Question

When borax is heated with $\text{CoO}$ on a platinum loop, blue coloured bead formed is largely due to :

• A. $\text{Co}{\left({\text{BO}}_2\right)}_2$
• B. ${\text{B}}_2{\text{O}}_3$
• C. $\text{Co}\left[{\text{B}}_4{\text{O}}_5\right(\text{OH}{)}_4]$
• D. ${\text{CoB}}_4{\text{O}}_7$

Answer 1

The correct option is A $\text{Co}{\left({\text{BO}}_2\right)}_2$

${\text{Na}}_2{\text{B}}_4{\text{O}}_7\cdot 10{\text{H}}_2\text{O}\stackrel{\Delta }{⟶}{\text{Na}}_2{\text{B}}_4{\text{O}}_7+10{\text{H}}_2\text{O}$

${\text{Na}}_2{\text{B}}_4{\text{O}}_7\stackrel{\Delta }{⟶}2{\text{NaBO}}_2$ (sodium meta borate) $+{\text{B}}_2{\text{O}}_3$

Mixture of $NaB{O}_2$ and $B_2{O}_3$ is borax bead.

${\text{B}}_2{\text{O}}_3+\text{CoO}\to \underset{\text{Blue bead}}{Co{\left(B{O}_2\right)}_2}$
(cobalt (II) meta borate)