Question

The most acidic hydrogen atoms are present in:
A.ethane
B. ethene
C. ethyne
D.benzene

Answer 1

The hybridisation of all the four compounds are needed to be taken out. After that, check for the s- character in each of them. Compound with the maximum percentage of s character will have most acidic hydrogen.

Complete answer:
In order to answer our question, we need to learn about the types of hybridisation:
i.sp-hybridisation: This is the simplest type of hybridisation involving s and p-orbitals. In this hybridisation one s and one p orbitals hybridise (or intermix) to produce two equivalent hybrid orbitals, known as sp hybrid orbitals. The orbitals that are suitable for sp hybridisation are s and ${{p}_{z}}$, if z-axis is the axis that the hybrid orbitals are to lie along. The two sp-hybrid orbitals are arranged in a line fashion which makes an angle of ${{180}^{0}}$ and therefore the geometry of the molecule is linear. Each of the hybrid orbitals has 50% s-character and 50% p-character. This type of hybridisation is referred to as diagonal hybridisation.
ii. $s{{p}^{2}}$hybridisation: In $s{{p}^{2}}$ hybridisation one s and two p (${{p}_{x}}$ and ${{p}_{y}}$) orbitals of one atom hybridize to give three equivalent $s{{p}^{2}}$ hybrid orbitals. These three $s{{p}^{2}}$ hybrid orbitals are directed towards the three corners of an equilateral triangle with an angle of ${{120}^{0}}$ and give a triangular geometry to the molecule. $s{{p}^{2}}$ hybrid orbitals are larger in size than sp-hybrid orbitals but slightly smaller than that of $s{{p}^{3}}$ hybrid orbitals. Each $s{{p}^{2}}$ hybrid orbitals has 1/3 (or 33.33%) s-character and 2/3 (or 66.7%) p-character.
iii.$s{{p}^{3}}$ hybridisation: In this hybridisation one s and three p-orbitals intermix to form $s{{p}^{3}}$ hybrid orbitals of equivalent energy and identical shape. These four $s{{p}^{3}}$ hybrid orbitals are directed towards the four corners of a tetrahedron separated by an angle of ${{109}^{0}}{{28}^{'}}$. $s{{p}^{3}}$ hybrid orbitals have 25% s-character and 75% p-character.
Now, more the s character, more the acidic character of hydrogen. Following table shows the data:

Compound	Formula	Hybridisation	% s character
Ethane	$C{{H}_{3}}-C{{H}_{3}}$	$s{{p}^{3}}$	25%
Ethene	$C{{H}_{2}}=C{{H}_{2}}$	$s{{p}^{2}}$	33.33%
Ethyne	$CH\equiv CH$	$sp$	50%
Benzene	${{C}_{6}}{{H}_{6}}$	$s{{p}^{2}}$	33.33%

As ethyne has the maximum s character, so it should have the most acidic hydrogen too. So,we obtain option C as the correct answer for this question.

Note:
In ethyne molecules, $C\equiv C$ bond consists of one $sp-sp$ $\sigma$-bond along with two $\pi$-bonds. The $C\equiv C$bond length is 120 pm. C-H bond is sp-s sigma bond. The H-C-C angle is ${{180}^{0}}$, i.e., the molecule is linear.