Solution.
SnCl5−SnCl5^-SnCl5−
e=4+(5∗7)+1=40e = 4 + (5*7) + 1 = 40e=4+(5∗7)+1=40
IF4+IF4^+IF4+
e=7+(4∗7)−1=34e = 7 + (4*7) - 1 = 34e=7+(4∗7)−1=34
Answer:
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