Predict the hybridization of each carbon in the molecule of the organic compound given below. Also,indicate the total number of sigma $(\sigma)$ and pi $(\pi)$ bonds in this molecule.
$HC \equiv C-CO-CH_2-COOH$

(N/A) The given molecule is $HC \equiv C-CO-CH_2-COOH$.
$1$. The first carbon (from left) is bonded to one $H$ atom and triple bonded to the second carbon. It has two $\sigma$ bonds and two $\pi$ bonds,so it is $sp$ hybridized.
$2$. The second carbon is triple bonded to the first carbon and single bonded to the third carbon. It has two $\sigma$ bonds and two $\pi$ bonds,so it is $sp$ hybridized.
$3$. The third carbon is part of a carbonyl group $(C=O)$. It is bonded to the second carbon,the fourth carbon,and double bonded to an oxygen atom. It has three $\sigma$ bonds and one $\pi$ bond,so it is $sp^2$ hybridized.
$4$. The fourth carbon is bonded to the third carbon,two $H$ atoms,and the fifth carbon. It has four $\sigma$ bonds,so it is $sp^3$ hybridized.
$5$. The fifth carbon is part of a carboxyl group $(-COOH)$. It is bonded to the fourth carbon,an oxygen atom (double bond),and an $OH$ group. It has three $\sigma$ bonds and one $\pi$ bond,so it is $sp^2$ hybridized.
Total number of $\sigma$ bonds = $11$
Total number of $\pi$ bonds = $4$