The empirical formula for a compound with a c | vedclass

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(B) In a cubic close packed $(CCP)$ arrangement,the number of anions $(A^-)$ per unit cell is $4$.The number of octahedral voids in a $CCP$ lattice is

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$1$

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(B) In a cubic close packed $(CCP)$ arrangement,the number of anions $(A^-)$ per unit cell is $4$.The number of octahedral voids in a $CCP$ lattice is equal to the number of atoms forming the lattice,which is $4$.Since cations $(B^+)$ occupy all octahedral sites,the number of cations per unit cell is $4$.The formula of the unit cell is $A_4B_4$,which simplifies to the empirical formula $AB$.Comparing $AB$ with $A_xB$,we get $x = 1$.

The empirical formula for a compound with a cubic close packed arrangement of anions and with cations occupying all the octahedral sites is $A_{x}B$. The value of $x$ is ..... .
(Integer answer)

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The empirical formula for a compound with a cubic close packed arrangement of anions and with cations occupying all the octahedral sites is $A_{x}B$. The value of $x$ is ..... .(Integer answer)