An element M forms three types of compounds with fluorine:
Compound (i) polar with complete octet Compound
(ii) polar with expanded octet Compound
(iii) non-polar with expanded octet then element 'M' can't be :

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answer is A, C, D.

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Detailed Solution

Because the Sulphur (2.58) and fluorine (3.98) atoms in the molecule have different electronegativity and the molecule has a bent geometric structure, is polar in nature. As a result, the S-F bond's dipoles do not cancel one another out, ${SF}_{2}$ and molecules become polar and contribute some dipole moment.

Around the centre atom, the geometry of ${SF}_{4}$ will have an asymmetric distribution of electron regions (S). This molecule is polar as a result.

According to the VSEPR theory, ${SF}_{6}$ is a nonpolar molecule because the six fluorine atoms aligned symmetrically with the Sulphur atom cancel out the dipole moment of the S-F bond.

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