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Chemical Bonding and Molecular Structure question

2013 · Q109
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Chemical Bonding and Molecular Structure question

2013 · Q109

NEETChemistryChemical Bonding and Molecular StructureMCQ+4 / −1
In which of the following ionization processes the bond energy increases and the magnetic behaviour changes from paramagnetic to diamagnetic.
  1. A
    O2 →\to→ O2+
  2. B
    C2 →\to→ C2+
  3. C
    NO →\to→ NO+
  4. D
    N2 →\to→ N2+
View written solutionFree

Correct answer: C

Molecular orbital configuration of

O2+ ⇒\Rightarrow⇒ σ1s2σ∗1s2σ2s2σ∗2s2σ2pz2π2py2π∗2px1π∗2py1\sigma 1{s^2}{\sigma ^*}1{s^2}\sigma 2{s^2}{\sigma ^*}2{s^2}\sigma 2p_z^2\pi 2p_y^2{\pi ^*}2p_x^1{\pi ^*}2p_y^1σ1s2σ∗1s2σ2s2σ∗2s2σ2pz2​π2py2​π∗2px1​π∗2py1​ ⇒\Rightarrow⇒ Paramagnetic

Bond order = 10−52=2.5{{10 - 5} \over 2} = 2.5210−5​=2.5

C2 ⇒\Rightarrow⇒ σ1s2σ∗1s2σ2s2σ∗2s2σ2pz2π2py2\sigma 1{s^2}{\sigma ^*}1{s^2}\sigma 2{s^2}{\sigma ^*}2{s^2}\sigma 2p_z^2\pi 2p_y^2σ1s2σ∗1s2σ2s2σ∗2s2σ2pz2​π2py2​ ⇒\Rightarrow⇒ Diamagnetic

Bond order = 8−42=2{{8 - 4} \over 2} = 228−4​=2

C2+ ⇒\Rightarrow⇒ σ1s2σ∗1s2σ2s2σ∗2s2σ2pz2π2py1\sigma 1{s^2}{\sigma ^*}1{s^2}\sigma 2{s^2}{\sigma ^*}2{s^2}\sigma 2p_z^2\pi 2p_y^1σ1s2σ∗1s2σ2s2σ∗2s2σ2pz2​π2py1​ ⇒\Rightarrow⇒ Paramagnetic

Bond order = 7−42=1.5{{7 - 4} \over 2} = 1.527−4​=1.5

NO ⇒\Rightarrow⇒ σ1s2σ∗1s2σ2s2σ∗2s2σ2pz2π2py2π∗2px1\sigma 1{s^2}{\sigma ^*}1{s^2}\sigma 2{s^2}{\sigma ^*}2{s^2}\sigma 2p_z^2\pi 2p_y^2{\pi ^*}2p_x^1σ1s2σ∗1s2σ2s2σ∗2s2σ2pz2​π2py2​π∗2px1​ ⇒\Rightarrow⇒ Paramagnetic

Bond order = 10−52=2.5{{10 - 5} \over 2} = 2.5210−5​=2.5

NO+ ⇒\Rightarrow⇒ σ1s2σ∗1s2σ2s2σ∗2s2σ2pz2π2px2π2py2\sigma 1{s^2}{\sigma ^*}1{s^2}\sigma 2{s^2}{\sigma ^*}2{s^2}\sigma 2p_z^2\pi 2p_x^2\pi 2p_y^2σ1s2σ∗1s2σ2s2σ∗2s2σ2pz2​π2px2​π2py2​ ⇒\Rightarrow⇒ Diamagnetic

Bond order = 10−42=3{{10 - 4} \over 2} = 3210−4​=3

N2 ⇒\Rightarrow⇒ σ1s2σ∗1s2σ2s2σ∗2s2π2px2π2py2σ2pz2\sigma 1{s^2}{\sigma ^*}1{s^2}\sigma 2{s^2}{\sigma ^*}2{s^2}\pi 2p_x^2\pi 2p_y^2\sigma 2p_z^2σ1s2σ∗1s2σ2s2σ∗2s2π2px2​π2py2​σ2pz2​ ⇒\Rightarrow⇒ Paramagnetic

Bond order = 10−42=3{{10 - 4} \over 2} = 3210−4​=3

N2+ ⇒\Rightarrow⇒ σ1s2σ∗1s2σ2s2σ∗2s2π2px2π2py2σ2pz1\sigma 1{s^2}{\sigma ^*}1{s^2}\sigma 2{s^2}{\sigma ^*}2{s^2}\pi 2p_x^2\pi 2p_y^2\sigma 2p_z^1σ1s2σ∗1s2σ2s2σ∗2s2π2px2​π2py2​σ2pz1​ ⇒\Rightarrow⇒ Paramagnetic

Bond order = 9−42=2.5{{9 - 4} \over 2} = 2.529−4​=2.5

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