Humboldt State University ® Department of Chemistry

Richard A. Paselk

	General Chemistry	Spring 2015
Exercise: Hybrid Molecular Orbitals	© R. Paselk 2015
	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. Methane, CH₄

• LS: from symmetry C will be central atom, therefore=
• Considering C as the central atom, have 4 bonded atoms and no lone-pairs, therefore
  • Steric Number (SN) = 4, so need to hybridize 4 orbitals, one s and three p, therefore have 4 sp³ orbitals = tetrahedral electronic geometry
  • No lone pairs so tetrahedral molecular geometry as well:

rotated to a different angle =

2. Hydrogen sulfide

• LS: from symmetry S will be central atom, therefore =
• Considering S as the central atom, have 2 bonded atoms and 2 lone-pairs, 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

3. Carbon monoxide, CO

• valence electrons = 4 + 6 = 10, 6y + 2 = 14, thus 4 fewer electrons than required for all single bonds, 4/2 = 2 multi-bonds (2 double or 1 triple)
• LS = :C:::O:
• Considering C as the central atom, have one bonded atom and one lone-pair, therefore
  • steric number = 2, so two hybridized orbitals, obe s and one p = linear electronic geometry
  • one lone pair, so linear molecular geometry

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.
• Carbon monoxide is polar as shown with the dipole arrow in the image:

4. Ammonia, NH₃

• valence electrons = 5 + 3 x 1= 8, 6y + 2 = 8, so single bonds:
• LS: from symmetry N will be central atom, therefore=
  • From LS have 3 bonded atoms and one 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

rotated to a different angle =

Polarity:

• H and N have significantly different Electronegativity values (2.1& 3.0), so the H-N bond will be polar, with N partially negative.
• Ammonia is polar as shown with the dipole arrow in the image below:



• The polar contributions of the angled polar H-N bonds along the central axis add up so overall the molecule is polar.

 

---

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

---

Hybrid Molecular Orbitals

© R A Paselk

Last modified 20 April 2015