Aromatic compounds are compounds that contain carbon-carbon multiple bonds.
The question did not mention that a heteroatom is present in the compound so we can assume that there is none of such. In that case, the compound contains only hydrogen and carbon.
So,
(CH)n = 78
where n is the number of each atom present.
(12 +1)n = 78
n = 78/13
n = 6
The molecular formula of the compound is C6H6
When C6H6 is treated with .conc.HNO3/conc.H2SO4 the compound shown in image 1 is formed. The reaction occurs at the C-C multiple bond.
When C6H6 is reacted with chlorine in the presence of sunlight, hexachlorobenzene (shown in image 2 attached) is formed.
Answer:
for the given reaction is -238.7 kJ
Explanation:
The given reaction can be written as summation of three elementary steps such as:
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Answer:
The required hybridization for a central atom that have:
a tetrahedral arrangement of electron pairs is
a trigonal planar arrangement of electron pairs is
a linear arrangement of electron pairs is
Explanation:
From the given information:
The required hybridization for a central atom that have:
a tetrahedral arrangement of electron pairs is
a trigonal planar arrangement of electron pairs is
a linear arrangement of electron pairs is
Hybridization is the mixing and blending of two or more pure atomic orbitals ( s, p and d) to forma two or more hybrid atomic orbitals that are identical in shape and energy e. sp ,sp² , sp³, sp³d hybrid orbitals.
Examples:
Tetrahedral
In CH₄ , carbon C is the central atom.
A 2s electron is excited from the ground state of boron 1s²2s²2p² to one of the empty orbitals to 2p to give the excited state 1s²2s²2p³.
In the excited state of carbon, the 2-s orbital can be mixed with the 2p orbitals in three ways : sp³, sp² and sp hybridization. For the formation of four sp³ hybrid orbitals, the 2s orbital are mixed with all the three p orbitals. The four sp³ hybrid orbitals are tetrahedrally arranged with a bond angle of 109.5⁰
Trigonal Planar
In BF₃ , Boron B is the central atom
A 2s electron is excited from the ground state of boron 1s²2s²2p¹ to one of the empty orbitals to 2p. The 2s orbital is then mixed with two orbitals of 2p to form three sp² hybrid orbitals that are trigonal planar arranged in the plane in order to minimize repulsion. They bonds between them form an equal strength and with bond angles of 120⁰
Linear arrangement:
In BeCl₂, Be is the central atom
To provide the unpaired electrons for covalent bonds a 2s electron is excited to a 2p orbital. Thereafter, the two atomic orbitals hybridize to give two identical orbitals called sp hybrid orbitals. The sigma bond for,ed are equal in bond lengths and form a bond angle of 180°