Na2B4O7⋅10H2O⟶ Heat X+NaBO2+H2O, X+Cr2O3⟶ Heat Y (Green coloured) X and Y are:

Na2B4O7⋅10H2O→-10H2O∆ Na2B4O→2NaBO2 +B2O33B2O3+Cr2O3⟶ heat 2Cr(BO2)3(Y)(Green coloured))

Na2B4O7⋅10H2O⟶ Heat X+NaBO2+H2O, X+Cr2O3⟶ Heat Y (Green coloured) X and Y are:

Na2B4O7⋅10H2O→-10H2O∆ Na2B4O→2NaBO2 +B2O33B2O3+Cr2O3⟶ heat 2Cr(BO2)3(Y)(Green coloured))

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${Na}_{2} B_{4} O_{7} \cdot 10 H_{2} O \overset{Heat}{⟶} X + {NaBO}_{2} + H_{2} O, X + {Cr}_{2} O_{3} \overset{Heat}{⟶} \begin{matrix} Y \\ (Green coloured) \end{matrix}$
X and Y are:

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${Na}_{3} {BO}_{3} and Cr {({BO}_{2})}_{3}$

${Na}_{2} B_{4} O_{7} and Cr {({BO}_{2})}_{3}$

$B_{2} O_{3} and Cr {({BO}_{2})}_{3}$

$B_{2} O_{3} and {CrBO}_{3}$

answer is C.

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Detailed Solution

${Na}_{2} B_{4} O_{7} \cdot 10 H_{2} O \overset{}{\to_{- 10 H_{2} O}^{∆}} {Na}_{2} B_{4} O \overset{}{\to 2 {NaBO}_{2}} + B_{2} O_{3}$
$3 B_{2} O_{3} + {Cr}_{2} O_{3} \overset{heat}{⟶} 2 \underset{\underset{(Green coloured))}{(Y)}}{Cr {({BO}_{2})}_{3}}$