A) \[M{{g}_{2}}B{{e}_{3.5}}{{O}_{2.5}}\]
B) \[M{{g}_{3}}B{{e}_{3}}{{O}_{3.75}}\]
C) \[M{{g}_{3}}B{{e}_{3}}{{O}_{3.5}}\]
D) \[M{{g}_{4}}B{{e}_{4}}{{O}_{3.5}}\]
Correct Answer: B
Solution :
[b] Number of \[{{O}^{2-}}=4\] Number of \[B{{e}^{2+}}=4\] Number of \[M{{g}^{2+}}=4\] (ii) \[B{{e}^{2+}}\] and \[M{{g}^{2+}}\] are present inTVs and they are present on each body diagonal of the cube. When single body diagonal is removed, Number of comer ions \[({{O}^{2-}})\] removed \[=2\times \frac{1}{8}\] per comer share \[=\frac{1}{4}=0.25\] Number of \[{{O}^{2-}}\]left \[=4-0.25=3.75\] (iii) Number of \[B{{e}^{2+}}\] removed = 1 Number of \[B{{e}^{2+}}\] left \[=4-1=3\] (iv) Number of \[M{{g}^{2+}}\] removed = 1 Number of \[M{{g}^{2+}}\]'left =3 Thus, formula is: \[M{{g}_{3}}B{{e}_{3}}{{O}_{3.75}}\].
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