A) 1
B) 6
C) 2
D) 5
Correct Answer: A
Solution :
\[CO\] is 2 electron donor and \[NO\] is 3electron donor so maximum 6 CO replaced by 4 \[NO\]. \[Cr{{(CO)}_{6}}\]\[6CO\] are present number of \[CO\] molecule \[\times 2=\] number of NO molecule \[\times \,36\times 2=x\times 3\] (\[x=\] number of NO molecule) \[x=4\]You need to login to perform this action.
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