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Quantum Theory and the Electronic Structure of Atoms
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Schrodinger Wave Equation
Y
= fn(n,
l
, m
l
, m
s
) Existence (and energy) of electron in atom is described by its
unique
wave function
Y
.
Pauli exclusion principle
- no two electrons in an atom can have the same four quantum numbers.
Each seat is uniquely identified (E, R12, S8) Each seat can hold only one individual at a time 7.6
For a certain value of
l , there are (2l +1)
integral values of m
l
And For a certain value of
n, there are n
2 orbital and 2
n
2 electron.
7.6
Schrodinger Wave Equation
Y
= fn(n,
l
, m
l
, m
s
) Shell – electrons with the same value of
n ( 2 s and 2 p)
Subshell – electrons with the same values of
n ( 2p x , 2p y and 2p z )
and
l
Orbital – electrons with the same values of
n, l
,
and
m l ( 2p x )
How many electrons can an orbital hold?
If
n, l,
and
m l
are fixed, then
m s
= ½ or - ½ Y
= (n,
l
, m
l
,
½
)
or Y
= (n,
l
, m
l
, -
½
)
An orbital can hold 2 electrons 7.6
How many 2p orbitals are there in an atom?
n=2 2p If
l
= 1, then m
l
= -1, 0, or +1 3 orbitals
l
= 1 How many electrons can be placed in the 3d subshell?
n=3 If
l
= 2, then m
l
= -2, -1, 0, +1, or +2 3d 5 orbitals which can hold a total of 10 e -
l
= 2 7.6
Worked Example 7.6
Worked Example 7.7
Energy of orbitals in a
single
electron atom
Energy only depends on principal quantum number
n
n=3 n=2
E
n
= -R
H
n
1
( )
n=1 7.7
Energy of orbitals in a
multi
-electron atom
Energy depends on
n
and
l
n=3
l
= 2 n=3
l
= 0 n=3
l
= 1 n=2
l
= 1 n=2
l
= 0 n=1
l
= 0 7.7
“Fill up” electrons in lowest energy orbitals (
Aufbau principle
) ? ?
B 1s 2 2s 2 2p 1 H 1s 1 2 7.9
The most stable arrangement of electrons in subshells is the one with the greatest number of parallel spins (
Hund’s rule
)
.
Ne 10 electrons Ne 1s 2 2s 2 2p 6 F 9 electrons O 8 electrons N 7 electrons C 6 electrons C 1s 2 2s 2 2p 2 F 1s 2 2s 2 2p 5 7.7
Order of orbitals (filling) in multi-electron atom
1s < 2s < 2p < 3s < 3p < 4s < 3d < 4p < 5s < 4d < 5p < 6s 7.7
Electron configuration
is how the electrons are distributed among the various atomic orbitals in an atom.
number of electrons in the orbital or subshell 1s 1 principal quantum number
n
angular momentum quantum number
l Orbital diagram
H 1s 1 7.8
What is the electron configuration of Mg?
Mg 12 electrons 1s < 2s < 2p < 3s < 3p < 4s 1s 2 2s 2 2p 6 3s 2 2 + 2 + 6 + 2 = 12 electrons Abbreviated as [Ne]3s 2 [Ne] 1s 2 2s 2 2p 6 What are the possible quantum numbers for the last (outermost) electron in Cl?
Cl 17 electrons 1s < 2s < 2p < 3s < 3p < 4s 1s 2 2s 2 2p 6 3s 2 3p 5 2 + 2 + 6 + 2 + 5 = 17 electrons Last electron added to 3p orbital n = 3
l
= 1 m
l
= -1, 0, or +1 m s = ½ or -½ 7.8
Outermost subshell being filled with electrons
7.8
7.8
Paramagnetic
unpaired electrons 2p
Diamagnetic
all electrons paired 2p 7.8
Worked Example 7.10
Worked Example 7.11
Chemistry Mystery: Discovery of Helium
In 1868, Pierre Janssen detected a new dark line in the solar emission spectrum that did not match known emission lines Mystery element was named Helium In 1895, William Ramsey discovered helium in a mineral of uranium (from alpha decay).
1.
What is the wavelength (in nanometers) of light having a frequency of 8.6 x 10 13 Hz?
A) B) C) D) 3.5 nm 3.5 x 10 3 nm 3.5 x 10 6 2.9 x 10 5 nm nm 2.
What is the frequency of light having a wavelength of 456 nm?
A) 1.37 x 10 2 Hz B) C) D) 6.58 x 10 5 6.58 x 10 14 Hz Hz 1.37 x 10 14 Hz 3.
A photon has a wavelength of 624 nm. Calculate the energy of the photon in joules.
A) B) C) D) 3.19 x 10 –16 3.19 x 10 –19 1.24 x 10 –22 3.19 x 10 –28 J J J J
5.
4.
