Draw the Lewis structure of [NH2]–
Draw the Lewis structure of [(CH3)3O]+
Label(or unlabel) the sp-hybridized atoms, sp2-hybridized atoms and sp3-hybridized atoms. Basically I need help determining which of those atoms are the sp, sp2, sp3 hybridized atoms
Picture this as written going straight across.
-H2C single bonded to N+ triple bonded to N
It is almost impossible to draw Lewis structures on these boards because of the spacing problem. Single bonds are sp3, double bonds are sp2 and triple bonds are sp.
To draw the Lewis structure of [NH2]–, we need to determine the number of valence electrons for each atom.
N - Nitrogen has 5 valence electrons (group 15), and since it has a negative charge, we add one more electron.
H - Hydrogen has 1 valence electron.
To draw the Lewis structure:
1. Place the atoms in the structure, connecting them with single bonds. Start by placing the N atom in the center and the H atoms surrounding it.
H
|
H-N-H
2. Fill the octet for each atom. Nitrogen can accommodate 8 electrons and hydrogen can accommodate 2 electrons, so place all the remaining electrons around the atoms.
H
|
H-N-H
:
3. Check if all atoms have an octet. In this case, nitrogen still has only 7 electrons around it.
To draw the Lewis structure of [(CH3)3O]+, we need to determine the number of valence electrons for each atom.
C - Carbon has 4 valence electrons.
H - Hydrogen has 1 valence electron.
O - Oxygen has 6 valence electrons (group 16), and since it has a positive charge, we subtract one electron.
To draw the Lewis structure:
1. Place the atoms in the structure, connecting them with single bonds. Start by placing the C atom in the center and the H and O atoms surrounding it.
H
|
H-C-H
|
H
2. Fill the octet for each atom. Carbon can accommodate 8 electrons, hydrogen can accommodate 2 electrons, and oxygen can accommodate 8 electrons.
H
|
H-C-H
|
H
:
3. Check if all atoms have an octet. In this case, all atoms have a complete octet.
Now let's determine the hybridization of each atom:
- sp hybridized atoms have two regions of electron density around the central atom (e.g., linear shape).
- sp2 hybridized atoms have three regions of electron density around the central atom (e.g., trigonal planar shape).
- sp3 hybridized atoms have four regions of electron density around the central atom (e.g., tetrahedral shape).
In [NH2]–, the nitrogen atom is bonded to two hydrogen atoms, resulting in three regions of electron density. Therefore, it is sp2 hybridized.
In [(CH3)3O]+, the carbon atom is bonded to three hydrogen atoms and one oxygen atom, leading to four regions of electron density. Therefore, it is sp3 hybridized.
Note that the hybridization of atoms can be determined by counting the regions of electron density around them, which can be obtained by summing the number of bonds and lone pairs on the central atom.