Question
Among the following, the paramagnetic compound is
A.
$$N{a_2}{O_2}$$
B.
$$\,{O_3}$$
C.
$${N_2}O$$
D.
$$K{O_2}$$
Answer :
$$K{O_2}$$
Solution :
(i) In $$N{a_2}{O_2},$$ we have $$O_2^{2 - }$$ ion. Number of valence electrons of the two oxygen in $$O_2^{2 - }$$ ion $$ = 8 \times 2 + 2 = 18$$ which are present as follows
$$\,\sigma 1{s^2},{\sigma ^*}1{s^2},\sigma 2{s^2},{\sigma ^*}2{s^2},\sigma 2p_x^2,$$ $$\left\{ {\pi 2p_y^2 = \pi 2p_z^2,\,\left\{ {{\pi ^*}2p_y^2 = {\pi ^*}2p_z^2} \right.} \right.$$
∴ Number of unpaired electrons $$\, = 0,$$ hence, $$O_2^{2 - }$$ is diamagnetic.
(ii) No. of valence electrons of all atoms in $$\,{O_3} = 6 \times 3 = 18.$$
Thus, it also, does not have any unpaired electron, hence it is diamagnetic.
(iii) No. of valence electrons of all atom in $$\,{N_2}O = 2 \times 5 + 6 = 16.$$ Hence, here also all electrons are paired. So it is diamagnetic.
(iv) In $$K{O_2}$$ we have $$O_2^ - $$ No. of valence electrons of all atoms in $$O_2^ - = 2 \times 6 + 1 = 13,$$
Thus it has one unpaired electron, hence it is paramagnetic.