Aluminium trifluoride is insoluble in anhydrous $HF$ but dissolves on addition of $NaF$. Aluminium trifluoride precipitates out of the resulting solution when gaseous $BF_3$, is bubbled through. Give reasons.
Hydrogen fluoride ($HF$) is a covalent compound and has a very strong intermolecular hydrogen-bonding. Thus, it does not provide ions and aluminium fluoride ($AlF$) does not dissolve in it.
Sodium fluoride (NaF) is an ionic compound and when it is added to the mixture, AlF dissolves. This is because of the availability of free $\mathrm{F}^{-}$. The reaction involved in the process is:
$3 \mathrm{NaF}+\mathrm{AlF}_{3} \rightarrow \mathrm{Na}_{3}\left[\mathrm{AlF}_{6}\right]$
Soluble complex
When boron trifluoride $\left(\mathrm{BF}_{3}\right)$ is added to the solution, aluminium fluoride precipitates out of the solution. This happens because the tendency of boron to form complexes is much more than that of aluminium. Therefore, when $\mathrm{BF}_{3}$ is added to the solution, $B$ replaces Al from the complexes according to the following reaction.
$\mathrm{Na}_{3}\left[\mathrm{AlF}_{6}\right] 3 \mathrm{BF}_{3} \rightarrow 3 \mathrm{Na}\left[\mathrm{BF}_{6}\right]+\mathrm{AlF}_{3}$
Aluminium is more reactive than iron. But aluminium is less easily corroded than iron because
In diborane, the two $H - B - H$ angles are nearly
Outermost electronic configuration of a group $13$ element, $E$, is $4 \,\mathrm{~s}^{2}, 4 p^{1}$. The electronic configuration of an element of $p$-block period-five placed diagonally to element, $E$ is :
The gaseous product $(s)$ expected at room temperature by reaction of sodium borohydride and boron trifluoride under anhydrous conditions is/are:
Mention isotops of boron.