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Answer 1 · 2017-11-17T10:07:38

The oxidation number of #Cr# is #+6#

We will rewrite the formula as an equation.

#2K + 2Cr + 7O = 0#

The zero comes from there being no overall charge on the compound.

Each oxidation number is represented by the element's symbol.

Since this compound is not a peroxide or superoxide, the oxidation value for #O# is #-2#.

#K# being in group 1 has a #+1# charge and therefore has an oxidation number of #+1#

Now, by putting known values in we have:
#2(+1) + 2Cr + 7(-2)=0#

#2+2Cr-14=0#

#2Cr=0+14-2=+12#

#Cr=+12/2=+6#

To make it clear, we must keep the positive sign even if we know it is positive.

Answer 2 · 2017-11-17T10:07:40

#+6#.

#"K"# has an oxidation number of #+1#. There are two #"K"# atoms, thus #+2# oxidation number for both #"K"#.

#"O"# has an oxidation number of #-2#. There are seven #"O"# atoms, thus #-14# oxidation number for all #"O"# atoms.

Overall, the oxidation numbers have to equal 0 as it is an uncharted molecule.

Thus

#-14 + 2 = -12#

So the #"Cr"# atoms have to have a combined oxidation number equaling #+12# to cancel out the #-12# number we just calculated.

There are #2# #"Cr"# atoms present, and so

#(+12) / 2 = +6#