Match the species in Column I with the bond order in Column II.

Column I

Column II

(i) NO

(a) 1.5

(ii) CO

(b) 2.0

(iii) O{_{2}}^{-}

(c) 2.5

(iv) O_{2}

(d) 3.0

Answers (1)

(i) →(c); (ii) →(d); (iii) →(a); (iv) →(b)

Explanation:

Column I

Column II

  1. NO

NO(7+8=15\; electrons)

\sigma 1s^{2}<\sigma ^{*}1s^{2}<\sigma 2s^{2}<\sigma ^{*}2s^{2}<\sigma 2pz^{2}<[\pi 2px^{2}=\pi 2px^{2}]<[\pi^{*} 2px^{1}=\pi^{*} 2px]<\sigma ^{*}2p_{z}

Bond order = (N_{b}-N_{a})/2=^{*}(\frac{(10-5)}{2})=2.5

  1. CO

CO (6+8 = 14\; electrons)

\sigma 1s^{2}<\sigma ^{*}1s^{2}<\sigma 2s^{2}<\sigma ^{*}2s^{2}<[\pi ^{*}2p{_{x}}^{2}=\pi wp{_{x}}^{2}]\sigma 2pz{_{z}}^{2}<[\pi^{*} 2px=\pi ^{*}2px]<\sigma ^{*}2p

Bond order = \frac{(N_{b}-N_{a})}{2}=^{*}(\frac{(10-4)}{2})=3

  1. O{_{2}}^{-}

(8+8+1=17\; electrons)

\sigma 1s^{2}<\sigma ^{*}1s^{2}<\sigma 2s^{2}<\sigma ^{*}2s^{2}<[\pi 2p{_{x}}^{2}=\pi 2p{_{z}}^{2}]<\sigma 2p{_{z}}^{2}<[\pi^{*}2p{_{x}}^{1}=\pi ^{*}2p{_{z}}^{1}]<\sigma ^{*}2p

Bond order = \frac{(N_{b}-N_{a})}{2}=(\frac{(10-7)}{2})=1.5

  1. O_{2}

(8+8=16\; electrons)

\sigma 1s^{2}<\sigma ^{*}1s^{2}<\sigma 2s^{2}<\sigma ^{*}2s^{2}<[\pi 2p{_{z}}^{2}=\pi 2p{_{z}}^{2}]<\sigma 2p{_{z}}^{2}<[\pi^{*}2p{_{x}}^{1}=\pi ^{*}2p{_{x}}^{1}]<\sigma ^{*}2p

Bond order = \frac{(N_{b}-N_{a})}{2}=(\frac{(10-6)}{2})=2

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