A) \[1F{{e}^{2+}}\] and \[1S_{2}^{2-}\]
B) \[2F{{e}^{2+}}\]and \[2S_{2}^{2-}\]
C) \[4F{{e}^{2+}}\]and \[4S_{2}^{2-}\]
D) \[6F{{e}^{2+}}\]and \[6S_{2}^{2-}\]
Correct Answer: C
Solution :
\[d=\frac{ZM}{{{a}^{3}}{{N}_{A}}}\] \[5.02=\frac{Z\times 120}{6.023\times {{10}^{23}}\times {{(5.4\times {{10}^{-8}})}^{3}}}\] Z = 4 i.e., \[4F{{e}^{2+}}\] and \[4S_{2}^{2-}\] ions are present in a single unit cell.
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