Give reasons for the following: (A) CCl4 is immiscible in water, whereas SiCl4 is easily hydrolysed. (B) Carbon has a strong tendency for catenation compared to silicon.

This is a Short Answers Type Questions as classified in NCERT Exemplar (A) sarbon in CCl4 does not have a vacant d-orbital to accommodate the electrons from OH of water molecules. Also CCl4 is nonpolar covalent compounds whereas H2O is polar. So, no strong interaction occurs between them. Hence CCl4 is miscible in water. Whereas in SiCl4, silicon has bigger size than carbon and have d-orbitals for accommodation of electrons donated by OH of water in the process of hydroxylation. This leads to a strong interaction and silicon acid Is formed as a product. SiCl4 is completely miscible in water. (B) As we move from carbon to silicon, the size increases and electronegativity decreases. This results in the decrease in M-M bond energy. Hence the catenation property decreases. Bond energy is the amount of energy required to fragment the atoms combined in a molecule are born into separate atoms. Carbon has maximum bond energy as compared to silicon. Therefore, shows maximum catenation property.

Give reasons for the following:

(A) CCl₄ is immiscible in water, whereas SiCl₄ is easily hydrolysed.

(B) Carbon has a strong tendency for catenation compared to silicon.

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Alok Kumar Singh | Contributor-Level 10

8 months ago

This is a Short Answers Type Questions as classified in NCERT Exemplar

(A) sarbon in CCl₄ does not have a vacant d-orbital to accommodate the electrons from OH of water molecules. Also CCl₄ is nonpolar covalent compounds whereas H₂O is polar. So, no strong interaction occurs between them. Hence CCl₄

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Assertion (A): Silicones are water repelling in nature.

Reason (R): Silicones are organosilicon polymers, which have (-R₂SiO-) as repeating unit.

(A) Both A and R are true and R is the correct explanation of A.

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(i)→ (b), (ii) → (c), (iii) → (b), (iv) → (a) (v)→ (b) (vi)→ (c)

Match the species given in Column I with properties given in Column II.

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(a) Used as a flux for soldering metals

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(v) Quartz

(e) Used as catalyst in petrochemical industries.

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(i)→ (c) (ii)→ (d) (iii)→ (a) (iv)→ (e) (v)→ (b)

Match the species given in Column I with the properties mentioned in Column II.

Column I

Column II

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(a) Oxidation state of central atom is +4

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(c) Lewis acid

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Chemistry NCERT Exemplar Solutions Class 11th Chapter Eleven 2025

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(i) Boron in [B(OH)₄]⁻	(a) sp²
(ii) Aluminium in [Al(H₂O)₆]³⁺	(b) sp³
(iii) Boron in B₂H₆	(c) sp³d²
(iv) Carbon in Buckminsterfullerene
(v) Silicon in SiO₄⁴⁻
(vi) Germanium in [GeCl₆]²⁻

Column I	Column II
(i) Diborane	(a) Used as a flux for soldering metals
(ii) Gallium	(b) Crystalline form of silica
(iii) Borax	(c) Banana bonds
(iv) Aluminosilicate	(d) Low melting, high boiling, useful for measuring high temperatures
(v) Quartz	(e) Used as catalyst in petrochemical industries.

Column I	Column II
(i) BF₄⁻	(a) Oxidation state of central atom is +4
(ii) AlCl₃	(b) Strong oxidising agent
(iii) SnO	(c) Lewis acid
(iv) PbO₂	(d) Can be further oxidised
	(e) Tetrahedral shape