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QJEEAdv23Ch6a48: JEE Advanced 2023 Chemistry - Class 11 JEEAdv23Ch6a48 Solution: The correct molecular orbital diagram for molecule in the…

The correct molecular orbital diagram for F2\mathrm{F}_2 molecule in the ground state is :

Multiple Choice Options

  • A https://ik.imagekit.io/novatra/photos/c29da09c250c1251d63052283c444db2c6015eb8fbfe03f70a4c5894d1ff6c9a.webp
  • B https://ik.imagekit.io/novatra/photos/78567bd749b08d3d79798bb9fc56a864b93958e5c6f46e15e22e31560c97a243.webp
  • C https://ik.imagekit.io/novatra/photos/d549d13ca06aa5d46ef130c499d0ae4771f8e57601f9d44deae065c81f7effa4.webp
  • D https://ik.imagekit.io/novatra/photos/806abd23255f27fd0ffaf350735f21cdf09b60051d4e96770a13a5a5fe8a8941.webp

Step-by-Step Verified Solution

(1) Upto 14 electrons, molecular orbital configuration is ->

JEE Advanced 2023 Paper 2 Online Chemistry - Chemical Bonding & Molecular Structure Question 5 English Explanation 1
Nb = No of electrons in bonding molecular orbital

Na == No of electrons in anti bonding molecular orbital

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

(2) After 14 electrons to 20 electrons molecular orbital configuration is ->

JEE Advanced 2023 Paper 2 Online Chemistry - Chemical Bonding & Molecular Structure Question 5 English Explanation 2

Here Na = 10

and Nb = 10

Now from question, in F atom 9 electrons present, so in F2, 9 ×\times 2 = 18 electrons present.

Molecular orbital configuration of F2(18 electrons) is

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

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