Question
- Explain, why Be2 molecule does not exist by using molecular orbital theory.
- Describe the state of hybridization in PCI5 Why are the axial bonds longer as compared to equatorial bonds?
✓
Answer
- Electronic configuration of Be = 1s2 2s2
M.O. configuration of $\text{Be}_2=(\sigma\text{ls})^2(\sigma*\text{ls})^2(\sigma\text{2s})^2(\sigma*2\text{s})$
Bond order of $\text{Be}_2=\frac12(4-4)=0$
Since bond of $\text{Be}_2=\frac{1}{2}(4-4)=0$
- $\text{P}(15)=\text{ls}^2\ \text{2s}^2\ \text{2p}^6\ \text{3s}^2\ \text{3p}^3$
$\therefore$ Hydridization in PCl5 is sp3d
Axial bonds experience more electronic repulsion from three equatorial bond pairs and equatorial bonds experience repulsion from only two axial bond pairs hence axial bonds are longer as compared to equatorial bonds.
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→- $\text{CH}_3-\text{CH}_2-\text{CH}_2-\text{CH}_2-\text{OH}$
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$\text{CH}_3-\text{O}-\text{CH}_2-\text{CH}_2-\text{CH}_3$
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$\text{CH}_3-\text{O}-\text{CH}-\text{CH}_3\\\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ |\\\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \text{CH}_3$
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$\text{CH}_3-\text{CH}_2-\text{CH}-\text{CH}_3\\\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ |\\\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \text{OH}$
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