Match List-I with List-II List-I (Si-Compounds): (A) $\left(\mathrm{CH}_{3}\right)_{4} \mathrm{Si}$ (B) $\left(\mathrm{CH}_{3}\right) \mathrm{Si}(\mathrm{OH})_{3}$ (C) $\left(\mathrm{CH}_{3}\right)_{2} \mathrm{Si}(\mathrm{OH})_{2}$ (D) $\left(\mathrm{CH}_{3}\right)_{3} \mathrm{Si}(\mathrm{OH})$ List-II (Si-Polymeric/other products): (I) Chain silicone (II) Dimeric silicone (III) Silane (IV) 2D - Silicone Choose the correct answer from the options given below:

(D) (A) - (III), (B) - (IV), (C) - (I), (D) - (II)

(A) (A) - (III), (B) - (II), (C) - (I), (D) - (IV)

(B) (A) - (IV), (B) - (I), (C) - (II), (D) - (III)

(C) (A) - (II), (B) - (I), (C) - (IV), (D) - (III)

JEE Main · 2022 · 27 Jun 2022 · Shift 1 · p-Block Elements

Match List-I with List-II

List-I (Si-Compounds):
(A)

(CH_{3})_{4} Si

(B)

(CH_{3}) Si (OH)_{3}

(C)

(CH_{3})_{2} Si (OH)_{2}

(D)

(CH_{3})_{3} Si (OH)

List-II (Si-Polymeric/other products):
(I) Chain silicone
(II) Dimeric silicone
(III) Silane
(IV) 2D - Silicone

Choose the correct answer from the options given below:

(A)(A) - (III), (B) - (II), (C) - (I), (D) - (IV)
(B)(A) - (IV), (B) - (I), (C) - (II), (D) - (III)
(C)(A) - (II), (B) - (I), (C) - (IV), (D) - (III)
(D)(A) - (III), (B) - (IV), (C) - (I), (D) - (II)✓ Correct

Answer: (D)

Solution

Matching silicon compounds with their polymerization products:

$(CH_{3})_{4} Si$ is tetramethylsilane, a silane (no OH groups, cannot polymerize)
$(CH_{3}) Si (OH)_{3}$ has 3 OH groups → polymerizes to form 2D cross-linked silicone
$(CH_{3})_{2} Si (OH)_{2}$ has 2 OH groups → polymerizes to form linear chain silicone
$(CH_{3})_{3} Si (OH)$ has 1 OH group → forms dimer $(CH_{3})_{3} Si - O - Si (CH_{3})_{3}$

Therefore: (A)-(III), (B)-(IV), (C)-(I), (D)-(II)

JEE Main · 2022 · 27 Jun 2022 · Shift 1 · Single correct option

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JEE Main · 2022 · 27 Jun 2022 · Shift 1 · p-Block Elements

Solution

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