A molecule with the formula AX_4Y has all its | vedclass

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(A) Given that $A$ is the rarest,monoatomic,non$-$radioactive $p-$block element that forms an $AX_4Y$ type of molecule,it is concluded that $A$ is $Xe

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Square pyramidal

Vedclass · Answer

(A) Given that $A$ is the rarest,monoatomic,non$-$radioactive $p-$block element that forms an $AX_4Y$ type of molecule,it is concluded that $A$ is $Xe$.Given that the electronegativity of $A$ is less than $X$ and $Y$,and that $X$ and $Y$ have the highest and second highest electronegativity values among all elements,$X$ and $Y$ are $F$ and $O$ respectively.Thus,the compound is $XeOF_4$.In $XeOF_4$,the central atom $Xe$ has $8$ valence electrons. It forms $4$ single bonds with $F$ atoms and $1$ double bond with the $O$ atom,leaving $1$ lone pair.The steric number is $5 + 1 = 6$,which corresponds to $sp^3d^2$ hybridization with an octahedral electron geometry.Due to the presence of one lone pair,the molecular shape is square pyramidal.

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