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Chemical Bonding and Molecular Structure — Chemistry STD 11 Science — Question

Gujarat BoardEnglish MediumSTD 11 ScienceChemistryChemical Bonding and Molecular Structure3 Marks
Question
i. Explain, why $\mathrm{Be}_2$ molecule does not exist by using molecular orbital theory.
ii. Describe the state of hybridization in $\mathrm{PCl}_5$ Why are the axial bonds longer as compared to equatorial bonds?
✓

Answer

  1. i. Electronic configuration of $\mathrm{Be}=1 \mathrm{~s}^2 2 \mathrm{~s}^2$
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 $\text{Be}_2=\frac{1}{2}(4-4)=0$
  1. $\text{P}(15)=\text{ls}^2\ \text{2s}^2\ \text{2p}^6\ \text{3s}^2\ \text{3p}^3$
$\therefore$ Hydridization in $\mathrm{PCl}_5$ is $\mathrm{sp}^3 \mathrm{~d}$


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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