The blue color of the sky results from the scattering of sunlight by air molecules. The blue light has a frequency of about 7.5 x 10 14 Hz. Calculate the energy, in joules, of a single photon associated with this frequency.
A) 2.6 x 10 –31 J B) C) D) 2.6 x 10 –22 5.0 x 10 –19 5.0 x 10 –16 J J J A photon has a frequency of 6.0 x10 4 photons all with this frequency.
A) 2.4 x 10 –5 J/mol B) 4.0 x 10 –10 J/mol C) D) 6.6 x 10 –15 4.0 x 10 –20 J/mol J/mol Hz. Calculate the energy (in joules) of 1 mole of 6.
What is the wavelength, in nm, of radiation that has an energy content of 1.0 x 10 3 kJ/mol?
In which region of the electromagnetic spectrum is this radiation found?
A) 1.2 x 10 –1 nm, X-ray B) C) D) 2.0 x 10 1 1.2 x 10 2 2.0 x 10 3 nm, ultraviolet nm, ultraviolet nm, infrared
7.
Some copper compounds emit green light when they are heated in a flame. How would you determine whether the light is of one wavelength or a mixture of two or more wavelengths?
A) B) C) D) Observe the emitted light with green tinted glasses.
Pass the emitted light through a beaker of water.
Pass the emitted light through a prism.
Pass the emitted light through green tinted glasses.
8.
Calculate the wavelength (in nanometers) of a photon emitted by a hydrogen atom when its electron drops from the A) 1.28 x 10 –6 nm B) 1.46 x 10 –6 nm C) D) 1.46 x 10 3 1.28 x 10 3 nm nm
n =
5 state to the
n =
3 state.
9.
The first line of the Balmer series occurs at a wavelength of 656.3 nm. What is the energy difference between the two energy levels involved in the emission that results in this spectral line?
A) 3.367 x 10 –36 J B) C) D) 3.027 x 10 –28 1.299 x 10 –22 3.027 x 10 –19 J J J
10.
Calculate the frequency (Hz) of the emitted photon when an electron drops from the A)
n =
4 to the 2.74 x 10 14 Hz B) C) D) 6.17 x 10 14 1.62 x 10 15 3.65 x 10 15 Hz Hz Hz
n =
2 level in a hydrogen atom.
11.
An electron in the hydrogen atom makes a transition from an energy state of principal quantum numbers
n i
to the
n =
2 state. If the photon emitted has a wavelength of 434 nm, what is the value of
n i
?
A) 3 B) C) D) 4 5 6 12.
An electron in a certain atom is in the
n =
2 quantum level. List the possible values of A) B) C)
l l l
= 0
l
= 0, = 0, , and
, m m m l l l m l
, = 0; = 0;
l l
that it can have.
= 1, = 1, = –1, 0, 1
m m l l
= –1, 0, 1; = –1, 0, 1 D)
l
= 1,
m l
= –1, 0, 1
l
= 2;
m l
= –2, –1, 0, 1, 2
13.
Give the values of the quantum numbers associated with the 2
p
subshell.
A) B) C) D)
n n n n
= 2, = 2, = 2, = 2,
l l
= 2, = 1,
l
= 1,
l
= 1,
m l m l m l m l
= –2, –1, 0, 1, 2 = 0 = 1 = –1, 0, 1 14.
15.
Calculate the total number of electrons that can occupy: (A) one
s
orbital, (B) three
p
orbitals, (C) five
d
orbitals, (D) seven
f
orbitals.
A) B) (A)2; (B)9; (C)10, (D)14 (A)2; (B)6; (C)8, (D)14 C) D) (A)2; (B)6; (C)10, (D)14 (A)2; (B)6; (C)10, (D)16 What is the total number of electrons that can be held in all orbitals having the same principal quantum number A) 4
n
2 B) 2
n
2
n
?
C) D) 2
n
2
16.
17.
18.
Determine the maximum number of electrons that can be found in each of the following subshells: 3
s
, 3
d
, 4
p
, 4
f
, 5
f
.
A) B) 3
s
(2); 3
d
(8); 4
p
(6); 4
f
(14); 5
f
(14) 3
s
(2); 3
d
(10); 4
p
(6); 4
f
(14); 5
f
(16) C) D) 3
s
(2); 3
d
(8); 4
p
(6); 4
f
(14); 5
f
(14) 3
s
(2); 3
d
(10); 4
p
(6); 4
f
(14); 5
f
(14) State the total number of: electrons in S (
Z =
16).
p
electrons in N (
Z =
7);
s
electrons in Si (
Z =
14); and 3
d
A) B) C) D) N, 3 N, 2 N, 3 N, 6
p p p p
electrons; Si, 6 electrons; Si, 6 electrons; Si, 6 electrons; Si, 6
s s s s
electrons; S, 5 electrons; S, 5 electrons; S, 0 electrons; S, 0
d d d d
electrons electrons electrons electrons Why do the 3
s
, 3
p
, and 3
d
orbitals have the same energy in a hydrogen atom but different energies in a many-electron atom?
