Humboldt State University ® Department of Chemistry

Richard A. Paselk

General Chemistry
Spring 2015

Exercise: Hybrid Molecular Orbitals
© R. Paselk 2008

Discussion Modules

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*Humboldt State University* ® *Department of Chemistry*

Richard A. Paselk
	General Chemistry	Spring 2015
Exercise: Hybrid Molecular Orbitals	© R. Paselk 2008
	Discussion Modules
	Use Back Button to Return to Notes

Hybrid Molecular Orbitals

Periodic Table of the Elements

IA IIA IIIA IVA VA VIA VIIA VIIIA

H He

2 Li Be B C N O F Ne

3 Na Mg IIIB IVB VB VI VIIB VIIIB IB IIB Al Si P S Cl Ar

4 K Ca Sc Ti V Cr Mn Fe Co Ni Cu Zn Ga Ge As Se Br Kr

5 Rb Sr Y Zr Nb Mo Tc Ru Rh Pd Ag Cd In Sn Sb Te I Xe

6 Cs Ba Lu Hf Ta W Re Os Ir Pt Au Hg Tl Pb Bi Po At Rn

1. Formaldehyde, CH₂O

Valence electrons = 4 + 2 x 1 + 6 = 12, 6y + 2 = 6 x 2 + 2 = 14; so molecule has 2 fewer electrons than required for all single bonds, 1 double bond

LS: from symmetry C will be central atom, therefore=

From LS have 3 bonded atoms and no lone-pairs, therefore

SN = 3, so need to hybridize 3 orbitals, one s and two p, therefore have 3 sp² orbitals = trigonal planar electronic geometry, and 3 atoms so

trigonal planar molecular geometry

Three sigma (single) bonds, one pi (double) bond.

or, showing the double bond

Polarity:

C and O have significantly different Electronegativity values (2.5 & 3.5), so the C-O bond will be polar, with O partially negative. C and H differ only slightly in EN (2.5 & 2.1), so the C-H bonds will be only slightly polar.

Formaldehyde is polar as shown with the dipole arrow in the image below:

2. Carbon dioxide, CO₂

Valence electrons = 4 + 2 x 6 = 16, 6y + 2 = 20, thus 4 fewer electrons than required for all single bonds, 4/2 = 2 multi-bonds (2 double or 1 triple)

LS: from symmetry C will be central atom, therefore= :O::C::O:

Considering C as the central atom, have 2 bonded atoms and no lone-pairs, therefore

steric number = 2, so so need to hybridize 2 orbitals, one s and one p, therefore have 2 sp orbitals = linear electronic geometry, and 2 atoms and no lone pairs, so

linear molecular geometry

Two sigma bonds, and two pi bonds.

Polarity:

C and O have significantly different Electronegativity values (2.5 & 3.5), so the C-O bond will be polar, with O partially negative.

However, the two polar bonds exactly cancel each other since they point in opposite directions, so the molecule is not polar.

3. Chloromethane, CH₃Cl

**Periodic Table of the Elements**
	IA	IIA		IIIA	IVA	VA	VIA	VIIA	VIIIA
H	He
2	Li	Be		B	C	N	O	F	Ne
3	Na	Mg	IIIB	IVB	VB	VI	VIIB	VIIIB	IB	IIB	Al	Si	P	S	Cl	Ar
4	K	Ca	Sc	Ti	V	Cr	Mn	Fe	Co	Ni	Cu	Zn	Ga	Ge	As	Se	Br	Kr
5	Rb	Sr	Y	Zr	Nb	Mo	Tc	Ru	Rh	Pd	Ag	Cd	In	Sn	Sb	Te	I	Xe
6	Cs	Ba	Lu	Hf	Ta	W	Re	Os	Ir	Pt	Au	Hg	Tl	Pb	Bi	Po	At	Rn

Valence electrons = 4 + 3x1 + 7 = 14, 6y + 2 = 6(2) + 2 = 14, no differenceso 8 electrons in single bonds around central atom.
LS: from symmetry C will be central atom, therefore =
Considering C as the central atom, have 4 bonded atoms and no lone-pairs,
SN = 4, so need to hybridize 4 orbitals, one s and three p, therefore have 4 sp³ orbitals = tetrahedral electronic geometry.

