Correct Answer - A::B::C
are correct options.
(a)`._(24)Cr=1s^(2)2s^(2)2p^(2),3s^(2)3p^(2)3d^(5),4s^(1)`
`=[Ar]3d^(5)4s^(1)`
(b)For magnetic quantum number `(m_(l))`negative value is possible.
For s-sub-shell , l=0, hence m=0
for p-sub-shell , l=1, hence m=-1, 0, +1
(c)`._(47)Ag=1s^(2)2s^(2)2p^(6)3s^(2)33p^(6)3d^(10)4s^(2)4p^(6)4d^(10)5s^(1)`
Hence 23 electrons have a spin of one type and 24 of the opposite type.
(d) Oxidation state of N in `HN_(3)`is `-(1)/(3)`.