p-Block elementsEasy
Question
In [Cr(O2)(NH3)4H2O]Cl2 ; oxidation number of Cr is + 3, then oxygen will be in the form :-
Options
A.Dioxo
B.Peroxo
C.Superoxo
D.Oxo
Solution
Let the oxidation number of oxygen is x
+ 3 + 2x + 0 x 4 + 0 - 2 = 0
2x + 1 = 0 ⇒ 2x = - 1 [x = - 1/2]
Thus, oxygen will be in superoxoform
+ 3 + 2x + 0 x 4 + 0 - 2 = 0
2x + 1 = 0 ⇒ 2x = - 1 [x = - 1/2]
Thus, oxygen will be in superoxoform
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