A) super oxide
B) peroxide
C) dioxide
D) oxide
E) Nacent oxygen
Correct Answer: B
Solution :
\[[\underset{m}{\mathop{\underset{\downarrow }{\mathop{Co}}\,}}\,\underset{0}{\mathop{\underset{\downarrow }{\mathop{{{(N{{H}_{3}})}_{2}}}}\,}}\,\underset{n}{\mathop{\underset{\downarrow }{\mathop{{{O}_{2}}}}\,}}\,\underset{0}{\mathop{\underset{\downarrow }{\mathop{(en)}}\,}}\,\,\underset{-1}{\mathop{\underset{\downarrow }{\mathop{Cl}}\,}}\,]\] Let us try hit and trial method \[m+0+n+0-1=0\] Generally, Co shows + 2, + 3 oxidation state \[\therefore \] If \[m=+\,2\] then \[y=-1\] and EAN = 27 - 2 + 12 = 37 If x = + 3 then y = - 2 EAN = 27 - 3 + 12 = 36
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