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(B) In a cubic close-packed $(CCP)$ arrangement,the number of atoms per unit cell is $4$. Since $Y$ forms the $CCP$ arrangement,the number of $Y$ atom

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$MY_2$

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(B) In a cubic close-packed $(CCP)$ arrangement,the number of atoms per unit cell is $4$. Since $Y$ forms the $CCP$ arrangement,the number of $Y$ atoms $= 4$. The number of octahedral voids in a $CCP$ arrangement is equal to the number of atoms,which is $4$. Given that $M$ occupies half of the octahedral voids,the number of $M$ atoms $= \frac{1}{2} 	imes 4 = 2$. Thus,the ratio of $M:Y$ is $2:4$,which simplifies to $1:2$. Therefore,the chemical formula of the ionic compound is $MY_2$.

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

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