A solid is made up of two type of atoms $X$ and $Y$ . Atoms of $X$ occupied all tetrahedral sites while atoms of $Y$ have hcp arrangement . Its formula is:

X Y

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X_{2} Y

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X Y_{2}

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X Y_{4}

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Solution

The correct option is B $X_{2} Y$
No. of atoms of $X$ occupied tetrahedral rates $= 12$
No. of atoms of $Y$ having $h c p$ arrangement $= 6$
Their ratio $= 12 : 6 = 2$ .
$∴$ Formula will be $X_{2} Y$ .
Correct answer is $B$ .

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A solid is made up of two type of atoms X and Y. Atoms of X occupied all tetrahedral sites while atoms of Y have hcp arrangement . Its formula is:

The correct option is B X2YNo. of atoms of X occupied tetrahedral rates =12No. of atoms of Y having hcp arrangement =6Their ratio =12:6=2.∴ Formula will be X2Y.Correct answer is B.

A solid is made up of two type of atoms $X$ and $Y$ . Atoms of $X$ occupied all tetrahedral sites while atoms of $Y$ have hcp arrangement . Its formula is:

The correct option is B $X_{2} Y$
No. of atoms of $X$ occupied tetrahedral rates $= 12$
No. of atoms of $Y$ having $h c p$ arrangement $= 6$
Their ratio $= 12 : 6 = 2$ .
$∴$ Formula will be $X_{2} Y$ .
Correct answer is $B$ .