Explanation
The number of lone pairs of electron on $$Xe$$ in $$XeO{F}_{4}$$ is 1. $$Xe$$ in $$XeO{F}_{4}$$ has $${ sp }^{ 3 }{d}^{2}$$-hybridization having one lone pair on $$Xe$$-atom and is square pyramidal in shape.
The ratio of lone pair and bond pair electrons on central $$I$$ atom in $${I}_{3}^{-}$$ is $$1.5$$.
Thus, No. of lone pair of electrons $$=3$$
No. of bond pair of electrons $$=2$$
The ratio of lone pair and bond pair electrons $$= 3/2 = 1.5$$
In the cyanide ion the formal negative charge resonates between $$C$$ and $$N$$. Cyanide ion has resonating structure, $$-\overset { \_ }{ C } \equiv N\longleftrightarrow \overset { \_ }{ N } \underrightarrow { = } C$$
Explanation:
$$ Structure \space of \space ClO_{3}^{-}$$$$ClO_3^-$$ has central $$Cl$$ as $$sp^3$$ hybridized and $$1$$ lone pair present on the central $$Cl$$.It is trigonal pyramidal in shape.
$$Structure \space of \space XeF_4$$$$XeF_4$$ has central $$4 \ Xe$$ as $$sp^3d^2$$ hybridisation and there are $$2$$ lone pair present on the central $$Xe$$.It has square planar shape.
$$Structure of SF_4$$$$SF_4$$ has central $$S$$ as $$sp^3d$$ hybridization and there is $$1$$ lone pair present on the central $$S$$.It has a distorted tetrahedral shape.
$$Structure \space of \space I_{3}^{-}$$$$I_3^-$$ has central $$I$$ having $$sp^3d$$ hybridization and there are $$3$$ lone pair present on the central $$I$$.It has linear shape.So, $$I_3^-$$ has maximum lone pair present on the central atom.
Correct Option : $$D$$.
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