A) B) C) Many-electron atoms have shielding from the lower orbitals.
The many electrons excite each other to higher energies.
The orbitals are shaped differently with many-electron atoms.
19.
For each of the following pairs of hydrogen orbitals, indicate which is higher in energy: (A) 1
s
, 2
s
; (B) 2
p
, 3
p
; (C) 3
d xy
, 3
d yz
; (D) 3
s
, 3
d
; (E) 4
f
, 5
s
.
A) (A) 2
s
; (B) 3
p
; (C) equal; (D) 3
d
; (E) 5
s
B) (A) 2
s
; (B) 3
p
; (C) equal; (D) equal; (E) 5
s
C) D) (A) 2
s
; (B) 2
p
; (C) equal; (D) equal; (E) 5
s
(A) 2
s
; (B) 3
p
; (C) equal; (D) equal; (E) 4
f
20.
Indicate which of the following sets of quantum numbers in an atom are unacceptable: (A) (1, 0, ½, ½); (B) (3, 0, 0, +½); (C) (2, 2, 1, +½); (D) (4, 3, –2, +½); (E) (3, 2, 1, 1).
A) B) C) D) (A) and (E) are unacceptable.
(B), (C) and (E) are unacceptable.
(A), (B), (C) and (E) are unacceptable.
(A), (C) and (E) are unacceptable.
21.
Indicate the number of unpaired electrons present in each of the following atoms: B, Ne, P, Sc, Mn, Se.
A) B) B(1); Ne(0); P(3); Sc(1); Mn(5); Se(2) B(0); Ne(0); P(3); Sc(1); Mn(5); Se(2) C) D) B(1); Ne(0); P(2); Sc(2); Mn(5); Se(2) B(1); Ne(0); P(3); Sc(2); Mn(4); Se(2) 22.
Use the Aufbau principle to obtain the ground-state electron configuration of selenium.
A) Se: [Ar]4
s
2 3
d
10 4
p
3 B) C) D) Se: [Ar]4
s
2 3
d
10 4
p
4 Se: [Ar]4
s
2 3
d
10 4
p
5 Se: [Ar]4
s
2 3
d
10 4
p
6
25.
23.
24.
What is the maximum number of electrons in an atom that can have the following quantum numbers: (1) 2,
l
= 0,
m s
= ½ ; (5) A)
n n
= 2, = 4,
l m s
= 3, = +½ ; (2)
m l
= 2.
(1)4; (2)5; (3)8; (4)2; (5)2
n
= 4,
m l
=
+
1; (3)
n
= 3,
l
= 2; (4)
n
= B) C) D) (1)4; (2)6; (3)8; (4)1; (5)2 (1)4; (2)6; (3)10; (4)1; (5)2 (1)4; (2)6; (3)10; (4)2; (5)2 The He + ion contains only one electron and is therefore a hydrogen-like ion.
Calculate the wavelengths of the first four transitions in the Balmer series of the He + ion. (The Rydberg constant for He + is 8.72 ´ 10 18 J.) Which of the following is not one of these transitions?
A)
n
= 3 to 2; l = 164 nm; UV B) C) D)
n n n
= 4 to 2; = 5 to 2; = 6 to 2; l l l = 121 nm; UV = 107 nm; UV = 103 nm; UV The electron configurations described in this chapter all refer to gaseous atoms in their ground states. An atom may absorb a quantum of energy and promote one of its electrons to a higher-energy orbital. When this happens, we say that the atom is in an excited state.
The electron configurations of some excited atoms are given. Identify the species. (A) 1
s
1 2
s
1 ; (B) 1
s
2 2
s
2 2
p
2 3
d
1 ; (C) 1
s
2 2
s
2 2
p
6 4
s
1 A) B) C) D) (A) He; (B) C; (C) Ne (A) He + ; (B) N + ; (C) Na + (A) He; (B) N; (C) Na (A) He; (B) O; (C) Na
26.
Scientists have found interstellar hydrogen atoms with quantum number
n
in the hundreds.
Calculate the wavelength of light emitted when a hydrogen atom undergoes a transition from
n =
236 to
n =
235. In what region of the electromagnetic spectrum does this wavelength fall?
A) B) C) D) 0.596 nm, Xray 9.12 ´ 10 1 nm, Ultraviolet 5.96 ´ 10 8 9.12 ´ 10 12 nm, Microwave nm, Radiowave
Answer Key
1- B 2- C 3- B 4- C 5- A 6- C 7- C 8- D 9-D 10- B 11- C 12- B 13- D 14- C 15- B 16- D 17- C 18- A 19- B 20- D 21- A 22- B 23- C 24- C 25- C 26- C