According to MO theory, number of species/ions from the following having identical bond order is ___.
CN-, NO+, O2, O+2, O2+2
CN-, NO+, O2, O+2, O2+2
Correct Answer: 3
Approach Solution - 1
To find the number of species/ions with an identical bond order using Molecular Orbital (MO) theory, we need to calculate the bond order for each given species using the formula: Bond Order = (Number of bonding electrons - Number of antibonding electrons) / 2.
- CN-:
The electronic configuration is (σ1s)2(σ1s*)2(σ2s)2(σ2s*)2(π2p)4(σ2p)2.
Bonding electrons: 10, Antibonding electrons: 4.
Bond Order = (10 - 4) / 2 = 3. - NO+:
The electronic configuration is (σ1s)2(σ1s*)2(σ2s)2(σ2s*)2(π2p)4(σ2p)2.
Bonding electrons: 10, Antibonding electrons: 4.
Bond Order = (10 - 4) / 2 = 3. - O2:
The electronic configuration is (σ1s)2(σ1s*)2(σ2s)2(σ2s*)2(π2p)4(π2p*)2.
Bonding electrons: 10, Antibonding electrons: 6.
Bond Order = (10 - 6) / 2 = 2. - O2+:
The electronic configuration is (σ1s)2(σ1s*)2(σ2s)2(σ2s*)2(π2p)4(π2p*)1.
Bonding electrons: 10, Antibonding electrons: 5.
Bond Order = (10 - 5) / 2 = 2.5. - O22+:
The electronic configuration is (σ1s)2(σ1s*)2(σ2s)2(σ2s*)2(π2p)4.
Bonding electrons: 10, Antibonding electrons: 4.
Bond Order = (10 - 4) / 2 = 3.
Species with identical bond order of 3: CN-, NO+, O22+.
Number of such species = 3, which fits within the provided range (3,3).
Approach Solution -2
CN-, NO+ and \(O_2^{2+} \)have bond order of `3'.
O2 has bond order of 2,
\(O_2^+\) has bond order of 2.5.
∴ Three species have similar bond order.
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