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

2 2 2 2

2 2 2 2

8 m(E V ) 0

x y z h

¶ y ¶ y ¶ y p+ + + - y =

¶ ¶ ¶

Describes the probability of finding an electron in a given

volume element.

2 Probability offi nding the electron in a given region.y ®

Describes the wave m otion of the electron.y ®

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 whichelectron 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 = lSpherical nodes= (n –l -1)

Quantum Numbers

Quantum Numbers

Principal

Azimuthal Magnetic

Spin

They specify the address of each electron in an atom. These are four types.

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.

In 3d orbital,principal quantum number n is 3.

For example:

Azimuthal or Angular Momentum Quantum number:(l)

It describes

Energy sub-level

Shape of the orbital

p

d

s

f

Quantum Numbers

Orbital angular momentum of an

electron is given by

)1l(l2

h+

p=

Subshell

Values of l

s p d f

0 1 2 3

Illustrative example

The orbital angular momentum of an electron is 4s orbital is

+p

1 h(a )

4 2(b) zero

h(c)

2p ph

(d) 42

AIEEE 2003

SolutionOrbital angular momentum = +

ph

l l 1 .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 the exact orbital within each sublevel

Orbitals combine to form a spherical shape:

2s

2pz

2py

2px

Spin Quantum Number(s)

Describes the direction of spin of an electron

Clockwise or anticlockwise (+1/2,-1/2)

p+

2

h)1s(sSpin angular momentum=

n(n 2)+Magnetic moment =

n= no. of unpaired electrons

ms = +½ ms = -½

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 =

12

1

2

Quantum Numbers

New Delhi n New Delhi (shell)

Area l Okhla PhaseI (sub shell)

Street m B Block (orbital)

House number s 52 (spin)

Illustrative example

Oxygen (Z=8) having configuration i.e., 8O

16 = 1s2 2s2 2p4

hence, for the last electron n=2, l=1, m=-1, s=-1/2

Write down all the four quantum numbers for the last electron of oxygen.

Solution:

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) (b) (ii) < (iv) < (i) < (iii)

(c) (i) < (iii) < (ii) < (iv) (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 ExampleA compound of vanadium has a magnetic moment of 1.73 B.M. Find out the oxidation state of vanadium in the compound.

Magnetic moment = + =n n 2 1.73

The compound contains only one unpaired electron, electronic configuration of V will be [Ar] 3d1 4s0

n = 1

Electronic configuration of vanadium (23) is [Ar] 3d3 4s2 (five unpaired electrons)

V is present as V4+-

Solution

Aufbau’s principle

“Fill up” electrons in lowest energy orbitals

Order of orbitals (filling) in multi-electron atom

1s < 2s < 2p < 3s < 3p < 4s < 3d < 4p < 5s < 4d < 5p < 6s

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

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

It is impossible for two electrons in a given atom to have same set of four quantum numbers.

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

The above electronic configuration is for ground state of chromium (3d5 4s1)

Solution

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

= 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

Solution

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

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

Solution

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

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

Solution

The number of electrons that can be accommodated by p-orbitals are

(a) two electrons with parallel spins(b) six electrons

(c) four electrons (d) eight electrons

b p orbital can accomodate a total of six electrons (two each in px, py, pz).

Hence correct option is (b)

Class exercise-5

Solution

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

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

Solution

Which orbital has equal probability of finding an electron in all directions?

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

Solution

What is the angular momentum of 4s orbital?

Orbital angular momentum =

In case of 4s orbital,

= 0

Orbital angular momentum is zero.

+p

h

12

Class exercise-8

Solution

Why do some atoms exhibit exceptional electronic configuration?

Some atoms exhibit exceptional electronic configuration because half-filled orbitals and fully filled orbitals are more stable than partially filled orbitals.

Class exercise-9

Solution

Write down the electronic configuration for Si (atomic number 14), V (atomic number 23), Zn (atomic number 30), Kr (atomic number 36).

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

Solution

An electron is in a 4d orbital. What possible values of the quantum

numbers can it have?

n = 4

l = 2

m = ±2, ±1, 0

s = 1

2

Thank you