Correct option is D)
Let the total weight of the compound is 100 g.
So the amount of Xe present is 53.5 g and of F is 46.7 g.
Now we find moles of Xe is
133
53.5
=0.4 and of F is
19
46.7
is 2.45.
Divide both by 0.4 so that we get a simple whole-number ratio, we get
Moles of Xe=
0.4
=1
Moles of F=
2.45
=6
So the formula is XeF
6
.
The oxidation state of Xe is +6.
Hence, the correct option is D
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Answers & Comments
Correct option is D)
Let the total weight of the compound is 100 g.
So the amount of Xe present is 53.5 g and of F is 46.7 g.
Now we find moles of Xe is
133
53.5
=0.4 and of F is
19
46.7
is 2.45.
Divide both by 0.4 so that we get a simple whole-number ratio, we get
Moles of Xe=
0.4
0.4
=1
Moles of F=
0.4
2.45
=6
So the formula is XeF
6
.
The oxidation state of Xe is +6.
Hence, the correct option is D