IronIIoxide crystallise in cubic structure with unit cell edge length of 5 A- If the density of oxide is 3-8 g cm-3- Calculate the number of Fe2- ions present in each unit cell- Molar mass of Fe-56 g-mol -

Given: Edge length, a= nbsp;5 A^∘=5 × 10^ 8cm Density, D = nbsp;3.8 g cm^ 3 Molecular mass of FeO =56+16=72 u We have the equation, D= ZMN_Aa^3 nbsp; N_A ...

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Standard XII

Chemistry

Ionic Compounds and Radius Ratios

IronIIoxide c...

Question

Iron $(II)$ oxide crystallise in cubic structure with unit cell edge length of $5 A^{\circ}$ . If the density of oxide is $3.8 {g cm}^{- 3}$ . Calculate the number of $F e^{2 +}$ ions present in each unit cell.
[ Molar mass of $F e = 56 g/mol$ ]

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Solution

The correct option is D $4$
Given:
Edge length, $a$ = $5 A^{\circ} = 5 \times 10^{- 8} cm$
Density, $D$ = $3.8 {g cm}^{- 3}$
Molecular mass of $F e O = 56 + 16 = 72 u$
We have the equation,
$D = \frac{Z M}{N_{A} a^{3}}$
$N_{A} \to 6.023 \times 10^{23}$
$Z \to Number of F e O per unitcell$
Substituting the values,
$Z = \frac{D N_{A} a^{3}}{M} = \frac{3.8 \times 6.023 \times 10^{23} \times (5 \times 10^{- 8})^{3}}{72} = 4$
So there are four $F e O$ per unit cell. Hence $4 F e^{2 +}$ per unit cell.

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Standard XII Chemistry

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Iron(II)oxide crystallise in cubic structure with unit cell edge length of 5 A∘. If the density of oxide is 3.8 g cm−3. Calculate the number of Fe2+ ions present in each unit cell. [ Molar mass of Fe=56 g/mol ]

Iron $(II)$ oxide crystallise in cubic structure with unit cell edge length of $5 A^{\circ}$ . If the density of oxide is $3.8 {g cm}^{- 3}$ . Calculate the number of $F e^{2 +}$ ions present in each unit cell.
[ Molar mass of $F e = 56 g/mol$ ]