Lecture 9 covalent bonding part 1 lewis structures ch 7
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Lecture 9: Covalent Bonding: Part 1 Lewis Structures (Ch. 7). Dr. Harris Suggested HW: 2, 7, 9, 13, 14, 17, 27d. . Recap. Ionic compounds are formed by the transfer of electrons from a metal to a nonmetal

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Lecture 9: Covalent Bonding: Part 1 Lewis Structures (Ch. 7)

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Lecture 9 covalent bonding part 1 lewis structures ch 7

Lecture 9: Covalent Bonding: Part 1 Lewis Structures (Ch. 7)

Dr. Harris

Suggested HW: 2, 7, 9, 13, 14, 17, 27d.


Recap

Recap

  • Ionic compounds are formed by the transfer of electrons from a metal to a nonmetal

  • The result is a cation-anion pair. Each ion has reached its nearest noble gas configuration

Na ([Ne] 3s1) + Cl ([Ne] 3s2 3p5) ---> Na+ Cl-

Cl-

Na+

[Ar]

[Ne]

  • The Coulombic attraction between the oppositely charged ions is what holds the molecule together. This is why ionic compounds are solids at room temperature.


Covalent bonding

Covalent Bonding

  • Covalent bonding is electron sharing (not transfer) between nonmetals

  • Consider Cl2(g)

    • Each Cl has 7 valence electrons

    • Chlorine is 1 electron short of a full octet ([Ar] configuration)

    • Since both atoms are the same, they have the same electron affinities and ionization energies, so 1 chlorine will not donate an electron to the other.

[Ne]3s23p5

[Ne]3s23p5

Cl

Cl


Covalent bonding1

Covalent Bonding

  • For both atoms to achieve an [Ar] configuration (full octet), they share a pair of electrons between them. An element in a covalent bond will react so that eight electrons occupy its valence shell. This is the OCTET RULE.

  • To indicate the covalent bond, we use a solid line. This is a single bond (2 electrons). Bonding represents an overlap of the valence orbitals.

  • The electrons not involved in bonding are called lone pairs.

[Ar]

[Ar]

Covalent bond

Cl

Cl

Cl

Cl


Covalent bonding2

Covalent Bonding

  • All elements in a covalent bond will achieve an octet

  • Ex. OF2

    • Oxygen has 6 valence electrons (needs 2)

    • Fluorine has 7 (each F needs 1)

    • To achieve octets, each F will share its electron with O

[He]2s22p5

[He]2s22p5

[He]2s22p4

O

F

F

F

O

F

F O F

[Ne]

[Ne]

[Ne]


Electronegativity

Electronegativity

  • Before we learn about how to draw a Lewis structure, it is very important to consider what makes ionic and covalent bonds so different: electronegativity

  • Electronegativity is the ability of an atom to attract electrons to itself.

  • Therefore, electrons are drawn more towards the more electronegative atom in a molecule.

  • Thus, when there is a difference in electronegativities between atoms in a molecule, the electrons are NOT equally shared.


Table of electronegativities

Table of Electronegativities

Electronegativity of atoms increases up and to the right.


Electronegativity1

Electronegativity

  • Although we have discussed ionic and covalent bonds, most chemical bonds are neither purely ionic or purely covalent

  • Most compounds are an intermediate between the two.


Electronegativity2

Electronegativity

  • Let’s consider NaCl. The difference in electronegativity between Na and Cl is:

  • Because the difference in electronegativity is so big, the Na electron is completely pulled away by the Cl atom. So, the molecule is totally polar (ionic)

  • This is why there is no actual bond in ionic compounds, only coulombic attraction.

3.16 – 0.93 = 2.23 ionic

Cl-

Na+


Electronegativity3

Electronegativity

  • Let’s look at another molecule, like HCl:

  • The HClmolecule is not purely ionic or covalent, but BOTH.

    • The electron density is unevenly distributed, such that more of the electron density is on the Cl than on the H.

  • Therefore, the H and Cl have partial charges. The arrow depicts the direction of electron “pull”, or dipole. Any molecule with a net dipole has polarity. We will discuss dipoles later.

