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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

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

Dr. Harris

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


Recap
Recap 7)

  • 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 7)

  • 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 7)

  • 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 7)

  • 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 7)

  • 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 7)Electronegativities

Electronegativity of atoms increases up and to the right.


Electronegativity1
Electronegativity 7)

  • 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 7)

  • 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 7)

  • 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 7)

  • 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 7)

  • 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 7)

  • 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 7)

  • 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 7)

  • 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 7)

  • 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 7)

  • Draw the following Lewis Structures

    • CH3Cl

    • NH3

    • H2O

    • CO2

    • HCN


Formal charges
Formal Charges 7)

  • 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 7)

- 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 7)

  • 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 7)

  • 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 7)

  • 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 7)

  • 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 7)

  • 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 7)

  • 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)

O N O

-

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


Resonance1
Resonance 7)

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

O N O

O N O

-

-

-

O N O

  • 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 7)

  • 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 7)

  • 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 7)

  • 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 7)Electronegativities

Electronegativity of atoms increases up and to the right.


Here, electronegativity is labeled as 7)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 7)

  • 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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