Ferrous oxide has a cubic structure and each | vedclass

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(A) The density formula for a unit cell is given by: $d = \frac{Z 	imes M}{N_A 	imes a^3}$.Here,$d = 4.0 \ g \ cm^{-3}$,$M (FeO) = 56 + 16 = 72 \ g

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

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(A) The density formula for a unit cell is given by: $d = \frac{Z 	imes M}{N_A 	imes a^3}$.Here,$d = 4.0 \ g \ cm^{-3}$,$M (FeO) = 56 + 16 = 72 \ g \ mol^{-1}$,$a = 5.0 \ \mathring{A} = 5.0 	imes 10^{-8} \ cm$,and $N_A = 6.022 	imes 10^{23} \ mol^{-1}$.Substituting the values: $4.0 = \frac{Z 	imes 72}{6.022 	imes 10^{23} 	imes (5.0 	imes 10^{-8})^3}$.$4.0 = \frac{Z 	imes 72}{6.022 	imes 10^{23} 	imes 125 	imes 10^{-24}}$.$4.0 = \frac{Z 	imes 72}{0.075275}$.$Z = \frac{4.0 	imes 0.075275}{72} \approx 4.18 \approx 4$.Since the stoichiometry of $FeO$ is $1:1$,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:

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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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