A) \[Mn\]
B) \[Cu\]
C) \[Sn\]
D) \[Sc\]
Correct Answer: A
Solution :
Number of oxidation states = unpaired\[{{e}^{-}}\]in \[(n-1)d\]orbital + ns electrons \[_{25}Mn=[Ar]3{{d}^{5}},4{{s}^{2}}\] Oxidation states\[=5+2=7\] \[_{29}Cu=[Ae]3{{d}^{10}},4{{s}^{1}}\] Oxidation states \[=0+1=1\] \[_{50}Sn=[Kr]4{{d}^{10}},5{{s}^{2}},5{{p}^{2}}\] Oxidation states =4 \[_{21}Sc=[Ar]3{{d}^{1}},4{{s}^{2}}\] Oxidation states \[=1+2=3\] Hence, the oxidation states is maximum for \[Mn\](ie, 7 ).
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