What is the oxidation number of #Sn# in the compound #Na_2SnO_2#?

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Answer 1 · 2016-05-29T12:14:46

#+2#

To find the oxidation number of #Sn# in #Na_2SnO_2# we should know that:

The oxidation number for #Na# is always #+1#.

The oxidation number of #O# is always #-2# except in #H_2O_2# it is #-1# and in #OF_2# it is #+2#.

Let the oxidation number of #Sn# be #x#, therefore,

#2xx(+1)+x+2xx(-2)=0#

#=>2+x-4=0#

#=>x-2=0#

#=>x=+2#

Thus, the oxidation number of #Sn# in #Na_2SnO_2# is #+2#.