Download Quantum Number - Career Launcher

Chemistry
Atomic structure
Session Objectives
Session objectives
Schrodinger wave equation
Shapes of orbitals
Nodal Plane
Quantum Numbers
Rules for electronic configuration of an atom
Schrodinger wave equation
Describes the probability of finding an electron in a given
volume element.
¶2y
¶x
2
+
¶ 2y
¶y
2
+
¶ 2y
¶z
2
+
8 p 2m
2
h
(E - V) y = 0
y2 ® Probability of finding the electron in a given region.
y ® Describes the wave motion of the electron.
Quantum mechanical model of atom
The energy of electrons in atoms is quantised.
The number of possible energy levels for electrons in atoms of different
elements is a direct consequence of wave-like properties of electrons.
The position and momentum of an electron cannot be determined
simultaneously.
Electrons of different energies are likely to be found in different regions.
The region in which an electron with a specific energy will most probably
be located is called an atomic orbital
Orbit and Orbital
Orbit is a fixed circular path around the nucleus in which
electron moves(proposed by Bohr) whereas orbital is the
quantum mechanical concept and refers to the wave function.
Nodal Plane
The plane where the probability
of finding the electron is almost
zero.
Total nodes in a shell = (n -1)
Angular nodes = l
Spherical nodes= (n –l -1)
Quantum Numbers
They specify the address of each electron in an atom. These
are four types.
Quantum Numbers
Azimuthal
Principal
Magnetic
Spin
Principal Quantum Number (n)
•
Average distance of the electron from
the nucleus
•
Energy Level of electron
•
Possible values (n=1,2,3…..)
•
Maximum number of electrons in any
shell is 2n2.
For example:
In 3d orbital,principal quantum number n is 3.
Azimuthal or Angular Momentum
Quantum number:(l)
It describes
Energy sub-level
Shape of the orbital
p
s
f
d
Quantum Numbers
Orbital angular momentum of an
electron is given by
h

l(l  1)
2
s
p
d
f
Values of l 0
1
2
3
Subshell
Illustrative example
The orbital angular momentum of an electron is 4s orbital is
1 h
(a) +
4 2p
(c)
h
2p
Solution
(b) zero
(d)
h
4
2p
Orbital angular momentum =
AIEEE 2003
ll  1 .
h
2
For s electrons, l = 0
For 4s electrons, orbital angular momentum is zero.
Hence, answer is (b)
Magnetic quantum
number(m)
Magnetic
Quantum Number (ml)

Orientations of an orbital in space.

Explains the Stark and Zeeman effect.

Takes (2l+1) values,m=- l to + l.
Specifies
Orbitals
the exact orbital within each sublevel
combine to form a spherical shape:
2s
2px
2py
2pz
Spin Quantum Number(s)
Describes the direction of spin of an electron
ms = +½
ms = -½
Clockwise or anticlockwise (+1/2,-1/2)
Spin angular momentum=
Magnetic moment =
s(s1
)
h
2
n(n2)
n= no. of unpaired electrons