Since no lone pairs, molecular geometry = electronic geometry.
All sigma bonds.

4. Acetic acid, CH₃CO₂H

Valence electrons = 2x4 + 4x1 + 2x6 = 24, 6y + 2 = 6(4) + 2 = 26, thus 2 fewer electrons than required for all single bonds, 2/2 = 1 double bond.
LS: from formula can see that three hydrogens and one carbon are bonded to first carbon, with the two oxygens bonded to the second giving =
Consider the two carbons individually as central atoms.
- C-1 as the central atom, have 4 bonded atoms (3H & C) and no lone-pairs, therefore
  - SN = 4, so need to hybridize 4 orbitals, one s and three p, therefore have 4 sp³ orbitals = tetrahedral electronic geometry.
  - Since no lone pairs, molecular geometry = electronic geometry
  - All sigma bonds.

C-2 as central atom, SN = 3, so need to hybridize 3 orbitals, one s and two p, therefore have 3 sp² orbitals = trigonal planar electronic geometry, and 3 atoms so

trigonal planar molecular geometry. Overall then have a tetrahedron connected to a trigonal plane:

rotated =

5. Ozone, O₃

Valence electrons = 3x6 = 18, 6y + 2 = 6(3) + 2 = 20, thus 2 fewer electrons than required for all single bonds, 2/2 = 1 double bond.
LS: =
- However, since three atoms identical the bonding must also be identical - need resonance structures:
- Have 2 sigma or single bonds and one "resonant" pi bond.
Considering the middle O as the central atom have 2 bonded atoms and one lone pair in both resonance forms,
- steric number = 3, so trigonal planar electronic geometry,
- But one lone pair, so V-shape or Bent geometry,

6. Sulfite ion, SO₃^2-

Valence electrons = 6 + 3x6 + 2= 26, 6y + 2 = 6(4) + 2 = 26, thus atoms have enough electrons for octets with no multibonding.
LS: from symmetry S will be central atom, therefore =
Considering S as the central atom, have 4 bonded atoms and 1 lone-pair, therefore
steric number = 4, so need to hybridize 4 orbitals, one s and three p, therefore have 4 sp³ orbitals = tetrahedral electronic geometry
One lone pair so trigonal pyramidal molecular geometry with three sigma bonds:

rotated =

Polarity:

S and O have significantly different Electronegativity values (2.5 & 3.5), so the S-O bond will be polar, with O partially negative.

Sulfite ion is polar as shown with the dipole arrow in the image below:

The polar contributions of the angled polar S-O bonds along the central axis add up so overall the molecule is polar. (Note that the dipole is in the opposite direction to that seen earlier in ammonia.)

7. Water, H₂O

Looking at the Periodic Table, H is Intermediate and O is Hi, therefore covalent;

valence electrons = 6 + 2x1= 8

only 4 pairs, single bonds because only have H's bound to O, 2 bonds, since only 2 H's

LS: from symmetry O will be central atom, therefore=

SN = 4, so need to hybridize 4 orbitals, one s and three p, therefore have 4 sp³ orbitals = tetrahedral electronic geometry

two lone pairs, so bent molecular geometry with two sigma bonds:

rotated =

Polarity:

H and O have significantly different Electronegativity values (2.1& 3.5), so the H-O bond will be polar, with O partially negative.

The two dipoles add together in the water molecule which is polar as shown with the dipole arrow in the image:

8. Carbocation, CH₃⁺

Valence electrons = 4 + 3x1 - 1 = 6

Three sigma bonds possible, since only three pairs, single bonds because only have H's bound to C.

LS: from symmetry C will be central atom, therefore=

SN = 3, so need to hybridize 3 orbitals, one s and two p, therefore have 3 sp² orbitals = trigonal planar electronic geometry, and 3 atoms and no lone pairs so

trigonal planar molecular geometry

In addition to these exercises you should familiarize yourself with the text materials.