The oxidation number of sulphur in $H_{2} S_{2} O_{7}$ and iron in $K_{4} F e (C N)_{6}$ is respectively

+ 6 and + 2

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

+ 2 and + 2

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

+ 8 and + 2

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

+ 6 and + 4

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

Open in App

Solution

The correct option is A
+ 6 and + 2

$H_{2} S_{2} O_{7}$

2 $\times$ (+1)+2 $\times$ x +7x(-2)=0

+2+2x-14=0

2x =14 -2=12

x = $\frac{12}{2}$ =+6 for s

$K_{4} F e (C N)_{6}$

4 $\times$ x +6 $\times$ (-1) =-4

x =6-4 =+2 for Fe

Suggest Corrections

Similar questions

H_{2} S_{2} O_{7}

K_{4} F e (C N)_{6}

The oxidation number of sulphur in sulphur dioxide $S O_{2}$ and sulphuric acid $H_{2} S O_{4}$ is respectively

H_{2} S_{2} O_{7}

K_{4} F e (C N)_{6}

The oxidation number of sulphur in $S_{8}$ , $S_{2}$ $F_{2}$ , $H_{2} S$ respectively, are

The oxidation number of sulphur in $S_{8}, S_{2} F_{2} a n d H_{2} S$ respectively are

Join BYJU'S Learning Program