The highest oxidation state of manganese in f | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(D) Step $1$: Manganese can exhibit a maximum oxidation state of $+7$ by using all its valence electrons $(3d^5 4s^2)$.Step $2$: Oxygen has the unique

Vedclass · Accepted Answer

In covalent compounds, fluorine can form single bond only, while oxygen forms double bond

Vedclass · Answer

(D) Step $1$: Manganese can exhibit a maximum oxidation state of $+7$ by using all its valence electrons $(3d^5 4s^2)$.Step $2$: Oxygen has the unique ability to form multiple bonds (specifically $p\pi - d\pi$ multiple bonds) with transition metals, which stabilizes the higher oxidation states of the metal.Step $3$: Fluorine, being highly electronegative and small, can only form single covalent bonds $(Mn-F)$. It cannot form multiple bonds with manganese.Step $4$: Therefore, oxygen can satisfy the high valency of manganese in $Mn_2O_7$ through multiple bonding, whereas fluorine is restricted to $MnF_4$.

The highest oxidation state of manganese in fluorides is $+4$ $(MnF_4)$, but the highest oxidation state in oxides is $+7$ $(Mn_2O_7)$, because

Similar Questions

Addition of high proportions of manganese makes steel useful in making rails of railroads,because manganese

All $Cu(II)$ halides are known except $CuI_2$,because ....

The highest Mn fluoride is $MnF_4$ whereas the highest oxide is $Mn_2O_7$, because . . . . . . .

The highest oxidation state of manganese in fluoride is $+4$ $(MnF_4)$,but the highest oxidation state in oxides is $+7$ $(Mn_2O_7)$ because __

The nature of oxides $V_{2}O_{3}$ and $CrO$ is indexed as '$X$' and '$Y$' type respectively. The correct set of $X$ and $Y$ is:

Vedclass Test Series

Exam Paper Generator

Online Exam Module