energy unit learning goal 2: examine the placement of electrons in orbitals

Energy Unit

Learning Goal 2: Examine the Placement of Electrons in Orbitals.

Copyright © by McDougal Littell. All rights reserved. 2

Bohr’s Atomic Model

1. Atom has small positive nucleus.2. Electrons orbit like planets orbit the sun.

3. Electrons orbit in certain allowed energy levels.4. Electrons can jump to different orbits but only by absorbing or emitting a photon of light with the correct energy content.

Good start but some basic problems:Electrons do not “orbit” in circular paths.

Could not explain why negative electrons didn’t get attracted into the nucleus

This model only worked for Hydrogen


DeBroglie & the Wave Mechanical Model of the Atom

DeBroglie & Schrodinger suggested that electrons exhibited both wave and particle characteristics.

This is referred to the wave/particle

duality Schrodinger came up with a model (wave mechanical model) that worked with atoms in addition to hydrogen.

Wave Mechanical Model

• Electron states are described as orbitals.• Electrons are more like fireflies than planets.– An orbital is described as the probability map of

an electron’s motion.


Orbitals vs. Orbits

Orbitals are nothing like orbits.To picture orbitals, imagine a single male firefly in a room. In the center of the room is a vial of nectar.

The room is dark with a camera with an open shutter in the corner.The developed picture will look something the diagram to the left.


Suppose while watching the room, you see a flash here.

XWhere will the firefly flash next?

There’s no way of really knowing, but the likely-hood is that it will flash where the “film” had the densest concentration of flashes.

Shrodinger’s model cannot predict the

path of the electron, but can predict the

probability of find the electron in a certain

region.


The first quantum number: n

Describes the energy level

n = 1, 2, 3, 4, etc.

Notice how the rows are numbered on the periodic table!


Sublevels

The principle energy levels are divided into sublevels. The second quantum number describes the shape of these sublevels.

l = 0, 1, 2, 3, 4, 5, 6, 7

However, we rarely refer to these sublevels by number. We usually use a letter:

s p d f


Notice the energy level determines the # of sublevels!

nl

Principle Energy Levels

• Discrete Energy levels that are labeled with integers.– 1, 2, 3, 4, 5, 6, 7.

• Sublevels– Each Principle Energy Level is subdivided into

sublevels and labeled with a letter.• s (holds 2 e- )• p (holds 6 e- ) • d (holds 10 e- ) • f (holds 14 e- ) • The letters tell the shape of the orbital.


Figure 11.20: The hydrogen 1s orbital.


Shapes of Sublevels

Every energy level has an s sublevel. The only difference being the diameter!

s sublevels are “spherical” in shape.


The p sublevels are dumbell shaped and are made of 3 orbitals or “lobes”.Each orbital can hold 2 electrons.

p sublevels are found on energy levels 2 or greater


The d sublevels are four-petaled and are made of 5 orbitals or “lobes”.Each orbital can hold 2 electrons.

d sublevels are found on energy levels 3 or greater


The f sublevels are made of 7 orbitals or “lobes”.

Each orbital can hold 2 electrons.

f sublevels are found on energy levels 4 or greater


Summary ....so far

sublevel # orbitals # electrons

s

p

d

f

1

3

5

7

2

6

10

14


Notice the # of columns in each group!


Pauli Exclusion Principle:

• An atomic orbital can hold a maximum of two electrons, and those two electrons must have opposite spin.

Now let’s put it all together.....


H

H

1s1s1

He

He1s

He has 2 electrons 1s2

1234567


H

H

1s1

He

He1s

has 3 electrons

1s2

Li

Li1s 2s

1s22s1

1234567


H He

has 4 electrons

Li

Li1s 2s

1s22s1

Be

Be2s1s

1s22s2

B

B

2s1s 2p1s22s22p1

1234567


H He

Li Be B

C2s1s 2p

1s22s22p2

Li1s 2s

1s22s1

Be 1s 2s1s22s2

B2p 1s22s22p1

1s 2s

C

Stop! Before adding the next electron, we have to know about Hund’s Rule:

Hund’s Rule: we put 1 electron in each orbital

before we pair them up!

1234567


H He

Li Be B

C

1s22s22p3

B2p 1s22s22p1

1s 2s

C N

N2s1s 2p

1s22s22p22p1s 2s

1234567


H He

Li Be B

C

1s22s22p4

B2p 1s22s22p1

1s 2s

C N

1s22s22p2

O2s1s 2p

2p1s 2s

N 2p1s 2s1s22s22p3

O

1234567


H He

Li Be B

C

1s22s22p5

C N

1s22s22p22p1s 2s

N 2p1s 2s1s22s22p3

F2s1s 2p

O F

O 2p1s 2s1s22s22p4

1234567


H He

Li Be B

1s22s22p6

C N

N 2p1s 2s1s22s22p3

O F

O 2p1s 2s1s22s22p4

Ne2s1s 2p

F 2p1s 2s1s22s22p5

Ne

This is a very important arrangement! With 8 electrons in the valence shell, we have a stable octet. Notice neon is a noble gas, very inert, and is at the end of its row!

