A) \[Os\]
B) \[Mn\]
C) \[Co\]
D) \[Cr\]
Correct Answer: A
Solution :
Maximum oxidation number exhibited by a d-block element, = number of d electrons (in penultinate shell) + number of s electrons \[Os=[Xe]\,4{{f}^{14}},5{{d}^{6}},6{{s}^{2}}\] Maximum oxidation number\[~=6+2=8\] \[Mn=[Ar]3{{d}^{5}},4{{s}^{2}}\] Maximum oxidation number \[=5+2=7\] \[Co=[Ar]3{{d}^{5}},4{{s}^{2}}\] Maximum oxidation number\[~=3+2=5\] \[Cr=[Ar]\,3{{d}^{5}},4{{s}^{1}}\] Maximum oxidation number \[=5+1=6\] Hence, Os shows the maximum oxidation state.
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