the terminal two-centre B-H bonds.

   Remember this

   Diborane has 4 terminal B-H bonds and 2 B-H-B banana bonds.