Experts can i please have the answer for the question below 'OR'

Dear Student,

(i) Structure of B₂H₆ molecule : 


Industrial preparation of diborane :
$2{\mathrm{BF}}_3 + 6\mathrm{NaH} \stackrel{180°\mathrm{C} }{\to } {\mathrm{B}}_2{\mathrm{H}}_6 + 6\mathrm{NaF}$

(ii) It should be BF₄^- instead of BH₄^-. 
In BF_​3 boron has sp² hybridisation and is planar molecule. The F atom has 3 lone pairs of electrons in the 2p orbital. Due to the small size of fluorine atom and strong repulsions pπ-​pπ takes place. Due to this back bonding,B-F acquires some double bond character. 
In BF_​4^- boron has sp³ hybridisation and all B-F bonds are purely single bond. There is no back bonding in this case. The double bonds have shorter length than the single bond, therefore, the bond length in BF₃ is 130 ppm while in BF₄^- it is 143 ppm. 

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