The dipole moment of BF(3) is zero becau...

The dipole moment of `BF_(3)` is zero because

The electronegativity difference between boron and fluorine molecule

It is a covalent molecule

It is a tetra atomic molecule

It is having trigonal planar geometry.

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Covalent molecules formed by heteroatoms bound to have some ionic character. The ionic character is due to shifting of the electron pair towards A or B in the molecule AB. Hence, atoms acquire small and equal charge but opposite in sign. Such a bond which has some ionic character is described as polar covalent bond. Polar covalent molecules can exhibit dipole moment. Dipole moment is equal to the product of charge separation, q and the bond length, d for the bond. The unit of dipole moment is Debye. One Debye is equal to 10^(-18) esu cm. Dipole moment is a vector quantity. It has both magnitude and direction. Hence, dipole moment of molecules depends upon the relative orientation of the bond dipoles, but not on the polarity of bonds alone. A symmetrical structure shows zero dipole moment. Thus, dipole moments help to predict the geometry of the molecules. Dipole moment values can be used to distinguish between cis-and traps-isomers, ortho-, meta-and para-forms of a substance, etc. The percentage of ionic character of a bond can be calculated by the application of the following formula : % " ionic character " = ("Experimental value of dipole moment ")/("Theoretical value of dipole moment ") xx 100 The dipole moment of NF_(3) is very much less than that of NH_(3) because :

Number of lone pairs in `NF_(3)` is much less greater than in `NH_(3)`

unshared electron pair is not present in NF as in `NH_(3)`

both have different shapes

of different directions of moments of N-H and N-F bonds

1, 3

1, 2

2, 3

1, 2,3

Both (A) and (R) are true and (R) is the correct explanation of (A)

Both (A) and (R) are true and (R) is not the correct explanation of (A)

(A) is true but (R) is false

(A) is false but (R) is true