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# The structural formula of a compound is $C{{H}_{3}}-CH=C=C{{H}_{2}}$. The type of hybridization at the four carbons from left to right are:A. $s{{p}^{3}},sp,s{{p}^{2}},s{{p}^{3}}$ B. $s{{p}^{2}},s{{p}^{3}},s{{p}^{2}},sp$ C. $s{{p}^{3}},s{{p}^{2}},sp,s{{p}^{2}}$ D. $s{{p}^{3}},s{{p}^{2}},s{{p}^{2}},s{{p}^{2}}$

Last updated date: 20th Jun 2024
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Hint: The hybridization is given by the value of the total number of hybrid orbitals taking part in hybridization. This is found by knowing the total sigma bonds present in the atom as well as the total number of lone pairs on that atom.

Complete step by step solution:
- Many theories have been formulated to know the nature of bonds in compounds like Valence bond theory, Molecular orbital theory, Crystal field theory, Valence shell electron pair repulsion theory.
-The hybridization of any molecule is determined by VSEPR theory; i.e; valence shell electron pair repulsion theory.
-VSEPR theory states that the electron pairs repel each other, whether they are in the form of lone pair or bond pair. This is done in order to minimise the repulsion, thus increasing the stability of the compound.
-The knowledge of both, bond pair and lone pair of electrons gives us the idea of the hybridization of the molecule. It is found by counting the number of electron pairs as well as the number of lone pairs.
-When there are no lone pairs, the denotation is simply$A{{X}_{n}}$ . When lone pairs are added, the denotation becomes $A{{X}_{n}}{{E}_{m}}$
where n=no. of bond pairs
m=no. of lone pairs
-Hybridization is counted by the number of bond pairs and lone pairs and can be shown as

 hybridization Total no. of bond pairs and lone pairs sp 2 $s{{p}^{2}}$ 3 $s{{p}^{3}}$ 4

-The lone pairs are placed apart from each other while the bond pairs are placed apart from each other to ensure maximum stability.
-There are no lone pairs in carbon atoms. So hybridization is found simply by counting the number of bond pairs which represents the number of sigma bonds in the structure and not the pi bonds.
-In any compound, a single bond is always a sigma bond. In double and triple bonds, there are 1 sigma bond and the others are pi bonds.
-Here in the compound we can see that the first carbon atom has 4 sigma bonds, the second carbon atom has 3 sigma bonds. Third carbon atom has 2 sigma bonds and the last carbon atom has 3 sigma bonds.

Therefore the hybridization of the carbon atoms will become $s{{p}^{3}},s{{p}^{2}},sp,s{{p}^{2}}$ and the correct option is C.

Note: The hybrid orbitals are different from the atomic orbitals. They have different strengths, different electronegativities and different percentages of s character. VSEPR theory decides the hybridization and the molecular geometry of the compounds.