Sodium borate on reaction with $H_2SO_4$ (conc.) and $C_2H_5OH$ gives a compound '$A$' which burns with a green flame. The compound '$A$' is
- A$H_2B_4O_7$
- B$(C_2H_5)_2B_4O_7$
- C$H_3BO_3$
- D$(C_2H_5)_3BO_3$
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Similar Questions
The element which exists in liquid state for a wide range of temperature and can be used for measuring high temperature is :-
Easy
View SolutionWhich is not true about borax?
$BCl_3$ and $AlCl_3$ exist in $:-$
Medium
View SolutionConsider the following reactions.
$Na_{2}B_{4}O_{7} \xrightarrow{\Delta} 2 X + Y$
$CuSO_{4} + Y \xrightarrow{\text{Non-Luminous flame}} Z + SO_{3}$
$2Z + 2X + \text{Carbon} \xrightarrow{\text{Luminous flame}} 2Q + Na_{2}B_{4}O_{7} + CO$
The oxidation states of $Cu$ in $Z$ and $Q$, respectively are:
$Na_{2}B_{4}O_{7} \xrightarrow{\Delta} 2 X + Y$
$CuSO_{4} + Y \xrightarrow{\text{Non-Luminous flame}} Z + SO_{3}$
$2Z + 2X + \text{Carbon} \xrightarrow{\text{Luminous flame}} 2Q + Na_{2}B_{4}O_{7} + CO$
The oxidation states of $Cu$ in $Z$ and $Q$, respectively are:
DifficultJEE MAIN 2026
View SolutionThe geometry around boron in the product '$B$' formed from the following reaction is
$BF_{3} + NaH \xrightarrow{450 \ K} A + NaF$
$A + NMe_{3} \rightarrow B$
$BF_{3} + NaH \xrightarrow{450 \ K} A + NaF$
$A + NMe_{3} \rightarrow B$
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