An ionic compound is formed between a metal M and a non-metal Y. If M occupies half the octahedral voids in the cubic close-packed arrangement formed by Y, the chemical formula of the ionic compound is :

M Y

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M Y_{2}

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M_{2} Y

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M Y_{3}

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Solution

The correct option is C $M Y_{2}$
Number of atoms of $Y = \frac{1}{8} \times 8 + \frac{1}{2} \times 6 = 4$

Number of atoms of $M = \frac{1}{2} \times 4 = 2$

Formula: $M Y_{2}$

Hence option $B$ is correct.

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An ionic compound is formed between a metal M and a non-metal Y. If M occupies half the octahedral voids in the cubic close-packed arrangement formed by Y, the chemical formula of the ionic compound is :

The correct option is C MY2Number of atoms of Y=18×8+12×6=4Number of atoms of M=12×4=2Formula: MY2Hence option B is correct.

The correct option is C $M Y_{2}$
Number of atoms of $Y = \frac{1}{8} \times 8 + \frac{1}{2} \times 6 = 4$

Number of atoms of $M = \frac{1}{2} \times 4 = 2$

Formula: $M Y_{2}$

Hence option $B$ is correct.