Find the n-factor in the following chemical changes.
$K M n O_{4} O H^{-} - --- \to M n^{6 +}$

1

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

4

No worries! We‘ve got your back. Try BYJU‘S free classes today!

2

No worries! We‘ve got your back. Try BYJU‘S free classes today!

3

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Open in App

Solution

The correct option is A $1$
Steps for finding valency factor/n-factor:
1. Find the oxidation state (O.S.) of metal in reactant side.
2. Find the oxidation state of metal in product side.
3. Take the magnitude of difference of two oxidation state $| O . S ._{P r o d u c t} - O . S ._{R e a c t a n t} \times number of atom$ for the valency factor
$K M n O_{4} O H^{-} - --- \to M n^{6 +}$
O.S. of Mn in reactant side
$= + 1 + x + 4 \times (- 2) = 0$
$x = + 7$
O.S. of Mn in product side=+6
$valency factor = number of Mn atom \times | O . S ._{P r o d u c t} - O . S ._{R e a c t a n t} |$
$valency factor = 1 \times | + 6 - 7 |$
$valency factor = 1$

Suggest Corrections

Similar questions

K M n O_{4} H^{+} - -- \to M n^{2 +}

K M n O_{4} H_{2} O - --- \to M n^{4 +}

K M n O_{4} H_{2} O \to M n^{4 +}

K_{2} C r_{2} O_{7} H^{+} - -- \to C r^{3 +}

(3 C u + 8 H N O_{3} (d i l) \to 3 C u (N O_{3})_{2} + 4 N O + 2 H_{2} O)

Join BYJU'S Learning Program

Explore more

Join BYJU'S Learning Program