$\mathrm{PH}_3$ forms bubbles when passed slowly in water but $\mathrm{NH}_3$ dissolves. Explain why?
Dissolution of $\mathrm{NH}_3$ and $\mathrm{PH}_3$ in water can be explained on the basis of H -bonding. $\mathrm{NH}_3$ forms H -bond with water so it is soluble but $\mathrm{PH}_3$ does not form H -bond with water so it remains as gas and forms bubble in water.
In $\mathrm{PCl}_5$, phosphorus is in $\mathrm{sp}^3 \mathrm{d}$ hybridised state but all its five bonds are not equivalent. Justify your answer with reason.
It has trigonal bipyramidal geometry, in which two Cl atoms occupy axial position while three occupy equatorial positions. All five $\mathrm{P}-\mathrm{Cl}$ bonds are not identical. There are two types of bond lengths (i) Axial bond lengths (ii) Equatorial bond lengths
Thus, difference in bond length is due to fact that axial bond pairs suffer more repulsion as compared to equatorial bond pairs.
Why is nitric oxide paramagnetic in gaseous state but the solid obtained on cooling it is diamagnetic?
In gaseous state, $\mathrm{NO}_2$ exists as a monomer which has one unpaired electron but in solid state, it dimerises to $\mathrm{N}_2 \mathrm{O}_4$ so no unpaired electron left. Therefore, $\mathrm{NO}_2$ is paramagnetic in gaseous state but diamagnetic in solid state.
Give one reason to explain why $\mathrm{ClF}_3$ exists but $\mathrm{FCl}_3$ does not exist?
Existance of $\mathrm{ClF}_3$ and $\mathrm{FCl}_3$ can be explained on the basis of size of central atom. Because fluorine is more electronegative as compared to chlorine and has smaller size. Thus, one large Cl atom can accomodate three smaller F atoms but reverse is not true.
Out of $\mathrm{H}_2 \mathrm{O}$ and $\mathrm{H}_2 \mathrm{S}$, which one has higher bond angle and why?
Bond angle of $\mathrm{H}_2 \mathrm{O}\left(\mathrm{H}-\mathrm{O}-\mathrm{H}=104.5^{\circ}\right)$ is larger than that of $\mathrm{H}_2 \mathrm{S}\left(\mathrm{H}-\mathrm{S}-\mathrm{H}=92^{\circ}\right)$ because oxygen is more electronegative than sulphur therefore, bond pair electron of $\mathrm{O}-\mathrm{H}$ bond will be closer to oxygen and there will be more bond pair-bond pair repulsion between bond pairs of two $\mathrm{O}-\mathrm{H}$ bonds.
