Answer:
Explanation:
For a compound to have at least one polar covalent bond, but being a not polar compound, the geometric configuration of the molecule must lead to a symmetrical distribution of the charges. This is because, in that way, the dipoles will cancel each other, producing a null net dipole, which makes a non-polar molecule.
Consider the four given molecules:
a. SeBr₄
Since, Se has six valence electrons, and it is bonded to four Br atoms, there will be one lone electron pair which will make that the charge is not symmetrically distributed, leading to a polar molecule. so this is not correct choice.
b. HCN
HCN's structure is linear: H-C≡N.
Nitrogen is more electronegative than carbon an so it is pulling the electrons on in the triple bond with carbon (C≡N) more than the hydrogen on the other side of carbon (H-C)
Therefore, there is a net a dipole moment towards N and the molecule is polar.
c. ICl₃
This is quite evident non-symmtrical and so you quickly can predict that is is polar.
d. CF₄
This molecule has four F atoms around a C atom.
Since, F is more electronegative than F, each C - F bond has a dipole moment (negative charge is closer to F atoms). Then, whe have four dipole moments (covalent bonds) on each CF₄ molecule.
Now, since the molecule is symmetric the four dipoles will cancel each other and the compund in nonpolar.
In conclusion, the correct choice is the letter d. CF₄ is a compound that contains at least one polar covalent bond (four in fact), bust is nonpolar.
The compound that contains at least one polar covalent bond, but is nonpolar is CF4
From the above we can therefore known that CF4 which carbon and fluorine that differ in their electronegativity is non polar.
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