Solved: CHEMISTRY Question for NEET

NEET CHEMISTRYMedium

Match List-I with List-II: List-I (Compounds) (a) Borax (b) Kernite (c) Orthoboric acid (d) Borax bead List-II (Molecular formula) (i) $NaBO_2 + B_2O_3$ (ii) $Na_2B_4O_7 \cdot 4H_2O$ (iii) $H_3BO_3$ (iv) $Na_2B_4O_7 \cdot 10H_2O$ Choose the correct answer from the options given below:

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

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

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

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

Step-by-Step Solution

Let us match the compounds with their molecular formulas: (a) Borax: It is sodium tetraborate decahydrate, $Na_2B_4O_7 \cdot 10H_2O$ . (b) Kernite: It is sodium tetraborate tetrahydrate, $Na_2B_4O_7 \cdot 4H_2O$ . (c) Orthoboric acid: It is a weak monobasic Lewis acid with the formula $H_3BO_3$ . (d) Borax bead: On strong heating, borax loses its water of crystallization and forms a transparent glassy bead consisting of sodium metaborate ( $NaBO_2$ ) and boric anhydride/oxide ( $B_2O_3$ ). Therefore, the correct matching is (a)-(iv), (b)-(ii), (c)-(iii), (d)-(i).

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