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NEET 2022
Exam
NEET
Year
2022
Subject
Chemistry
Chapter
p Block Elements (Group 13 & 14)

NEET 2022 · Chemistry · p Block Elements (Group 13 & 14)

Match List-I with List-II : ( List-I List-II (Compounds) (Molecular...

Match List-I with List-II : (
List-IList-II
(Compounds)(Molecular formula)
(a) Borax(i) NaBO_2
(b) Kernite(ii) Na_2 B_4 O_7 \cdot 4 H_2 O
(c) Orthoboric acid(iii) H_3 BO_3
(d) Borax bead(iv) Na_2 B_4 O_7 \cdot 10 H_2 O
) Choose the correct answer from the options given below :

Options

  1. A(a)-(iv), (b)-(ii), (c)-(iii), (d)-(i)
  2. B(a)-(ii), (b)-(iv), (c)-(iii), (d)-(i)
  3. C(a)-(iii), (b)-(i), (c)-(iv), (d)-(ii)
  4. D(a)-(i), (b)-(iii), (c)-(iv), (d)-(ii)

Correct answer

A. (a)-(iv), (b)-(ii), (c)-(iii), (d)-(i)

Step-by-step solution

(a) Borax −Na2 B4O7⋅10H2O-\mathrm{Na}_2 \mathrm{~B}_4 \mathrm{O}_7 \cdot 10 \mathrm{H}_2 \mathrm{O} or Na2[ B4O5(OH)4]⋅8H2O\mathrm{Na}_2\left[\mathrm{~B}_4 \mathrm{O}_5(\mathrm{OH})_4\right] \cdot 8 \mathrm{H}_2 \mathrm{O} (b) Kernite −Na2 B4O7⋅4H2O-\mathrm{Na}_2 \mathrm{~B}_4 \mathrm{O}_7 \cdot 4 \mathrm{H}_2 \mathrm{O} (c) Orthoboric acid −H3BO3-\mathrm{H}_3 \mathrm{BO}_3 (d) Borax bead −NaBO2-\mathrm{NaBO}_2 (obtained by heating of borax) Na2 B4O7⋅10H2O⟶−10H2OΔNa2 B4O7⟶Δ2NaBO2+B2O3\mathrm{Na}_2 \mathrm{~B}_4 \mathrm{O}_7 \cdot 10 \mathrm{H}_2 \mathrm{O} \stackrel{\Delta}{\stackrel{-10 \mathrm{H}_2 \mathrm{O}}{\longrightarrow}} \mathrm{Na}_2 \mathrm{~B}_4 \mathrm{O}_7 \stackrel{\Delta}{\longrightarrow} 2 \mathrm{NaBO}_2+\mathrm{B}_2 \mathrm{O}_3 Sodium metaborate
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