
Hybridization of carbon atom in ${\rm{C}}{{\rm{H}}_{\rm{3}}}^ + $ is:
A. ${\rm{sp}}$
B. ${\rm{s}}{{\rm{p}}^{\rm{2}}}$
C. ${\rm{s}}{{\rm{p}}^{\rm{3}}}$
D. ${\rm{s}}{{\rm{p}}^{\rm{3}}}{\rm{d}}$
Answer
575.1k+ views
Hint: We know that hybridization is the process in which two or more than two orbitals overlap and generate a bond between them. Usually, these bonds may be single bond, double bond and the triple bond.
Complete step by step answer:
The main orbitals which participated in the overlapping of electrons are ${\rm{s,}}\;{\rm{p,}}\;{\rm{d}}$ and ${\rm{f}}$. With the help of these orbitals, we can identify the hybridization of any element or molecules. The main hybridization are ${\rm{sp,}}\;{\rm{s}}{{\rm{p}}^{\rm{2}}}{\rm{,}}\;{\rm{s}}{{\rm{p}}^{\rm{3}}}{\rm{,}}\;{\rm{s}}{{\rm{p}}^{\rm{3}}}{\rm{d}}$ and ${\rm{s}}{{\rm{p}}^{\rm{3}}}{{\rm{d}}^{\rm{2}}}$. All the hybridizations have different representations for the structural geometry.
As we all know, the molecule ${\rm{C}}{{\rm{H}}_{\rm{3}}}^ + $ has three pairs of electrons nearby the carbon atom. The structural geometry of t${\rm{C}}{{\rm{H}}_{\rm{3}}}^ + $ is trigonal planar. The total hybrid orbitals of ${\rm{C}}{{\rm{H}}_{\rm{3}}}^ + $ can be calculated is shown below.
${\rm{Total}}\;{\rm{hybrid}}\;{\rm{orbitals}} = {\rm{Number}}\;{\rm{of}}\;{\rm{bonds}} + {\rm{Number}}\;{\rm{of}}\;{\rm{lone}}\;{\rm{pair}}\;{\rm{of}}\;{\rm{electrons}}$
So, the number of bond of carbon in ${\rm{C}}{{\rm{H}}_{\rm{3}}}^ + $ molecule is three and the number of lone pair of electrons is zero.
Substitute all the respective values in the above mathematical equation.
$\begin{array}{c}
{\rm{Total}}\;{\rm{hybrid}}\;{\rm{orbitals}} = {\rm{3}} + {\rm{0}}\\
= {\rm{3}}
\end{array}$
Thus, we get three total hybrid orbitals which means it has one ${\rm{s}}$ orbital and two ${\rm{p}}$ orbitals. The carbon is one of the central atoms in it. So, the hybridization of carbon in ${\rm{C}}{{\rm{H}}_{\rm{3}}}^ + $ molecule is ${\rm{s}}{{\rm{p}}^{\rm{2}}}$
Hence, the correct answer for this question is B.
Note:
Nowadays, the hybridization I used for many research purposes in the branch of biochemical biology. The hybridization can generally be affected by the increasing or decreasing of temperature.
Complete step by step answer:
The main orbitals which participated in the overlapping of electrons are ${\rm{s,}}\;{\rm{p,}}\;{\rm{d}}$ and ${\rm{f}}$. With the help of these orbitals, we can identify the hybridization of any element or molecules. The main hybridization are ${\rm{sp,}}\;{\rm{s}}{{\rm{p}}^{\rm{2}}}{\rm{,}}\;{\rm{s}}{{\rm{p}}^{\rm{3}}}{\rm{,}}\;{\rm{s}}{{\rm{p}}^{\rm{3}}}{\rm{d}}$ and ${\rm{s}}{{\rm{p}}^{\rm{3}}}{{\rm{d}}^{\rm{2}}}$. All the hybridizations have different representations for the structural geometry.
As we all know, the molecule ${\rm{C}}{{\rm{H}}_{\rm{3}}}^ + $ has three pairs of electrons nearby the carbon atom. The structural geometry of t${\rm{C}}{{\rm{H}}_{\rm{3}}}^ + $ is trigonal planar. The total hybrid orbitals of ${\rm{C}}{{\rm{H}}_{\rm{3}}}^ + $ can be calculated is shown below.
${\rm{Total}}\;{\rm{hybrid}}\;{\rm{orbitals}} = {\rm{Number}}\;{\rm{of}}\;{\rm{bonds}} + {\rm{Number}}\;{\rm{of}}\;{\rm{lone}}\;{\rm{pair}}\;{\rm{of}}\;{\rm{electrons}}$
So, the number of bond of carbon in ${\rm{C}}{{\rm{H}}_{\rm{3}}}^ + $ molecule is three and the number of lone pair of electrons is zero.
Substitute all the respective values in the above mathematical equation.
$\begin{array}{c}
{\rm{Total}}\;{\rm{hybrid}}\;{\rm{orbitals}} = {\rm{3}} + {\rm{0}}\\
= {\rm{3}}
\end{array}$
Thus, we get three total hybrid orbitals which means it has one ${\rm{s}}$ orbital and two ${\rm{p}}$ orbitals. The carbon is one of the central atoms in it. So, the hybridization of carbon in ${\rm{C}}{{\rm{H}}_{\rm{3}}}^ + $ molecule is ${\rm{s}}{{\rm{p}}^{\rm{2}}}$
Hence, the correct answer for this question is B.
Note:
Nowadays, the hybridization I used for many research purposes in the branch of biochemical biology. The hybridization can generally be affected by the increasing or decreasing of temperature.
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