4.Chemical Bonding and Molecular Structure
hard

Explain $\mathrm{H}_{2}$ molecule by molecular orbitals theory.

Option A
Option B
Option C
Option D

Solution

$LCAO$ Method of molecular orbitals : Molecular orbital is not obtained by Schrodinger want equation but it can be obtain by $LCAO.$

$LCAO$ method for Hydrogen Molecule $\left(\mathrm{H}_{2}\right)$ :

– Hydrogen is homonuclear diatomic molecule consider the hydrogen molecule $\left(\mathrm{H}_{2}\right)$ consisting of two atoms $\mathrm{H}_{\mathrm{A}}$ and $\mathrm{H}_{\mathrm{B}}$.

– Mathematically, the formation of molecular orbitals may be described by the linear combination of atomic orbitals that can take place by addition an by subtraction of wave functions of individual atomic orbitals as shown below.

$\psi_{\mathrm{MO}}=\psi_{\mathrm{A}}+\psi_{\mathrm{B}} \quad$ OR $\quad \psi^{*} \mathrm{MO}=\psi_{\mathrm{A}}-\psi_{\mathrm{B}}$

Following fig. depicted the formation of $\mathrm{H}_{2}$ molecule from $\mathrm{H}_{2}$ atom and its energy.

Bonding molecular orbital $\left(\Psi_{\mathrm{MO}}\right)$ e.g. $\sigma:$ The molecular orbital $\sigma$ formed by the addition of atomic orbitals is called the bonding molecular orbital $\left(\psi_{\mathrm{MO}}\right)$. Here $\sigma$ type molecular orbital, $\psi_{\mathrm{MO}}=\sigma\left(\mathrm{H}_{2}\right)=\psi_{\mathrm{A}}+\psi_{\mathrm{B}}$

Antibonding molecular orbital $\left(\psi_{\text {MO }}^{*}\right)$ e.g. $\sigma^{*}$ : The molecular orbital $\sigma^{*}$ formed by the substraction of atomic orbital $\left(\psi_{\mathrm{A}}\right.$ and $\left.\psi_{\mathrm{B}}\right)$ is called antibonding molecular orbital. Here $\sigma^{*}$ type antibonding molecular orbital,

$\psi^{*}{ }_{\mathrm{MO}}\left(\mathrm{H}_{2}\right)=\sigma_{\left(\mathrm{H}_{2}\right)}^{*}=\psi_{\mathrm{A}}-\psi_{\mathrm{B}}$

Standard 11
Chemistry

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