Chemical Bonding and Molecular Structure - Result Question 104

####106. Four diatomic species are listed below. Identify the correct order in which the bond order is increasing in them:

[2008, 2012M ]

(a) NO<O2<C22<He2+

(b) O2<NO<C22<He2+

(c) C22<He2+<O2<NO

(d) He2+<O2<NO<C22

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Solution:

  1. (d) Calculating the bond order of various species.

O2:KKσ2s2σ2s2σ2pz2

$\pi 2 p_x^{2} \pi 2 p_y^{2} \pi^{} 2 p_x^{2} \pi^{} 2 p_y^{1}$

Number of electrons in bonding B.O. = Number of electrons in non-bonding 2 =852 or 1.5

NO: $KK \sigma 2 s^{2} \sigma^{} 2 s^{2} \pi 2 p_x^{2} \approx \pi 2 p_y^{2} \sigma 2 p_z^{2} \pi^{} 2 p_x^{1}$

B.O. =NbNa2=822 or 2.5

C22:kkσ2s2σ2s2π2px2π2py2σ2pz2

B.O. =NbNa2=822 or 3

He2+=σ1s2σ1s1

B. O. =NbNa2=212 or 0.5

From these values we find the correct order of increasing bond order is

He22+<O2<NO<C22



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