Answer
a) Ketone ($-CO-$), alcohol ($-OH$), tertiary amine ($-NR-$), primary amine ($-NH_{2}$), carboxylic acid ($-COOH$)
b) Only the C attached with $NH_{2}$ and the adjacent C bonded with N will have $sp^{3}$ hybridization. The rest C atoms will have $sp^{2}$ hybridization. See the below figure.
c) Total no. of $\sigma$ bonds = 24.
Total no. of $\pi$ bonds = 4.
Work Step by Step
To find out hybridization of C atom in the compound, we have to find out no. of $\sigma$ bonds the C atom attached with. If no. of $\sigma$ bond = 4, then hybridization is $sp^{3}$, 3 means $sp^{2}$ and 2 means $sp$.
Now, to find which bond is sigma and which one is pi bond we should know that a single bond is always sigma bond. If there is double bond, then one bond is sigma and other bond is pi bond. If there is a triple bond, then one is sigma and the other two bonds are pi bonds.
a) Functional groups present in mimosine are: Ketone ($-CO-$), Alcohol ($-OH$), Tertiary Amine ($-NR-$), Primary amine ($-NH_{2}$), carboxylic acid ($-COOH$)
b) Only the C attached with $NH_{2}$ and the adjacent C bonded with N will have $sp^{3}$ hybridization. The rest C atoms will have $sp^{2}$ hybridization. See the below figure.
c) Total no. of $\sigma$ bonds = 24.
Total no. of $\pi$ bonds = 4.
