Suggest a reason why the B - F bond lengths in BF3 (130 pm) and�
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What happens when:
BF3 is treated with ammonia?


BF3 being a Lewis acid accepts a pair of an electron from NH3 to form the corresponding compound.


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If B – Cl bond has a dipole moment, why does BCl3 has zero dipole moment ?


B and Cl have different electronegativities and chlorine (E.N. = 3.00) is more electronegative than B(E.N.  = 2.00). As a result B – Cl bond is polar and hence has a finite dipole moment. Now BCl3 is a planar molecule in which the three B – Cl bonds are inclined at an angle of 120°. Therefore, the resultant of two B – Cl bonds in cancelled by equal and opposite dipole moment of the bond B – Cl bond as shown.


Hence overall dipole moment of BCl3 is zero.

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Suggest a reason why the B - F bond lengths in BF3 (130 pm) and BF subscript 4 superscript minus (143 pm) differ?
Or
Why B - F bond length in BF3 is smaller than the expected value?


In BF3, boron is sp2 hybridised and therefore BF3 is a planar molecule. It has a vacant 2p-orbital. F-atom has three lone pairs of electrons. In BF3 molecule, one 2p-orbital of fluorine atom overlaps sidewise with empty 2p-orbtial of boron to form  back bonding (back donation) in which the lone pair is transferred from F to B as shown.



As a result of this back bonding (or black donation), the B-F bond acquires some double bond character. 
On the other hand in  ion,  boron is sp3 hybridised and therefore  is a tetrahedral molecule. B in  ion does not have vacant p-orbital available to accept the electrons donated by the F atom. Hence  ion, B -F is a purely single bond. Since double bonds are shorter than single bonds, therefore B-F bond length in BF3 is shorter (130 pm) than B-F bond length (143 pm) in  [BF4]–.

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What happens when:
Aluminium is treated with dilute NaOH.


Aluminium reacts with dilute NaOH and liberates dihydrogen.

  

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What happens when:
Hydrated alumina is treated with aqueous NaOH solution?


Alumina dissolves in aqueous NaOH solution to form sodium meta-aluminate.

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