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Dipole Moment

Boron trifluo...

Question

Boron trifluoride $(B F_{3})$ has no dipole moment $(μ = 0 D)$ . Explain how this observation confirms the geometry of $B F_{3}$ predicted by VSEPR theory.

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Solution

The $B F_{3}$ molecule has a symmetrical trigonal planar geometry, like the $S O_{3}$ molecule.

In such a structure, the resultant moment of any two $B - F$ dipoles is equal in magnitude but opposite in direction to the moment of the third one.

So, the net dipole moment of the $B F_{3}$ molecule is zero, and it is non-polar.

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Q. A reaction between ammonia and boron trifluoride is given below:

: N H_{3} + B F_{3} \to H_{3} N : B F_{3}

Identify the acid and base in this reaction. Which theory explains it? What is the hybridisation of

B

N

in the reactants?

$B F_{3} a n d N F_{3}$ both molecules are covalent, but $B F_{3}$ is non-polar and $N F_{3}$ is polar. Its reason is

Lewis's concept of covalency of an element involves octet rule. Later on, it was found that many elements in their compounds, e.g.,

B e F_{2}, B F_{3}

etc. have incomplete octet whereas

P C l_{5}, S F_{6}

, etc. have expanded octet. This classical concept also failed in predicting the geometry of molecules. The modern concept of covalence was proposed in terms of valence bond theory successfully explained the geometry of various molecules but failed in many molecules. The geometry of such molecules was explained by the VSEPR concept. Finally, the molecular orbital theory was proposed to explain many other molecules.

Which are true statements among the following?

I_{3}^{+}

has bent structure
(II) p

π -

π

bonds are present in

S O_{2}

S e F_{4}

C H_{4}

has same shape
(IV)

X e F_{2}

C O_{2}

Q. Boron trifluoride (

B F_{3}

) is a nonpolar molecule, whereas ammonia (

N H_{3}

) is a polar molecule. The difference in polarities is related to the fact that :

Q. The correct increasing order of dipole moments for the given molecules is:

(B F_{3}, N F_{3}

N H_{3})

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