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

Options

A. `Na` and excuss 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)rarrNa^(+)(NH_(3))_(x)+e^(-)(NH_(3))_(y)`
`K+underset("excess")(O_(2))rarrunderset("Paramagnetic")(KO_(2))`
`O_(2)` is paramagnetic due to unpaired electron in antibonding orbital
`3Cu+underset("diluted")(8HNO_(3))rarr3Cu(NO_(3))_(2)+2NO+4H_(2)O`
`NO` is paramagnetic due to unpaired electrons on `N`.