Anthracene is a polycyclic aromatic hydrocarbon with chemical formula C₁₄H₁₀. The number of fused rings in Anthracene are three in number. This compound is colorful and is used in the formation of different dyes due to its property of deloclization of pi electrons. All the carbon atoms in Anthracene are sp² hybridized with a trigonal planar structure hence, the Anthracene is planar in nature.
Number of Sigma Bonds:
There are 26 sigma bonds (colored in Blue) in Anthracene among which 10 sigma bonds are between carbon and hydrogen atoms while the remaining are between the carbon atoms.
Number of Pi-Bonds:
There are 7 pi bonds in Anthracene (colored in red). All pi bonds are present between carbon and carbon atoms.
Number of Electrons in Sigma Bonds:
As one sigma bond is formed by 2 electrons hence, 26 sigma bonds will be formed by 52 electrons.
Number of Electrons in Pi Bonds:
As one pi bond is formed by the side wise overlap of two p orbitals hence one pi bond is formed by two electrons so, 7 pi bonds will be formed by 14 electrons.
The hybrid orbital that is utilized by the carbon in anthracene is [tex]\boxed{{\text{s}}{{\text{p}}^2}}[/tex] .
The anthracene molecule contains [tex]\boxed{{\mathbf{26}}{\text{ }}{\mathbf{sigma}}}[/tex] bonds and [tex]\boxed{{\mathbf{7}}{\text{ }}{\mathbf{\pi }}}[/tex] bonds.
The number of valence electrons in sigma orbital is [tex]\boxed{{\mathbf{52}}}[/tex] and pi-orbitals is [tex]\boxed{{\mathbf{14}}}[/tex] .
Further explanation:
The anthracene is a crystalline compound and it is yellow in color. It contains three fused rings of benzene thus it is a polyaromatic compound.it has a molecular formula [tex]{{\text{C}}_{{\text{14}}}}{{\text{H}}_{{\text{10}}}}[/tex].
Prediction of hybridization:
The hybridization can be determined by calculating the number of hybrid orbitals (X) which is to be formed by the atom. The formula to calculate the number of hybrid orbitals (X) as follows:
[tex]\boxed{{\text{X}}={\text{Number of bond pair}}+{\text{Number of lone pair}}}[/tex]
Here X is a steric number.
When X is 2 then hybridization is sp.
When X is 3 then hybridization is [tex]{\text{s}}{{\text{p}}^2}[/tex].
When X is 4 then hybridization is [tex]{\text{s}}{{\text{p}}^3}[/tex] .
The structure of anthracene is attached in the image.
Since all carbon atom in anthracene contains three bond pairs and no lone pair thus, the hybridization can be calculated as follows:
[tex]\begin{aligned}{\text{X}}&={\text{Number of bond pair}}+{\text{Number of lone pair}}\\&=3+0\\&=3\\\end{aligned}[/tex]
The value of X is 3, therefore, the hybridization of each carbon in anthracene is [tex]{\text{s}}{{\text{p}}^2}[/tex]
The structure of anthracene contains 26 sigma bond and 7 pi bonds. (refer to the image attached).
The number of valance electron in sigma orbital is twice of the number of sigma bond present in the molecule because every sigma bond contains 2 electrons.
[tex]{\text{Valence electron in sigma orbital}} = 2\left({{\text{sigma bond}}}\right)[/tex]
In anthracene, number of sigma bond is 26, therefore,
[tex]\begin{aligned}{\text{Valence electron in sigma orbital}}&=2\left({{\text{sigma bond}}}\right)\\&=2\left({{\text{26}}}\right)\\&=52\\\end{aligned}[/tex]
The number of valance electron present in pi-orbital is twice of the number of pi bond present in the molecule because every pi-bond contains 2 electrons.
[tex]\begin{aligned}\text{Valence electron in }\pi\text{-orbital}&=2(\pi\text{-bond})\end{aligned}[/tex]
In anthracene, number of pi bond is 7, therefore,
[tex]\begin{aligned}\text{Valence electron in }\pi\text{-orbital}&=2(\pi\text{-bond})\\&=2(7)\\&=14\end{aligned}[/tex]
Learn more:
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Answer details:
Grade: Senior school
Subject: Chemistry
Chapter: Covalent bonding
Keywords: Anthracene, yellow crystalline, coal tar, structure of anthracene, C14H10, hybrid orbitals.
