Chemistry, asked by nnagababu726pb8ss4, 9 months ago

plz answer this question guys with explanation ​

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Answered by rocky200216
2

Option 1 : “3d³ 4s² ” is the configuration of “V” (vanadium) . Oxidation state's of “V”=+2,+3,+4,+5.

Option 2:“3d5 4s¹” is the configuration

of “Cr" . Oxidation state's of “Cr" = +2,+3,+4,+5,+6.

Option 3: “3d5 4s²” is the configuration of “Mn".

Oxidation state's of “Mn” = +2,+3,+4,+5,+6,+7.

Option 4: “3d6 4s²” is the configuration of “Fe".

Oxidation state's of “Fe" =+2,+3,+4,+5,+6.

So from above “Mn" have highest Oxidation state.

Correct option: “C".

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