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What happens when (a) Borax is heated strongly (b) Boric acid is added to water (c) Aluminum is treated with dilute NaOH (d) \mathrm{BF}_{3} is reacted with ammonia?
CBSE | Chemistry | Class 11 | Exercise 1 | NCERT | The p-Block Elements
What happens when (a) Borax is heated strongly (b) Boric acid is added to water (c) Aluminum is treated with dilute NaOH (d) \mathrm{BF}_{3} is reacted with ammonia?

Solution:

(a) Borax is warmed firmly 
Borax goes through different changes when warmed. It is losing atoms and expands of water right away. Then, at that point, it turns into a clear fluid, which sets to shape a glass-like material called a borax globule.
(b) Boric corrosive is added to water 
At the point when boric corrosive is added to water, it acknowledges electrons from – OH particle.
\mathrm{B}(\mathrm{OH})_{3}+2 \mathrm{H}_{2} \mathrm{O} \rightarrow\left[\mathrm{B}(\mathrm{OH})_{4}\right]^{-}+\mathrm{H}_{3} \mathrm{O}^{+}
(c) Aluminum is treated with weaken NaOH 
Aluminum responds to sodium tetrahydroxoaluminate(III) by framing weaken NaOH. In this cycle, hydrogen gas is freed.
2 \mathrm{Al}_{(s)}+2 \mathrm{NaOH}_{(a q)}+6 \mathrm{H}_{2} \mathrm{O}_{(l)} \rightarrow 2 \mathrm{Na}^{+}\left[\mathrm{Al}(\mathrm{OH})_{4}\right]_{(a q)}^{-}+3 \mathrm{H}_{2(g)}
(d) \mathrm{BF}_{3} is responded with alkali 
\mathrm{BF}_{3} (a Lewis corrosive) responds with \mathrm{NH}_{3} (a Lewis base) to frame an item. This outcomes in a total octet around B in \mathrm{BF}_{3}
B F_{3}+N H_{3} \rightarrow\left[H_{3} N \rightarrow B F_{3}\right]
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