Chemistry, asked by user1531, 11 months ago

Which of the following species has the highest bond order:
(1) 02+
(2) 02-
(3) O₂
(4) 02²-​

Answers

Answered by Anonymous
7

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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Answered by cutesmile590
5

Answer:

Option A is the answer ______

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