Which of the following compound is paramagnetic ? .

Options

A. `K_(2)O_(2)`
B. `O_(3)`
C. `KO_(2)`
D. `N_(2)O`

Explanation

Note Electronic configuration of `O (Z =8) = 1s^(2),2s^(2) ,2p^(4)` Number of valence `e^(-')` s in O-atom `=6`
(a) `K_(2)O_(2)` has `O_(2)^(2-)` (peroxide ion)
Number of valence `e^(-')s = (6 xx 2 + 2) = 14 e^(-')s`
Therefore Number of unpaired `e^(-')s =0` Hence `O_(2)^(2-)` is diamagnetic
(b) Number of valence `e^(-')` for all atoms in `O_(3) = 6 xx 3 =18`
Thus it also does not have unpaired `e^(-)` Hence diamagnetic
(c ) `KO_(2)` has `O_(2)^(Θ)` (superoxide ion)
Therefore number of valence `e^(-')s` of all atoms in `O_(2)^(Θ) = 6 xx 2 + 1 = 13` Thus it has one unpaired `e^(-)` Hence paramagnetic
(d) Number of valence `e^(-') s` of all atom in `N_(2)O =2 xx 5 + 6 =16`
Thus `e^(-')s` are paired and is diamagnetic .