In one of my books it is written that boron does not react with alkali but in another book it is written that it reacts with alkali.
Can any one help in clearing this confusion?
In one of my books it is written that boron does not react with alkali but in another book it is written that it reacts with alkali.
Can any one help in clearing this confusion?
Boron react with $\ce{NaOH}$ but under certain conditions. Boron exist in two form (1) crystalline form (2) amorphous form. Both forms are generally inert to chemical reactions and is unaffected by acids and alkali at normal conditions but at elevated temperature, concentrated oxidizing acids like nitric or sulfuric acids react with boron to yield boric acid. On the other hand, amorphous boron does react with fused alkali at high temperature to obtain trisodium borate:
$$\ce{2B(s) + 6NaOH(s) ->[t > 773K] 2Na3BO3(s) + 3H2(g) ^}$$
High purity, hydrated sodium metaborate(sodium tetrahydroxoborate, $\ce{Na[B(OH)4]}$) can also be produced from boron in a $\ce{NaOH-H2O}$ system. The process applied was called pressure alkaline leaching.
When fused in the presence of air at $\pu{350-400 ^\circ C}$, sodium metaborate is formed. You can find the reaction in this answer by @andselisk.
References
Your reaction is far from balanced but what you want seems to be trisodium borate.
According to this source (emphasis mine),
Boron reacts with fused sodium hydroxide to form sodium borate and hydrogen.
The cited reaction (already balanced) is the following:
$$\ce{2 B + 6 NaOH -> 2 Na3BO3 + 3 H2 ^}$$
This other source says something similar (again, emphasis mine),
It can also dissolve in alkalies and evolve hydrogen.
So, as it seems, boron reacts not with sodium hydroxide solutions, but with the pure, molten compound. This may be the source of confusion.