Question

$Mn$ shows the highest oxidation state of $+7$ with oxygen but with fluorine it shows the highest oxidation state of $+4$.

• A. True
• B. False

Answer 1

The correct option is A True

1)In case of oxygen, $Mn$ shows the highest oxidation state of $+7$. This is because $Mn$ forms $p\pi –d\pi$ multiple bonds using $2p$ orbitals of oxygen and $3d$ orbitals of $Mn$.

2) With $F,Mn$ displays an oxidation state of $+4$ because of the single bond formation caused by the unavailability of $2p$ orbitals in $F$ for multiple bonding.