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QJEEM20Ch68d8: JEE Main 2020 Chemistry - Class 11 JEEM20Ch68d8 Solution: Of the species, NO, NO + , NO 2+ and NO - , the one with…

Of the species, NO, NO+, NO2+ and NO- , the one with minimum bond strength is :

Multiple Choice Options

  • A NO–
  • B NO
  • C NO+
  • D NO2+

Step-by-Step Verified Solution

Note :

(1)     \,\,\,\, Bond strength ∝\propto Bond order

(2)     \,\,\,\, Bond length ∝\propto 1Bond  order{1 \over {Bond\,\,order}}

(3)  \, Bond order =12= {1 \over 2} [Nb −- Na]

Nb = No of electrons in bonding molecular orbital

Na == No of electrons in anti bonding molecular orbital

(4)     \,\,\,\, upto 14 electrons, molecular orbital configuration is

JEE Main 2020 (Online) 3rd September Morning Slot Chemistry - Chemical Bonding & Molecular Structure Question 157 English Explanation 1

Here Na = Anti bonding electron == 4 and Nb = 10

(5)     \,\,\,\, After 14 electrons to 20 electrons molecular orbital configuration is - - -

JEE Main 2020 (Online) 3rd September Morning Slot Chemistry - Chemical Bonding & Molecular Structure Question 157 English Explanation 2

Molecular orbital configuration of NO (15 electrons) is

σ1s2 σ1s2∗ σ2s2 σ2s2∗ σ2pz2 π2px2 = π2py2 π2px1∗ = π2pyo∗{\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^2}}\,\pi _{2p_x^1}^ * \, = \,\pi _{2p_y^o}^ *

∴    \therefore\,\,\,\, Nb = 10

Na = 5

∴    \therefore\,\,\,\, BO = 12[10−5]{1 \over 2}\left[ {10 - 5} \right] = 2.5

Similarly Molecular orbital configuration of NO+ (14 electrons) is

σ1s2 σ1s2∗ σ2s2 σ2s2∗ σ2pz2 π2px2 = π2py2 {\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^2}}\,

∴    \therefore\,\,\,\, Nb = 10

Na = 4

∴    \therefore\,\,\,\, BO = 12[10−4]{1 \over 2}\left[ {10 - 4} \right] = 3

Similarly Molecular orbital configuration of NO2+ (13 electrons) is

σ1s2 σ1s2∗ σ2s2 σ2s2∗ σ2pz2 π2px2 = π2py1 {\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^1}}\,

∴    \therefore\,\,\,\, Nb = 9

Na = 4

∴    \therefore\,\,\,\, BO = 12[9−4]{1 \over 2}\left[ {9 - 4} \right] = 2.5

Molecular orbital configuration of NO- (16 electrons) is

σ1s2 σ1s2∗ σ2s2 σ2s2∗ σ2pz2 π2px2 = π2py2 π2px1∗ = π2py1∗{\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^2}}\,\pi _{2p_x^1}^ * \, = \,\pi _{2p_y^1}^ *

∴    \therefore\,\,\,\, Nb = 10

Na = 6

∴    \therefore\,\,\,\, BO = 12[10−6]{1 \over 2}\left[ {10 - 6} \right] = 2

As Bond strength ∝\propto Bond order

∴\therefore NO– will have minimum bond strength.

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