JEE Main & Advanced Chemistry The Solid State / ठोस प्रावस्था Sample Paper Topic Test - Solid State

  • question_answer
    Ferrous oxide has a cubic structure and each edge of the unit cell is \[5.0\text{ }{{A}^{o}}\]. Assuming density of the oxide as \[4.0\text{ }g\text{ }c{{m}^{-3}}\] the number of \[F{{e}^{2+}}\] and \[{{O}^{2-}}\] ions present in each unit cell will be

    A) two \[F{{e}^{2+}}\] and four \[{{O}^{2-}}\]

    B) three \[F{{e}^{2+}}\] and three \[{{O}^{2-}}\]

    C) four \[F{{e}^{2+}}\] and two \[{{O}^{2-}}\]

    D) four \[F{{e}^{2+}}\] and four \[{{O}^{2-}}\]

    Correct Answer: D

    Solution :

    [d] Let the units of ferrous oxide in a unit cell = n
    Molecular weight of ferrous oxide \[(FeO)\]
    \[=56+16=72\text{ }g\text{ }mo{{l}^{-1}}\]
    Weight of n units \[=\frac{72\times n}{6.023\times {{10}^{23}}}\]
    Volume of one unit \[\text{= (length of corner}{{\text{)}}^{\text{3}}}\]
    \[={{(5{{A}^{0}})}^{3}}=125\times {{10}^{-24}}c{{m}^{3}}\]
    \[\text{Density =}\frac{\text{wt}\text{.of cell}}{\text{volume}}\]
    \[\therefore \]      \[4.09=\frac{72\times n}{6.023\times {{10}^{23}}\times 125\times {{10}^{-24}}}\]
    Hence, \[n=\frac{3079.2\times {{10}^{-1}}}{72}=42.7\times {{10}^{-1}}\]
    \[=4.27=4\]
    [d] They are comparatively soft and not very rigid.


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