Quantum Numbers
Posted: March 6th, 2011, 11:47 pmQN → solution to Schrodinger's wave equation (4 QN)
electron's address - most probable location
1. Energy level - n
n = 1 to ∞
The higher the n, the greater the energy and the farther from the nucleus
2. l - sublevel - indicates the shape
# of possible shapes for an energy level = n
each sublevel (shape) is assigned a letter
s < p < d < f < g ... go through the alphabet
→ Increasing E
3. m[sub]l[/sub] - indicates the orientation or # of orbitals per sublevel
count by ODD numbers
s = 1
p = 3
d = 5
f = 7
actually contain e[super]-[/super]
orbitals per EL = n²
4. m[sub]s[/sub] - indicates the spin of e[super]-[/super]
Must have OPPOSITE spins
max of 2 e[super]-[/super] in an ORBITAL; max # of e[super]-[/super] in an EL = 2n²
s = 1 × 2 = 2
p = 3 × 2 = 6
d = 5 × 2 = 10
f = 7 × 2 = 14
electron's address - most probable location
1. Energy level - n
n = 1 to ∞
The higher the n, the greater the energy and the farther from the nucleus
2. l - sublevel - indicates the shape
# of possible shapes for an energy level = n
each sublevel (shape) is assigned a letter
s < p < d < f < g ... go through the alphabet
→ Increasing E
| s = spherical p = peanut | d = double peanut f = flower |
3. m[sub]l[/sub] - indicates the orientation or # of orbitals per sublevel
count by ODD numbers
s = 1
p = 3
d = 5
f = 7
actually contain e[super]-[/super]
orbitals per EL = n²
4. m[sub]s[/sub] - indicates the spin of e[super]-[/super]
Must have OPPOSITE spins
max of 2 e[super]-[/super] in an ORBITAL; max # of e[super]-[/super] in an EL = 2n²
s = 1 × 2 = 2
p = 3 × 2 = 6
d = 5 × 2 = 10
f = 7 × 2 = 14