The pair(s) of reagents that yield paramagnetic species is/are

Options

A. Na and excess of `NH_(3)`
B. K and excess of `O_(2)`
C. Cu and dilute `HNO_(3)`
D. `O_(2)` and 2-ethylanthraquinol

Answer

A, B, C

Explanation

Dilute solution of Na in liquid ammonia is paramagnetic due to ammoniated `e^(-1)s`
`Na^(o+)+(x+y)NH_(3)rarr Na^(+)(NH_(3))_(x)+e.(NH_(3))_(y)`
`K+underset(("excess"))(O_(2))rarr underset("Paranagnetic")(KO_(2))`
`O_(bar2)` is paramagnetic due to unpaired electron in antibonding orbital
`3Cu+underset(("diluted"))(8HNO_(3))rarr 3Cu(NO_(3))_(2)+2NO+4H_(2)O`
NO is paramagnetic due to unpaired electrons on ''N''.

where `''H_(2)O_(2)''` is diamagnetic.