3.16 – 2.1 = 1.06 polar covalent

-

+

δ

δ

Cl

H

Partial positive character

Partial negative character


General rules of drawing covalent lewis structures

General Rules of Drawing Covalent Lewis Structures

  • Arrange atoms together. If possible, put the least electronegative atom in the center. Hydrogen is an exception to this rule. Hydrogen atoms are always terminal.

  • Compute the total number of valence electrons. Account for charges.

  • Represent bonds with solid lines. Ensure octets around all atoms, except hydrogen. Hydrogens can only accommodate 2 electrons.

  • Add in remaining lone pairs

  • If there are not enough electrons to complete an octet, consider multiple bonds or formal charges.


Example

Example

  • Draw the Lewis structure of CCl4

    • Carbon is the least electronegative atom (also, there is only one C), so C is the central atom

    • Compute the valence electrons

    • C: [He] 2s2 2p2 Cl: [Ne] 3s2 3p5

C

  • Each Cl needs 1, the C needs 4. Therefore, the C will share each of its 4 valence electrons with Cl

C

Cl

Cl

Cl

Cl

Cl

Cl

Cl

Cl

Cl

Cl

C

Cl

Cl


Hydrogen

Hydrogen

  • Hydrogen CAN ONLY ACCOMMODATE 2 ELECTRONS

    • Ex. H2(g) [He]-like core

    • HX where X = F, Cl, Br, I (halogens)

    • In chemical structures, hydrogens are always terminal atoms, meaning that they are at the ends of a molecule

H

H

H

X


Organic molecules

Organic Molecules

  • Ex. Methanol, CH3OH

Hydrogens are terminal, which means there is a C—O bond in the center

C

C

C

H

H

H

H

H

O

O

O

Now, C needs 3 more electrons around it to achieve an octet. O needs 1.

H

H

H

H

C

O

H

H

H


Double and triple bonds

Double and Triple Bonds

  • There are instances where single bonds (2 e- bonds) are not enough to satisfy the octet rule.

    • Ex. C2H4 (ethene)

C

C

C

C

C

C

  • As you can see, C only has 7 electrons around it, including a lone electron on each C. This is highly unfavorable.

  • Lone electrons migrate into the C-C bond to form a 4e-double bond.

H

H

H

H

H

H

H

H

H

H

H

H

H

H

H

H

H

H

C

C

C

C

H

H


Double and triple bonds1

Double and Triple Bonds

  • 6e-triple bonds are also possible, as in the case of N2 (g)

N

N

N

N

N

N

Two unpaired electrons each

N

N

6 electron Triple Bond

  • The N atoms are sharing 6 electrons, so each N now has a full octet.


Group examples

Group Examples

  • Draw the following Lewis Structures

    • CH3Cl

    • NH3

    • H2O

    • CO2

    • HCN


Formal charges

Formal Charges

  • So far, all of the atoms we’ve seen in covalent compounds have had a zero charge.

  • However, nonmetals can assume positive or negative charges in order to facilitate the formation of a covalent bond

  • Thus, to satisfy the octet rule in certain cases, formal chargeshave to be applied


Formal charges1

Formal Charges

- As is, N can only make 3 bonds.

X

N

N

N

H

H

H

H

H

H

H

H

H

H

H

H

- N loses an electron to accommodate the 4th H, attains a charge of 1+.

H

+

+

+

N

N

H

H

H

Consider ammonium, NH4+

-Lets start by looking at the valence electrons of the neutral atoms


Formal charge allows us to rule out structures

Formal Charge Allows Us to Rule Out Structures

  • Formal charges can help us determine if we’ve arranged a molecule incorrectly

    The preferred arrangement of atoms in a molecule is always the arrangement that requires the least amount of formal charge

  • If formal charges must exist, the more electronegative elements prefer negative formal charges, and less electronegative elements prefer positive ones.


Ambiguity in atomic arrangement

Ambiguity in Atomic Arrangement

  • Ex. Hydrogen cyanide, HCN

  • We know that C is the central atom in HCN. But how can we rule out HNC? Lets consider both possibilities

8

total

2

total

8

total

  • As we determined earlier, we have a CN triple bond and an HC bond. All octets are satisfied.