1234567


H He

O

1s22s22p63s1

1s22s22p42p1s 2s

F 2p1s 2s1s22s22p5

Ne 2p1s 2s1s22s22p6

Li Be B C N O F NeNa

Na2s1s 2p 3s

[Ne] 3s1

1234567


H He

Na [Ne] 3s[Ne] 3s1

Li Be B C N O F NeNa

Mg [Ne]3s

[Ne] 3s2

Ne 2p1s 2s 1s22s22p6

Mg

1234567


H He

Na [Ne] 3s[Ne] 3s1

Li Be B C N O F NeNa Mg

[Ne] 3s23p1

Ne 2p1s 2s 1s22s22p6

Al

Mg [Ne] 3s[Ne] 3s2

Al [Ne]3s 3p

1234567


H He

Li Be B C N O F NeNa Mg Al

[Ne] 3s23p2

3p

Si

Al [Ne] 3s[Ne] 3s23p1

Si [Ne]3s 3p

1234567


H He

Li Be B C N O F NeNa Mg Al Si

[Ne] 3s23p3

3p

P


3pSi [Ne] 3s[Ne] 3s23p2

P[Ne]3s 3p

1234567


H He

Li Be B C N O F NeNa Mg Al Si P

3p

S


3pSi [Ne] 3s[Ne] 3s23p2

S[Ne]3s 3p

3pP [Ne] 3s[Ne] 3s23p3

[Ne] 3s23p4

1234567


H He

Li Be B C N O F NeNa Mg Al Si P S Cl


Cl[Ne]3s 3p

[Ne] 3s23p5

3pS [Ne] 3s[Ne] 3s23p4

1234567


H He

Li Be B C N O F NeNa Mg Al Si P S Cl Ar


[Ne] 3s23p6

3pS [Ne] 3s[Ne] 3s23p4

Ar[Ne]3s 3p

3pCl [Ne] 3s[Ne] 3s23p5

Another Noble gas with a stable octet!

1234567


H He

Li Be B C N O F NeNa Mg Al Si P S Cl ArK

[Ar] 4s1K[Ar]4s

3pAr [Ne] 3s[Ne] 3s23p6

Notice that we have started filling the 4th energy level before even starting the 3d!

1234567


H He


Ca

[Ar] 4s2Ca[Ar]4s

K [Ar] 4s[Ar] 4s1

K

1234567


H He


Sc

[Ar] 4s23d1

K [Ar] 4s[Ar] 4s1

K Ca

Ca [Ar] 4s[Ar] 4s2

3dSc[Ar]

4s

1234567


H He


Ti

[Ar] 4s23d2

K [Ar] 4s[Ar] 4s1

K Ca Sc

Ca [Ar] 4s[Ar] 4s2

Sc [Ar] 4s[Ar] 4s23d1

3d

3dTi[Ar]

4s

1234567


H He


V

[Ar] 4s23d3

K Ca Sc Ti

Ti [Ar] 4s[Ar] 4s23d2

3d

Sc [Ar] [Ar] 4s23d13d4s

3dV[Ar]

4s

1234567


H He


Cr

[Ar] 4s23d4

K Ca Sc Ti V

Ti [Ar] 4s[Ar] 4s23d2

3d

Sc [Ar] [Ar] 4s23d13d4s

V [Ar] [Ar] 4s23d33d4s

3dCr[Ar]

4s

Stop! Chromium is a stealer!

A more stable arrangement is formed when all orbitals are half-filled than one full & one empty

[Ar] 4s13d5

1234567


H He


Mn

[Ar] 4s23d5

K Ca Sc Ti V Cr

Ti [Ar] 4s [Ar] 4s23d23d

V [Ar] [Ar] 4s23d33d4s

3dMn[Ar]

4s

Cr [Ar] 4s[Ar] 4s13d5

3d

1234567


H He


Cu

[Ar] 4s23d9

K Ca Sc Ti V Cr Mn Fe Co Ni

3dCu[Ar]

4s

Copper (and all in this column) steal an electron from the 4s orbital to fill

its 3d![Ar] 4s13d10

1234567


Zn

H He

Li Be B C N O F NeNa Mg Al Si P S Cl ArK Ca Sc Ti V Cr Mn Fe Co Ni Cu

Cu [Ar] 4s[Ar] 4s13d10

3d

[Ar] 4s23d10

3dZn[Ar]

4s

1234567


Ga

H He

Li Be B C N O F NeNa Mg Al Si P S Cl ArK Ca Sc Ti V Cr Mn Fe Co Ni Cu Zn

Cu [Ar] 4s[Ar] 4s13d10

3d

Zn[Ar] 4s[Ar] 4s23d10

3d

[Ar] 4s23d104p1

3dGa[Ar]

4s 4p

1234567

energy unit learning goal 2: examine the placement of electrons in orbitals

Documents