The oxidation numbers of Si in Be3Al2Si2O18 and U in UO2SO4 are respectively:
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a
+6,+4
b
+4,+6
c
+2,+3
d
+3,+2
answer is B.
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Detailed Solution
In Be3Al2Si6O18:3(O.N. of Be)+2(O.N. of Al)+6(O.N. of Si)+18(O.N. of O)=03(+2)+2(+3)+6(O.N. of Si)+18(−2) =0;+6+6+6 (O.N. of Si) −36=0O.N. of Si =36−6−66=+4In UO2SO4: O.N. of U+2(O⋅N⋅of O)+O⋅N of 5042−=0∴ O.N. of U+2(−2)+(−2)=0 ; O.N. of U−4−2=0∴ O.N. of U=+4+2=+6