Question

Out of \[S{F_4},Cl{F_3},Xe{F_4}\] and \[{H_2}O\] which one is non- polar?

Answer 1

Hint: In the case of polar compounds, which have defined regions of negative and positive charge. And the polar covalent bond is a bond which is formed between two non- identical atoms. In the case of non- polar compounds, the charge distribution is symmetrical and it does not exhibit polarity. Hence, the non- polar covalent bond formed by the electrons is shared equally between two molecules.

Complete answer:
Among the given compounds, xenon tetrafluoride is non- polar. Because, the bonds present in \[Xe{F_4}\] are symmetrically arranged. And the net effective dipole moment of the compound is equal to zero. And it has a symmetrical square planar structure. Hence, \[Xe{F_4}\] is non – polar in nature. But the individual xenon – fluorine bond is polar. Because, the electronegativity of xenon and fluorine is different.
But in the case of \[S{F_4},Cl{F_3},{H_2}O\], they are non- polar in nature. Because \[S{F_4}\] contains a lone pair of electrons, and the shape of the compound becomes asymmetric. And in water, there is an unequal distribution of charge. The \[Cl{F_3}\] have two lone pairs and it is asymmetric in shape. Hence, these three compounds are polar in nature.

Note:
In the case of polar compounds, there is an unequal distribution of charge, and the lone pair of electrons is present in the compound. Therefore the compound becomes polar in nature. But in the non- polar compounds, there are no lone pairs and the charge distribution is equal. Therefore, the compound becomes non- polar in nature. And the net electrical charge is zero in non- polar compounds.