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

Rajasthan BoardEnglish MediumSTD 11 ScienceChemistryChemical Bonding and Molecular Structure5 Marks
Question
Assign reason for the following:
  1. BaSO4 is insoluble although ionic in nature.
  2. ClF3 has only 90° bond angles.
  3. SO2 is angular but SO3 is planar.
  4. NH3 and PH3 have same hybridization but different bond angle.
  5. CuSO4.5H2O loses 4H2O on heating but not the fifth molecule.
✓

Answer

  1. BaSO4 is insoluble in water because lattice energy is more than hydration energy (energy released when $\text{Ba}^{2+}$ and $\text{SO}^{2-}_4$ get attracted by water molecules)
  2. ClF3 has three bonded pair which are at 90° but lone pair of electrons are at equatorial position at angle 120° so as to minimize repulsion.
  3. SO2 has 1 lone pair, therefore, it is bent molecule, SO3 does not have lone pair.

$\therefore$ SO2 is planar.

  1. 'N' is smaller in size and more electronegative than 'P' therefore in NH3 bond angle is 107° but in PH3 it is 94.5°. As the size of central atom increases, bond angle decreases.

$[\text{Cu}(\text{H}_2\text{O})_4]\text{SO}_4\dots\text{H}_2\text{O}\xrightarrow{\ \ \text{heat}\ \ }\text{CuSO}_4\dots\text{H}_2\text{O}+4\text{H}_2\text{O}$

  1. CuSO4.5H2O loses 4 molecules of water which are forming coordinate bond with Cu2+ but does not lose H2O molecule, which is H-bonded.

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