Answer
See answer below.
Work Step by Step
a) O: -2 (always except in fluorides)
$-1=nOx_{Br}-2\times3$
$nOx_{Br}=+5$
b) O: -2 (always except in fluorides)
$-2=2\times nOx_{C}-2\times4$
$nOx_{C}=+3$
c)Monoatomic ion, oxidation number equals the charge
F:-1
d) H:-1 ( hydride)
$0= nOx_{Ca}-1\times2$
$nOx_{Ca}=+2$
e) O: -2 (not a fluoride)
H: +1 (not a hydride)
$0=+1\times4+nOx_{Si}-2\times4$
$nOx_{Si}=+4$
f) O: -2 (not a fluoride)
H: +1 (not a hydride)
$-1=+1+nOx_{S}-2\times4$
$nOx_{S}=+6$
