The atoms of element Y from- hexagonal close packing and the atoms of element X occupies 2-3 portion of the number of tetrahedral voids- Write the formula of the compound formed by X and Y-

The correct option is A X_4 Y_3 Z_effective of Y=6(HCP) Z_effective of X=12×23=8 Number of tetrahedral void in HCP =12 #160; ...

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Standard XII

Chemistry

Close Packing

The atoms of ...

Question

The atoms of element Y from, hexagonal close packing and the atoms of element X occupies $\frac{2}{3}$ portion of the number of tetrahedral voids. Write the formula of the compound formed by X and Y.

X_{4} Y_{3}

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X_{2} Y_{1}

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X_{3} Y_{4}

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X_{4} Y_{5}

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Solution

The correct option is A $X_{4} Y_{3}$
$Z_{e f f e c t i v e}$ of $Y = 6 (H C P)$
$Z_{e f f e c t i v e}$ of $X = 12 \times \frac{2}{3} = 8$
Number of tetrahedral void in HCP $= 12$
Formula: $X_{8} Y_{6} = X_{4} Y_{3}$

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The atoms of element Y from, hexagonal close packing and the atoms of element X occupies 23 portion of the number of tetrahedral voids. Write the formula of the compound formed by X and Y.

The correct option is A X4Y3Zeffective of Y=6(HCP)Zeffective of X=12×23=8Number of tetrahedral void in HCP =12 Formula: X8Y6=X4Y3

The atoms of element Y from, hexagonal close packing and the atoms of element X occupies $\frac{2}{3}$ portion of the number of tetrahedral voids. Write the formula of the compound formed by X and Y.

The correct option is A $X_{4} Y_{3}$
$Z_{e f f e c t i v e}$ of $Y = 6 (H C P)$
$Z_{e f f e c t i v e}$ of $X = 12 \times \frac{2}{3} = 8$
Number of tetrahedral void in HCP $= 12$
Formula: $X_{8} Y_{6} = X_{4} Y_{3}$