# Use valence bond theory to write the hybridization and bonding scheme for NCCH3. Sketch the model with the right geometry?

Jul 15, 2017

Warning! Long Answer. Here's what I get.

#### Explanation:

Step 1. Draw the Lewis structure

The two $\text{C}$ atoms (least electronegative) will be the central atoms, with the $\text{N}$ attached to one of the carbons. (b) Attach the hydrogen atoms.

The question gives you a clue where they go.

The formula ${\text{NCCH}}_{3}$ tells you that the three $\text{H}$ atoms are attached to the terminal carbon atom. (c) Add electrons so that every atom gets an octet.

There is only one good way to do it: put a triple bond between the $\text{C}$ and the $\text{N}$, and a lone pair on the $\text{N}$. Step 2. Use VSEPR theory to predict the geometry about each atom.

• The terminal $\text{C}$ atom: four electron domains (3 $\text{C-H}$ bonds and a $\text{C-C}$ bond).
∴ Tetrahedral.
• The central $\text{C}$ atom: two electron domains (a $\text{C-C}$ bond and a $\text{C≡N}$ bond).
∴ Linear.
• The $\text{N}$ atom: two electron domains (a $\text{C≡N}$ bond and a lone pair)

Step 3. Assign hybridizations to each atom

• The terminal $\text{C}$ atom: tetrahedral. ∴ ${\text{sp}}^{3}$ hybridized.
• The central $\text{C}$ atom: linear. ∴ $\text{sp}$ hybridized.
• The $\text{N}$ atom: two electron domains. ∴ $\text{sp}$ hybridized.

Step 4. Sketch the orbitals involved

Here is my drawing (apologies! I'm not a graphics artist). My drawing may be confusing, so here is a description:

• $\text{H}$ atoms: unhybridized $\text{1s}$ orbitals
• Terminal $\text{C}$ atom: ${\text{sp}}^{3}$ hybridized; bond angles 109.5°
• $\text{C-H}$ bonds: σ bonds formed by overlap of $\text{H 1s}$ and ${\text{C sp}}^{3}$ orbitals
• Central $\text{C}$ atom: $\text{sp}$ hybridized
• $\text{C-C}$ bond: σ bond formed by overlap of ${\text{C sp}}^{3}$ and $\text{C sp}$ orbitals
• $\text{N}$ atom: $\text{sp}$ hybridized.
• $\text{C≡N}$ bond: σ bond formed by overlap of $\text{C sp}$ and $\text{N sp}$ orbitals, plus two bonds formed by side-on overlap of unhybridized $\text{2p}$ orbitals
• $\text{C-C-N}$ bond angle = 180°

Here's a model of the molecule for comparison: 