2.2.16 Hybridization

• Note that a carbon is tetravalent, meaning that it can form 4 covalent bond

• Note if true, then they are single bonds, and thus have 4 domains and thus has a Tetrahedral structure around the carbon atom approximately $109.5^{\degree}$

  • Look at $CH_4$!

However note the ground state electron configuration of carbon, $1s^22s^22p^2$, contradicts these observations,

• It only has two unpaired electrons so doesn't that mean they require only 2 electrons to form two bonds not 4$(a)$

• ** The two occupied $2p$are at $90 \degree$from one another not $109.5\degree$

  • Thus we conclude that the atomic orbitals must undergo certain changes when forming bonds.

This leads to hybridization, which is the concept of mixing atomic orbitals to form new hybrid orbitals for bonding. THere are two steps

1. Promotion

   In the case of figure 81, the 2s electron is promoted to be in the 2p orbitals

2. Hybridization

The singely occupied 2p orbitals are hybridized, meaning they combine and give rise to orbitals of new shapes. These resulting orbitals are called hybrid orbital and they all have the same energy

Hybridization is energetically favourable.

• The promotion step absorbs energy

• But the bonding formation that follows outweighs this

• Promotion does not use much energy because it relieves the 2s electron of the repulsion it experienced when paired

$sp^3$hybrid orbitals

• #of resulting hybrid orbitals is equal to the # of atomic orbitals required to make them

• Each hybrid orbital is a mixture of one part 2s and three parts 2p

  • 25% s character

  • 75% p character

• Tetrahedral shape

  • 109.5 bond angles

• Each orbitals is occupied by one electron and can form 4 sigma bonds

  • In $CH_4$, methane, each $sp^3$hybrid overlaps with a 1s atomic orbital on a hydrogen atom ( forming four covalent bonds ).

$sp^2$hybrid orbitals

• Combination of one 2s and two 2p atomic orbitals in carbon

  • To produce 3 $sp^2$hybrid orbitals

• Each orbitals contains one electron so can form sigma bonds

• The remaining unhybridized $p_z$orbital can then go to form a pi bond with a parallel $p_z$orbital in another atom.

sp hybrid orbitals

• Combination of 1 2s and 1 2p orbital

  • Two sp orbitals

• Linear arrangment w/ 180$\degree$between them

• Hybrid orbitals can form sigma bonds

  • The remaining unhybridized $p_y$and $p_z$orbitals can form two pi bonds with parallel p orbitals on a neighbouring atom

Hybridization in other atoms

• Also occurs in atoms other than carbon

  • number of electrons in the p orbitals will different, but general principle is the same

  • Consider the $sp^3$hybridized orbital

1. Oxygen ground state electron configuration is: $1s^22s^22p^4$

2. Distinguish between the 2p orbitals that contain one electron

$1s^2 2s^2 2p_x^2 2p_y^12p^1$

3. The 2s and 2p orbitals combine to form 4 equivalent $sp^3$hybrid orbitals containing 6 electrons

Two of the $sp^3$orbitals form bonded pairs and $\therefore$don't form bonds. The remaining two hybrid orbitals with one electrons each form sigma bonds with the s orbital on hydrogen atoms.

• $\because$the orbitals are tetrahedrally arranged--> bond angle 109.5$\degree$.

  • However, since the oxygen atom has 2 lone pairs, it has a slightly bent molecular geometry, instead of 109.5$\degree$, its 104.5$\degree$

    • Which suggest that the hybridization is close to but not exactly $sp^3$

Hybridiation and geometry

• Sigma bonding and Hybridization closely related to electron domain geometry

  • #of hybrid orbitals formed by an atom is equal to the number of its electron domains.

• Recall the discussion from the 2.2.4 The valence shell electron pair repulsion model( VSEPR ), that double bonds and triple bonds count as one domain in molecular shape

  • This is $\because$pi bonds do not contribute to hybridized orbitals --> not much effect on geometry of molecule

Hybridization and delocalization

• Consider ethanoate ion, $CH_3COO^-$

  • Forms when etanoic acid losesa hydrogen ion

• By doing so the carbon-oxygen bond inevitably changes

  • As seen from the figure, the bond order of C-O goes from 2 --> 1.5

    • $\because$the electrons in the double bond become delocalizaed across the two C-O domains

• NOte that in the C=O bond of $CH_3COOH$, etanoic acid, the carbon and oxygen are $sp^2$hybridized. The unhybridized 2p orbital electron in each atom forms the pi bond between them. The other oxygen atom in $-OH$has $sp^3$hybridization.

• After losing the $H$to form Ethanoate, the remaining oxygen atom inevitably adopts an $sp^2$hybridization.

  • Oxygen atoms, carbon atoms now have 3 electron domains around them

• Three electron domains correspond to $sp^2$hybridization and one unhybridized 2p orbital each.

  • $\because$the orbitals overlap, the p electrons in them become delocalizaed between the hybridized orbitals

  

Additional resources:

https://chem.libretexts.org/Bookshelves/Organic_Chemistry/Supplemental_Modules_(Organic_Chemistry)/Fundamentals/Hybrid_Orbitals