Explain $\sigma$ and $\pi$-bond,and planarity and their bond angles in the carbonyl group.

(N/A) Electronic configuration of carbon and oxygen is as follows:
$C^{*}[He] 2s^{1} 2p_{x}^{1} 2p_{y}^{1} 2p_{z}^{1}$
$O [He] 2s^{2} 2p_{x}^{2} 2p_{y}^{1} 2p_{z}^{1}$
$(b)$ $(i)$ The $C=O$ bond of the carbonyl group in aldehydes and ketones is made up of one $\sigma$-bond and one $\pi$-bond. The carbonyl carbon atom is $sp^{2}$-hybridised and forms three sigma $(\sigma)$ bonds.
$(ii)$ The fourth valence electron of carbon remains in its $p$-orbital and forms a $\pi$-bond with oxygen by overlap with the $p$-orbital of an oxygen atom.
$(iii)$ In addition,the oxygen atom also has two non-bonding electron pairs. Thus,the carbonyl carbon and the three atoms attached to it lie in the same plane,and the $p$-electron cloud is above and below this plane $\left(C \frac{\pi}{\sigma} \ddot{O}\right)$.
$(iv)$ In carbonyl group compounds,the carbonyl carbon atom is $sp^{2}$ hybridised and forms three $sp^{2}$ hybrid orbitals.
$(v)$ One of the $sp^{2}$ hybridised orbitals of carbon overlaps with the $p$-orbital of the oxygen atom forming a $\sigma$ bond,while the $p$-orbital overlap forms the $\pi$ bond.
$(vi)$ The remaining two hybridised $sp^{2}$ orbitals of the carbon atom form two additional $\sigma$-bonds by overlapping with $1s$-orbitals of two hydrogen atoms in formaldehyde.
$(vii)$ For any other aldehyde,two hybridised $sp^{2}$ orbitals of the carbon atom form two $\sigma$-bonds by overlapping with the $1s$-orbital of one hydrogen atom and one $sp^{3}$ hybrid orbital of an alkyl group carbon.
$(viii)$ For any ketone,two hybridised $sp^{2}$ orbitals of the carbon atom form two $\sigma$-bonds by overlapping with two $sp^{3}$ hybrid orbitals of two alkyl group carbons.
$(ix)$ All three $\sigma$-bonds lie in the same plane,inclined to one another by $120^{\circ}$ as shown in the figure.