A) + 7
B) + 8
C) + 6
D) + 5
Correct Answer: B
Solution :
Maximum oxidation state of transition metals = number of electrons in \[(n-1)\,d\]orbital + number of electrons in \[ns\]orbital. The electronic cofiguration of \[OS=[Xe]\,4{{f}^{14}},5{{d}^{6}},6{{s}^{2}}\] \[\therefore \]Maximum oxidation state \[=6+2=8\] \[\therefore \]The highest oxidation state exhibited by transition metal is \[+\,8\,eg,\,Os{{O}_{4}}.\]
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