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Explain formation of $\mathrm{H}_{2}$ and energy level diagram of $\mathrm{H}_{2}$ molecule.
Solution
$1 s$ atomic orbitals on two atoms (e.g. hydrogen form two molecular orbitals designated as $\sigma 1 s$ and $\sigma^{*} 1 s$ is.
$\sigma 1 s$ is bonding molecular orbital (BMO) and $\sigma 1 s$ is Antibonding molecular orbital (ABMO). Energy of
$\sigma 1 s$ is $<$ Energy of atomic orbital is $1 s<\sigma^{*} 1 s$.
(Energy of $\sigma 1 s+$ Energy of $\sigma^{*} 1 s$ ) $=($ Edition of energy of two $1 s)$
The energy diagram of $1 s, \sigma 1 s$ and $\sigma^{*} 1 s$ is as under.
Where, $\mathrm{MO}=$ molecular orbitals, $\sigma 1 s=\mathrm{BMO}$
$\mathrm{AO}=\text { Atomic orbitals, } \sigma 1 s=\mathrm{ABMO}$
Two $MO$, $\sigma 1 s$ and $\sigma^{*} 1 s$ are formed by overlapping of two $1 s$. Its figure is as under.
Similar Questions
Match $List-I$ with $List-II$.
| $List-I$ | $List-II$ |
| $(A)$ $\Psi_{ MO }=\Psi_{ A }-\Psi_{ B }$ | $(I)$ Dipole moment |
| $(B)$ $\mu=Q \times I$ | $(II)$ Bonding molecular orbital |
| $(C)$ $\frac{N_{b}-N_{a}}{2}$ | $(III)$ Anti-bonding molecualr orbital |
| $(D)$ $\Psi_{ MO }=\Psi_{ A }+\Psi_{ B }$ | $(IV)$ Bond order |
