Be sure to answer all parts. use mo diagrams and the bond orders you obtain from them to answer the following questions. (a) is o2− stable? yes no (b) is o2− paramagnetic? yes no (c) what is the outer (valence) electron configuration of o2−? (σ1s)2(σ*1s)2(σ2s)2(σ*2s)2(σ2p)2(π2 p)2(π2p)2(π*2p)2(π*2p)1 (σ2s)2(σ*2s)2(σ2p)2(π2p)2(π2p)2(π*2 p)2(π*2p)1 (σ1s)2(σ*1s)2(σ2s)2(π2p)2(π2p)2(π*2 p)2(π*2p)1 (σ2s)2(σ*2s)2(σ2p)2(σ2p)2(π2p)2(π*2 p)2(π*2p)1

a) yes b) no c) (σ2s)2(σ*2s)2(σ2p)2(σ2p)2(π2p)2(π*2p)2(π*2p)1

Explanation:

The stability of a molecule depends on the number of bonding and antibonding electrons present. In the O22- ion, there are 10 bonding and 8 antibonding electrons. From the formula bond order= bonding electrons- antibonding electrons/2, we realize that the bond order is 1 meaning that the ion is stable. All the electrons are paired so it isn't paramagnetic. The valence shell is ordered as shown in the answer.

i think so it’s will be b yes i believe its