Question

Give the bond order, stability a d magnetic nature of O2 and O2+
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Answer 1

Answer:

bond order is directly proportional to stability. magnetic nature depends paired and unpaired electrons. ... It has two unpaired electrons.so, O2 molecule is paramagnetic

Answer 2

Answer:

Explanation:

Bond order = (bonding electrons - antibonding electrons)/2

configuration $O_{2}$

$\sigma_{1s}^{2} , \sigma_{1s}^{2*}, \sigma_{2s}^{2}, \sigma_{2s}^{2*},\sigma_{2p_{z}}^{2}(\pi_{2p_{x}}^{2},\pi_{2p_{y}}^{2}),(\pi_{2p_{x}}^{1*},\pi_{2p_{y}}^{1*})$

Bond Order  = $\frac{10-6}{2} = 2$

Since it has two unpaired electrons it is Paramagnetic.

Configuration of $O^{+}_{2}$

$\sigma_{1s}^{2} , \sigma_{1s}^{2*}, \sigma_{2s}^{2}, \sigma_{2s}^{2*},\sigma_{2p_{z}}^{2}(\pi_{2p_{x}}^{2},\pi_{2p_{y}}^{2}),(\pi_{2p_{x}}^{1*},\pi_{2p_{y}}^{0*})$

Bond Order = $\frac{10-5}{2} = 2.5$

Since it has one unpaired electrons it is Paramagnetic.

$O^{+}_2 > O_{2}\\\\BO_{O^{+}_2}>BO_{O_{2}}$

Thus  $O^{+}_{2}$ is more stable then  $O_{2}$

Might help! Thanks!