Which of the following species has the highest bond order:
(1) 02+
(2) 02-
(3) O₂
(4) 02²-
Answers
Highest bond order electron is O2 +
Option 1
Answer with full explanation ❣✔
O2:
O2:valence electrons=6+6=12
O2:valence electrons=6+6=12(σ2s)2(σ∗2s)2(σ2p)2(π2p)4(π∗2p)2
O2:valence electrons=6+6=12(σ2s)2(σ∗2s)2(σ2p)2(π2p)4(π∗2p)2BO=1/2×(6−2)=2;
O2:valence electrons=6+6=12(σ2s)2(σ∗2s)2(σ2p)2(π2p)4(π∗2p)2BO=1/2×(6−2)=2;O2+:
O2:valence electrons=6+6=12(σ2s)2(σ∗2s)2(σ2p)2(π2p)4(π∗2p)2BO=1/2×(6−2)=2;O2+:valence electrons=6+6−1=11
O2:valence electrons=6+6=12(σ2s)2(σ∗2s)2(σ2p)2(π2p)4(π∗2p)2BO=1/2×(6−2)=2;O2+:valence electrons=6+6−1=11(σ2s)2(σ∗2s)2(σ2p)2(π2p)4(π∗2p)1
O2:valence electrons=6+6=12(σ2s)2(σ∗2s)2(σ2p)2(π2p)4(π∗2p)2BO=1/2×(6−2)=2;O2+:valence electrons=6+6−1=11(σ2s)2(σ∗2s)2(σ2p)2(π2p)4(π∗2p)1BO=1/2×(6−1)=2.5;
O2:valence electrons=6+6=12(σ2s)2(σ∗2s)2(σ2p)2(π2p)4(π∗2p)2BO=1/2×(6−2)=2;O2+:valence electrons=6+6−1=11(σ2s)2(σ∗2s)2(σ2p)2(π2p)4(π∗2p)1BO=1/2×(6−1)=2.5;O2−:
O2:valence electrons=6+6=12(σ2s)2(σ∗2s)2(σ2p)2(π2p)4(π∗2p)2BO=1/2×(6−2)=2;O2+:valence electrons=6+6−1=11(σ2s)2(σ∗2s)2(σ2p)2(π2p)4(π∗2p)1BO=1/2×(6−1)=2.5;O2−:valence electrons=6+6+1=13
O2:valence electrons=6+6=12(σ2s)2(σ∗2s)2(σ2p)2(π2p)4(π∗2p)2BO=1/2×(6−2)=2;O2+:valence electrons=6+6−1=11(σ2s)2(σ∗2s)2(σ2p)2(π2p)4(π∗2p)1BO=1/2×(6−1)=2.5;O2−:valence electrons=6+6+1=13(σ2s)2(σ∗2s)2(σ2p)2(π2p)4(π∗2p)3
O2:valence electrons=6+6=12(σ2s)2(σ∗2s)2(σ2p)2(π2p)4(π∗2p)2BO=1/2×(6−2)=2;O2+:valence electrons=6+6−1=11(σ2s)2(σ∗2s)2(σ2p)2(π2p)4(π∗2p)1BO=1/2×(6−1)=2.5;O2−:valence electrons=6+6+1=13(σ2s)2(σ∗2s)2(σ2p)2(π2p)4(π∗2p)3BO=1/2×(6−3)=1.5;
O2:valence electrons=6+6=12(σ2s)2(σ∗2s)2(σ2p)2(π2p)4(π∗2p)2BO=1/2×(6−2)=2;O2+:valence electrons=6+6−1=11(σ2s)2(σ∗2s)2(σ2p)2(π2p)4(π∗2p)1BO=1/2×(6−1)=2.5;O2−:valence electrons=6+6+1=13(σ2s)2(σ∗2s)2(σ2p)2(π2p)4(π∗2p)3BO=1/2×(6−3)=1.5;O22−:
O2:valence electrons=6+6=12(σ2s)2(σ∗2s)2(σ2p)2(π2p)4(π∗2p)2BO=1/2×(6−2)=2;O2+:valence electrons=6+6−1=11(σ2s)2(σ∗2s)2(σ2p)2(π2p)4(π∗2p)1BO=1/2×(6−1)=2.5;O2−:valence electrons=6+6+1=13(σ2s)2(σ∗2s)2(σ2p)2(π2p)4(π∗2p)3BO=1/2×(6−3)=1.5;O22−:valence electrons=6+6+2=14
O2:valence electrons=6+6=12(σ2s)2(σ∗2s)2(σ2p)2(π2p)4(π∗2p)2BO=1/2×(6−2)=2;O2+:valence electrons=6+6−1=11(σ2s)2(σ∗2s)2(σ2p)2(π2p)4(π∗2p)1BO=1/2×(6−1)=2.5;O2−:valence electrons=6+6+1=13(σ2s)2(σ∗2s)2(σ2p)2(π2p)4(π∗2p)3BO=1/2×(6−3)=1.5;O22−:valence electrons=6+6+2=14(σ2s)
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Answer:
Option A is the answer ______