Ferrous oxide has a cubic structure and each | vedclass

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(A) The density formula for a unit cell is $d = \frac{Z 	imes M}{N_A 	imes a^3}$.Here,$d = 4.0 \, g \, cm^{-3}$,$M = 72 \, g \, mol^{-1}$,$a = 5.0 \

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Four $Fe^{2+}$ and four $O^{2-}$

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(A) The density formula for a unit cell is $d = \frac{Z 	imes M}{N_A 	imes a^3}$.Here,$d = 4.0 \, g \, cm^{-3}$,$M = 72 \, g \, mol^{-1}$,$a = 5.0 \, \mathring{A} = 5.0 	imes 10^{-8} \, cm$,and $N_A = 6.022 	imes 10^{23} \, mol^{-1}$.Rearranging for $Z$:$Z = \frac{d 	imes a^3 	imes N_A}{M}$$Z = \frac{4.0 	imes (5.0 	imes 10^{-8})^3 	imes 6.022 	imes 10^{23}}{72}$$Z = \frac{4.0 	imes 125 	imes 10^{-24} 	imes 6.022 	imes 10^{23}}{72}$$Z = \frac{500 	imes 0.6022}{72} = \frac{301.1}{72} \approx 4.18 \approx 4$.Since the formula is $FeO$,the ratio of $Fe^{2+}$ to $O^{2-}$ is $1:1$. Thus,there are $4$ $Fe^{2+}$ ions and $4$ $O^{2-}$ ions per unit cell.

Ferrous oxide has a cubic structure and each edge of the unit cell is $5.0 \, \mathring{A}$. Assuming the density of the oxide is $4.0 \, g \, cm^{-3}$,the number of $Fe^{2+}$ and $O^{2-}$ ions present in each unit cell will be ($M_w$ of $FeO = 72$)

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