Assign reason for the following: i. BaSO _4 is insoluble although… - Vidyadip

Question

Assign reason for the following:
i. $BaSO _4$ is insoluble although ionic in nature.
ii. $ClF _3$ has only $90^{\circ}$ bond angles.
iii. $SO _2$ is angular but $SO _3$ is planar.
iv. $NH _3$ and $PH _3$ have same hybridization but different bond angle.
v. $CuSO _4 \cdot 5 H _2 O$ loses $4 H _2 O$ on heating but not the fifth molecule.

✓

Answer

$BaSO_4$ 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)
$ClF_3$ has three bonded pair which are at $90^\circ $ but lone pair of electrons are at equatorial position at angle $120^\circ $ so as to minimize repulsion.
$SO_2$ has 1 lone pair, therefore, it is bent molecule, $SO_3$ does not have lone pair.

$\therefore$ $SO_2$ is planar.

' N ' is smaller in size and more electronegative than ' P ' therefore in $NH _3$ bond angle is $107^{\circ}$ but in $PH _3$ it is $94.5^{\circ}$. As the size of central atom increases, bond angle decreases.

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

$CuSO_4.5H_2O$ loses $4$ molecules of water which are forming coordinate bond with $Cu^{2+}$ but does not lose $H_2O$ molecule, which is H-bonded.

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