  • There are no formal charges required

H

C

N


Ambiguity in atomic arrangement1

Ambiguity in Atomic Arrangement

  • If we rearrange N and C, we have the following structure. As shown, C is 1 electron short of an octet.

    • N can not make anymore bonds because it already has an octet. N has a 1+ formal charge because it loses a valence electron.

7

total

2

total

8

total

+

H

H

N

N

C

C

8

total

8

valence

2

valence

  • C must take on a 1- formal charge to reach an octet by accepting the N electron. Now, all the atoms are satisfied, and the molecule is neutral.

-

+


Ambiguity in atomic arrangement2

Ambiguity in Atomic Arrangement

  • Now, our two possibilities are:

  • Both structures satisfy the octet rule, and both are charge neutral, but the HNC structure has formal charges on both N and C. Thus, the HCN structure is preferred.

8

total

2

total

8

total

H

H

N

C

C

N

8

total

8

total

2

total

-

+


Group example

Group Example

  • The correct structure of the NOCl is a central N with a double bond to O, a single bond to Cl, and a lone pair on the N.

    • Use formal charges to show why O is notthe central atom.


Resonance

Resonance

  • When a covalent molecule has an overall charge, the charge is said to be delocalized, meaning that the charge is spread over the whole molecule, as opposed to being localized on a single atom

  • Ex. Nitrite NO2-

    • To draw this molecule, we put N as the central atom. We satisfy an octet on one O by making a N=O double bond. We can apply the 1- charge to the other O to complete its octet (O is more electronegative than N, so it will not go on N)

ONO

-

The blue electron represents the electron added by applying the formal charge


Resonance1

Resonance

  • You can’t distinguish between one O and the other. So, you could write either of the following:

ONO

ONO

-

-

-

ONO

  • These structure are called resonance structures. We can account for both structures by writing the resonance form, as shown below. All resonance structures have a double bond.


Expansion into the d orbital

Expansion into the d-orbital

  • Nonmetals with available d-orbitals (n> 3) can use these orbitals to hold more than 8 electrons. Unless Carbon is present, assume that these atoms are the central atoms.

  • When drawing Lewis structures of molecules with expanded octets, put octets around the outer atoms first, then account for any remaining electrons on the central atom.

  • Ex. SF4

F

  • After completing octets on the F atoms, there is a lone pair remaining on S

10 electrons around Sulfur

F

S

F

F


Group examples1

Group Examples

  • Draw the Lewis structures for the following “expanded octet” molecules

    • PCl5

    • SF6

    • IF5

Always be mindful when dealing with atoms of n>3 of the possibility of surpassing 8 valence electrons.

  • Draw the resonance structures of:

    • ClO4-

    • PO43-


Revisiting dipoles polar vs nonpolar molecules

Revisiting Dipoles: Polar vs. Nonpolar Molecules

  • Dipole moments describe the coulombic attraction that is formed between the partial positive and negative charges on nonmetals when electrons are unevenly shared.

  • Dipole moments are vectors, quantities with both magnitude and direction. Imagine pulling on a rope. The force is in the direction of the pull.

  • If two people pull on opposing ends of a rope with equal force, the net forceis zero.

  • Polar molecules have an overall dipole moment, nonpolar molecules don’t.


Table of electronegativities1

Table of Electronegativities

Electronegativity of atoms increases up and to the right.


Lecture 9 covalent bonding part 1 lewis structures ch 7

Here, electronegativity is labeled as β

-

+

O

O

C

C

S

O

-

δ

δ

δ

Equal force of attraction in opposite directions: Non polar

+

= 0

Δβ = 0.89

Δβ = 0.89

Unequal dipole moments. Thus, there is an overal dipole: Polar

-

-

+

δ

δ

δ

+

=

Δβ = 0.03

Δβ = 0.89


Group example1

Group Example

  • Draw a water molecule. Then draw the dipoles and determine the net dipole. Is the molecule polar?

O

H

H

+

=

net dipole =

polar


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