Question

What is the action of following on boron $(B)$ ?
A.Fused with $NaOH$
B.Conc. $HN{O_3}$

Answer 1

Boron is a chemical element with an atomic number $5$ . It is present in very low amounts. It is generally produced using cosmic ray spallation techniques. It is one of the atoms which reacts with silica, a rare chemical reaction.

Complete answer:
Nitric acid is a strong oxidizing agent and it reacts vigorously with many non- metallic compounds including boron $(B)$ .
So let’s look at our reaction i.e. reaction of boron $(B)$ with fused sodium hydroxide $NaOH$
So, the complete balanced reaction is:
$B + 6NaOH \to 2N{a_3}B{O_3} + 3{H_2}$
Hence, boron reacts with fused sodium hydroxide to produce sodium borate and the hydrogen gas is evolved out.
Now, coming to our next option i.e. when boron reacts with concentrated nitric acid. We know that conk nitric oxide is a strong oxidising agent so, it will reduce itself and will oxidize boron.
So, have a look at the complete reaction:
$B + HN{O_3} + {H_2}O \to {H_3}B{O_3} + NO$
So, in the reaction between boron and conc. Nitric acid, boron is oxidised to boric acid $({H_3}B{O_3})$ and nitric acid is reduced to nitric oxide $(NO)$ . Well, we can say that this is a redox reaction.

Note:
Nitric acid is very popular for its use in the production of explosives like TNT and nitro-glycerine. However, nitric acid is also useful in the production of many nitrogen based fertilizers and in the production of various plastics. One important fact is that Nitric acid is also useful in the formation of the Aqua regia in combination with hydrochloric acid. Aqua regia is actually a powerful reagent that removes or dissolves the precious metals like gold, platinum and palladium. It is used to remove gold, platinum and palladium from substrates such as micro fabrications and microelectronics labs.