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