Question

Sodium borohydride, NaBH4, and boron trifluoride, BF3, are compounds of boron. What are the shapes around boron in the borohydride ion and in boron trifluoride?

Answer 1

Compound 1: Sodium borohydride
In sodium borohydride (NaBH4), B is a central metal.
Electronic configuration of B is 1s2 2s2 2p1.
B undergoes sp3 hybridization in NaBH4, to generate 4 hybrid orbitals. These hybrid orbitals, forms sigma bond with 4 'H' atoms. Due to this, the structure of sodium borohydride in tetrahedral.
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Compound 2: Boron trifluoride
In boron trifluoride (BF3), B is a central metal.
Electronic configuration of B is 1s2 2s2 2p1.
B undergoes sp2 hybridization in NaBH4, to generate 3 hybrid orbitals. These hybrid orbitals, forms sigma bond with 3 'H' atoms. Due to this, the structure of boron trifluoride is triangular planner.

Answer 2

Answer: The shape of boron in borohydride ion $BH_4^-$ is tetrahedral and in boron trifluoride $BF_3$ is trigonal planar.

Explanation: Formula used

$:{\text{Number of electron pairs}} =\frac{1}{2}[V+N-C+A]$

where, V = number of valence electrons present in central atom

N = number of monovalent atoms bonded to central atom

C = charge of cation

A = charge of anion

1. $BH_4^{-}:$

${\text{Number of electrons}} =\frac{1}{2}[3+4-0+1]=4$

The number of electron pairs is 4 that means the hybridization will be $sp^3$ and the electronic geometry of the molecule will be tetrahedral.

2. $BF_3:$

${\text{Number of electron pairs}} =\frac{1}{2}[3+3-0+0]=3$

The number of electron pairs is 3 that means the hybridization will be $sp^2$ and the electronic geometry of the molecule will be trigonal planar.