1
2
Relation between
quantum numbers
•For every value of n, l = 0 to (n-1)
•For every value of l, m = -l to +l
•For every value of m, s = 
1
2
Quantum Numbers
New Delhi
n
Area
l
Street
m
House number
s
New Delhi
(shell)
Okhla PhaseI (sub shell)
B Block
52
(orbital)
(spin)
Illustrative example
Write down all the four quantum numbers for the last
electron of oxygen.
Solution:
Oxygen (Z=8) having configuration i.e.,
16 = 1s2 2s2 2p4
8O
hence, for the last electron
n=2, l=1, m=-1, s=-1/2
Illustrative example
The electrons, identified by quantum numbers n and l
(i) n = 4, l = 1, (ii) n = 4, l = 0,
(iii) n = 3, l = 2, (iv) n = 3, l = 1
can be placed in order of increasing energy from the lowest to the
highest as
(a) (iv) < (ii) < (iii) < (i)
(c) (i) < (iii) < (ii) < (iv)
(b) (ii) < (iv) < (i) < (iii)
(d) (iii) < (i) < (iv) < (ii)
Solution
(i)n = 4, l = 1(4p)
(ii)n = 4, l = 0(4s)
(iii)n = 3, l = 2(3d)
(iv)n = 3, l = 1(3p)
According to Aufbau’s rule order of increasing energy of subshells is
3p < 4s < 3d < 4p or (iv) < (ii) < (iii) < (i)
Hence, answer is (a)
Illustrative Example
A compound of vanadium has a magnetic moment of 1.73 B.M.
Find out the oxidation state of vanadium in the compound.
Solution
Magnetic moment =
nn2 1.73
n=1
Electronic configuration of vanadium
(23) is [Ar] 3d3 4s2 (five unpaired electrons)
The compound contains only one unpaired electron, electronic
configuration of V will be [Ar] 3d1 4s0
V is present as V4+-
Aufbau’s principle
1s < 2s < 2p < 3s < 3p < 4s < 3d < 4p < 5s < 4d < 5p < 6s
“Fill up” electrons in lowest
energy orbitals
Order of orbitals (filling)
in multi-electron atom
Aufbau’s principle
For example:
Consider 3d and 4s orbitals,
the electron will first enter the orbital
having minimum value of (n+l).
Electron will therefore enter 4s orbital (4+0=4) before
entering 3d orbital (3+2=5).
Aufbau’s principle
Incase (n+l) values are same!!!
Incase of 3d orbital (3+2=5) and 4p orbital (4+1=5), the (n+l)
values are same.
In such a case,electron enters the orbital for which n is minimum.
The electron will thus enter 3d orbital before entering 4p orbital.
Pauli’s exclusion principle
It is impossible for two electrons in a
given atom to have same set of four
quantum numbers.
For example:
Incase of 1s2,there are two electrons
in the 1s orbital.
The quantum numbers of the two electrons are:
n=1 , l=0 , m=0 , s=+1/2
n=1 , l=0 , m=0 , s=-1/2
Hund’s rule
•The most stable arrangement of electrons in sub shells is
the one with the greatest number of parallel spins.
•Electron pairing starts only after all the degenerate orbitals
are filled with electrons having same direction of spin.
For example:
Nitrogen (Atomic number=7)
Electronic configuration
1s2
2s2
2p3
Degenerate refers to orbitals having same energy.
Illustrative example
If the nitrogen atom had electronic configuration 1s7, it would have
energy lower than that of the normal ground state configuration 1s2
2s2 2p3, because the electrons would be closer to the nucleus. Yet
1s7 is not observed because it violates
(a) Heisenberg’s uncertainty principle
(b) Hund’s rule
(c) Pauli exclusion principle
(d) Bohr postulate of
stationary orbits
Solution
According to Pauli’s exclusion principle, an orbital cannot have more
than two electrons and these two with opposite spin.
Hence, answer is (c)
Exceptional electronic configuration
Orbitals in the same sub
shell tend to become
completely filled or half filled
since such orbitals are more stable.
Such as electronic configuration of Cr(24):
[Ar]3d44s2
But actually it is [Ar]3d54s1
Illustrative example
The electronic configuration of an element is 1s2
2s2 2p6 3s2 3p6 3d5 4s1. This represents its
(a) excited state
(b) ground state
(c) cationic form
(d) anionic form
Solution
The above electronic configuration is
for ground state of chromium (3d5 4s1)
Hence, answer is (b)
Class exercise
Class exercise-1
An electron is in one of 4d orbitals. Which of the following
quantum number is not possible?
(a) n = 4
(b) l = 1
(c) m = 1
(d) m = 2
Solution
b
In 4d orbital,
n=4
l=2
m = –2, –1, 0, 1, 2
Therefore, the quantum number value which is not
possible is (b).
Hence correct option is (b)
Class exercise-2
An electron in an isolated atom may be described by four quantum
numbers: n, l, m and s. The value of m for the most easily removed
electron from a gaseous atom of an alkaline earth metal is
(a) same as the maximum value of n for the element
(b) any number from –(n – 1) to +(n – 1)
(c) any positive number from 1 to ( n – 1)
(d) zero
Solution
d Alkaline earth metal belongs to s block. The
value of n for s block elements is 1.
l = (n – 1) = 0
m=0
Hence correct option is (d)
Class exercise-3
The set of quantum numbers that represents the electron of
highest energy is
(a) n=1, l=0, m=0, s=+1/2
(b) n=2, l = 0, m=0, s=+1/2
(c) n=4, l=0, m=0, s=1/2
(d) n=3, l=2, m=0, s=+1/2
Solution
d The set of quantum number which represents an electron
of the highest energy has the maximum value of (n + l).
Hence correct option is (d)
Class exercise-4
The number of electrons that can be accommodated by porbitals are
(a) two electrons with parallel spins
(b) six electrons
(c) four electrons
(d) eight electrons
Solution
b p orbital can accomodate a total of six electrons
(two each in px, py, pz).
Hence correct option is (b)
Class exercise-5
Which among the following set of quantum numbers is not
possible?
(a) n = 3, l = 0, m = 0
(b) n = 3, l = 1, m = 1
(c) n = 2, l = 1, m = 0
(d) n = 2, l = 0, m = –1
Solution
d The set of quantum number which is not possible is
n = 2, l = 0, m = –. In this case, m = 0.
Hence correct option is (d)
Class exercise-6
Which orbital has equal probability of finding an electron in
all directions?
Solution
The s orbital has equal probability of finding an
electron in all directions as it is spherical in shape,
which is symmetrical around the nucleus.
Class exercise-7
What is the angular momentum of 4s orbital?
Solution
Orbital angular momentum =
h
2

 1
In case of 4s orbital,
l=0
 Orbital angular momentum is zero.
Class exercise-8
Why do some atoms exhibit exceptional electronic configuration?
Solution
Some atoms exhibit exceptional electronic configuration
because half-filled orbitals and fully filled orbitals are more
stable than partially filled orbitals.
Class exercise-9
Write down the electronic configuration for Si (atomic number 14),
V (atomic number 23), Zn (atomic number 30), Kr (atomic
number 36).
Solution
Si = 1s2 2s2 sp6 3s2 3p2
V = 1s2 2s2 2p6 3s2 3p6 3d3 4s2
Zn = 1s2 2s2 2p6 3s2 3p6 3d10 4s2
Kr = 1s2 2s2 2p6 3s2 3p6 3d10 4s2 4p6
Class exercise-10
An electron is in a 4d orbital. What possible values of the quantum
numbers can it have?
Solution
n=4
l=2
m = ±2, ±1, 0
s= 
1
2
Thank you