the oxidation state of I in H4IO-6
a.+1 b.-1 c.+7 d.+5
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hey mate!
let it's oxidation no. be x
then, oxidation state of I in
H4lO-6 =
+4 +x -12 =0
x =8
#Phoenix
let it's oxidation no. be x
then, oxidation state of I in
H4lO-6 =
+4 +x -12 =0
x =8
#Phoenix
Anonymous:
??
ans may be +7 or +5or-1or+1
but thanks for ur help
:-( sry dear.... no need to mark it as brainliest if it's wrong
See I have correction ... the -6 is the overall charge on the Hydrogen Ionate ... thus to find the Oxidation of I we will write .. (H4IO6)^-1 = 1 ×4 + X + (-2)6 = -1 thus we get as 4 -12+ X = -1 ==》 -8 + X = -1 thus we get .. X = -1 + 8 ==》 X = 7 ..thus oxidation of iodine is + 7 here
ohh tq.... charge missing tha ques me...
yah... u r ri8....
but I can't edit it now :-(
allen.... which building? ,
tanq
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