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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