What is the hybridization of the central atom of each of the following molecules? Match the…
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Question “What is the hybridization of the central atom of each of the following molecules? Match the…”
What is the hybridization of the central atom of each of the
following molecules? Match the atoms below with their number—sp,
sp2, sp3, sp3d, sp3d2
COH2
H2S
PBr5
Answer
Electronic geometry can help explain the process of hybridization in organic compounds. Electronic geometry provides information about both the bond pair and the lone electron pair. Valence Shell Electron Pair Respulsion (VSEPR), theory can describe the geometry of a molecule.
The electronic and molecular geometry of a molecule can be determined using a Lewis structure.
A Lewis structure is a skeletal arrangement that includes all elements and non-bonding electronic components. Two electrons signify one pair of electrons.
Below is the table that shows hybridization using electronic geometry with different combinations of bonds or lone pairs electrons on central atoms.
|
|
|
|
2 | 0 | Linear | {\rm{sp}} |
3 | 0 | Trigonal planar | {\rm{S}}{{\rm{p}}^{\rm{2}}} |
2 | 1 | Trigonal planar | {\rm{S}}{{\rm{p}}^{\rm{2}}} |
4 | 0 | Tetrahedral | {\rm{S}}{{\rm{p}}^3} |
3 | 1 | Tetrahedral | {\rm{S}}{{\rm{p}}^3} |
2 | 2 | Tetrahedral | {\rm{S}}{{\rm{p}}^3} |
5 | 0 | Trigonal bipyramidal | {\rm{S}}{{\rm{p}}^{\rm{3}}}{\rm{d}} |
4 | 1 | Trigonal bipyramidal | {\rm{S}}{{\rm{p}}^{\rm{3}}}{\rm{d}} |
3 | 2 | Trigonal bipyramidal | {\rm{S}}{{\rm{p}}^{\rm{3}}}{\rm{d}} |
2 | 3 | Trigonal bipyramidal | {\rm{S}}{{\rm{p}}^{\rm{3}}}{\rm{d}} |
6 | 0 | Octahedral | {\rm{S}}{{\rm{p}}^{\rm{3}}}{{\rm{d}}^{\rm{2}}} |
5 | 1 | Octahedral | {\rm{S}}{{\rm{p}}^{\rm{3}}}{{\rm{d}}^{\rm{2}}} |
4 | 2 | Octahedral | {\rm{S}}{{\rm{p}}^{\rm{3}}}{{\rm{d}}^{\rm{2}}} |
Ans:
\begin{array}{l}\\{\rm{The}}\,{\rm{hybridization}}\,{\rm{of}}\,{\rm{the}}\,{\rm{central}}\,{\rm{atom}}\,{\rm{of}}\,{\rm{the}}\,{\rm{molecules}}\,{\rm{is:}}\\\\{\rm{CO}}{{\rm{H}}_{\rm{2}}}\,{\rm{ = }}\,{\rm{s}}{{\rm{p}}^{\rm{2}}}\\\\{{\rm{H}}_{\rm{2}}}{\rm{S = s}}{{\rm{p}}^{\rm{3}}}\\\\{\rm{PB}}{{\rm{r}}_{\rm{5}}}{\rm{ = s}}{{\rm{p}}^{\rm{3}}}{\rm{d}}\\\end{array}Conclusion
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