# P block elements boron family

I am confused that in one of my book it is written that boron does not react with alkali but in other book it is written that it reacts with alkali

Can any one help in clearing this confusion I would be very thankful

• Could you show us those reactions? – Felipe S. S. Schneider Jan 31 '17 at 11:19
• When boron reacts with alkali the reaction is this. 3B+NaOH -> 2Na3BO3+3H2O – Sarla Riya Mishra Jan 31 '17 at 12:10
• That equation shouldn't make sense to you. Boron is being oxidised from 0 to +3 but nothing is being reduced. – orthocresol Jan 31 '17 at 13:14

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.

Yes, boron reacts with $\ce{NaOH}$ but only in fused state as noted by @Felipe Schneider. In a different link, it has stated that:

Boron dissolves only in fused alkali. However, aluminum and gallium dissolve in fused as well as the aqueous forms of alkali. Indium remains inert with alkalis.

$$\ce{2B + 6NaOH (fused) → 2Na3BO3 + 3H2}$$

But apart from fused state, boron also reacts with $\ce{NaOH}$ in aqueous medium and also in presence of oxygen.

$$\ce{2B + 2NaOH + 6H2O → 2Na[B(OH)4] + 3H2}$$

Boron react with sodium hydroxide and water to produce sodium tetrahydroxoborate(III) and hydrogen. Boron is amorphous. Sodium hydroxide - concentrated solution.(chemiday 1)

$$\ce{2B + 2NaOH + 2H2O ->[\Delta] 2NaBO2 + 3H2}$$

Amorphous boron react with sodium hydroxide and water to produce sodium metaborate and hydrogen. Sodium hydroxide - concentrated solution. This reaction takes place heating the reaction mixture.(chemiday 2)

$$\ce{4B + 4NaOH + 3O2 ->[350-400 C] 4NaBO2 + 2H2O}$$

Boron react with sodium hydroxide and oxygen to produce sodium metaborate and water. This reaction takes place at a temperature of 350-400°C.(chemiday 3)

• Thanx Felipe and Nilay it is a great help for me – Sarla Riya Mishra Jan 31 '17 at 15:39