Statement 1: The highest manganese fluoride is MnF4 and the highest oxide is Mn2O7? Statement 2: In Mn2O7, each Mn is tetrahedrally surrounded by Os including Mn-O-Mn bridge

Question

Statement 1: The highest manganese fluoride is MnF4 and the highest oxide is Mn2O7? Statement 2: In Mn2O7, each Mn is tetrahedrally surrounded by Os including Mn-O-Mn bridge (A) If both Statement 1 and Statement 2 are true and the statement 2 is the correct explanation o f the statement 1. (B) If both Statement 1 and Statement 2 are true but statement 2 is not the correct explanation o f the statement 1. (C) If Statement 1 is true but Statement 2 is false. (D) If Statement 1 is false but Statement 2 is true

Tardigrade Admin · Accepted Answer

Manganese with fluorine forms MnF4 in highest oxidation state because of high electronegativity of fluorine <img class="img-fluid question-image" alt="image" src="https://cdn.tardigrade.in/img/question/chemistry/f386af4e0416ca2f8003328c8eebc3a1-.png" /> Oxygen forms multiple bonds to metals which explain its superiority over fluorine.

Q. Statement 1: The highest manganese fluoride is $M n F_{4}$ and the highest oxide is $M n_{2} O_{7} ?$
Statement 2: In $M n_{2} O_{7},$ each $M n$ is tetrahedrally surrounded by $O s$ including $M n - O - M n$ bridge

If both Statement 1 and Statement 2 are true and the statement 2 is the correct explanation o f the statement 1.

If both Statement 1 and Statement 2 are true but statement 2 is not the correct explanation o f the statement 1.

If Statement 1 is true but Statement 2 is false.

If Statement 1 is false but Statement 2 is true

Manganese with fluorine forms $M n F_{4}$ in highest oxidation state because of high electronegativity of fluorine

Oxygen forms multiple bonds to metals which explain its superiority over fluorine.

Q. Statement 1: The highest manganese fluoride is MnF4​ and the highest oxide is Mn2​O7​? Statement 2: In Mn2​O7​, each Mn is tetrahedrally surrounded by Os including Mn−O−Mn bridge