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The oxidation state of manganese in the green compound $K X_{2} MnO X_{4}$ is

A $+ 7$
B $+ 4$
C $+ 2$
D $+ 6$

Answer: D

2 (+ 1) + x + 4 (- 2) = 0

gives

x = + 6

. Manganate(VI) is green, whereas permanganate(VII) is purple.

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