Chemistry Introduction to Dual Behaviour of Matter, de Broglie Equation, Heisenberg's Uncertainty Principle, Quantum Mechanical Model of Atom, Quantum Numbers, Shape of Atomic Orbitals, Filling of Atomic Orbitals and Electronic Configuration of Atoms

Topics to be covered :

This lecture covers following topics :

`●` Dual Behaviour of Matter
`●` de Broglie Equation
`●` Heisenberg's Uncertainty Principle and its Significance
`●` Quantum Mechanical Model of Atom
`●` Important Features of Quantum Mechanical Model
`●` Orbitals and Quantum Numbers
`●` Shapes of Atomic Orbitals
`●` Boundary Surface Diagram
`●` Radial and Angular Nodes
`●` Energy of Orbitals
`●` Degenerate Orbitals
`●` Aufbau Principle
`●` Pauli's Exclusion Principle
`●` Hund's Rule of Maximum Multiplicity
`●` Electronic Configuration of Atom
`●` Importance of Electronic Configuration
`●` Stability of Half Filled and Completely Filled Subshells

Foundation of the Quantum Mechanical Model of the Atom

Two important developments contributed to this model :-

(i) Dual behaviour of matter,

(ii) Heisenberg uncertainty principle.


Dual Behaviour of Matter :

de Broglie (1924) proposed that matter, like radiation also exhibit dual behaviour i.e. both particle and wave like properties. de Broglie gave the relation between wavelength (`lamda`) and momentum (`p`) of a material particle as

`lamda = h/(mv) = h/p`

`m` = mass of electron , `v` = its velocity , `p` = its momentum

Note : The wave nature of electron is used in the making of electron microscope.

According to de Broglie, every object in motion has wave character. The wavelength associated with ordinaryobjects are so short that their wave properties cannot be detected.
Q 2612234139

What will be the wavelength of a ball of mass `0.1` kg moving with a velocity of `10 m s^(–1) ?`

Solution:

According to de Brogile equation

`lamda = h/(mv) = (6.626xx10^(-34) Js)/((0.1kg)(10 ms^(-1)))`

` = 6.626xx10^(-34) m(J = kg m^2 s^(-2))`
Q 2622334231

The mass of an electron is `9.1xx10^(-31)kg.` If its K.E. is `3.0xx10^(-25) J` calculate its wavelength



Solution:

Since `K. E. = ½ mv^2`

`v = ((2K.E.)/m)^(1/2) = ((2xx3.0xx10^(-25) kgm^2 s^(-2))/(9.1xx10^(-31) kg))^(1/2)`

` = 812 ms^(-1)`

`lamda = h/(mv)`

` = (6.626xx10^(-34) Js)/((9.1xx10^(-31))(812 ms^(-1))`

` = 8967xx10^(-10) m = 896.7nm`
Q 2652334234

Calculate the mass of a photon with wavelength `3.6 A^0`.

Solution:

`lamda = 3.6A^0 = 3.6xx10^(-10)m`

Velocity of photon = velocity of light

`m = h/(lamda v)`

` = (6.626xx10^(-34) Js)/((3.6xx10^(-10)m)(3xx10^8 ms^(-1)))`

` = 6.135xx10^(-29) kg`
Q 2612645539

If `E_A, E_B` and `E_C` represents kinetic energies of an electron, alpha particle and proton respectively and each moving with same de-Broglie wavelength then choose the correct among following.

(A)

`E_A = E_B = E_C`

(B)

`E_A > E_B > E_C`

(C)

`E_B > E_C > E_A`

(D)

`E_A > E_C > E_B`

Solution:

As we know,

`lamda = h/(mv) = h/sqrt(2mKE)`

`because lamda prop 1/sqrtm`

mass of `e,alpha` and `p` are

`m_e < m_p < m_alpha`

the correct order is

`E_A < E_C < E_B`
Correct Answer is `=>` (D) `E_A > E_C > E_B`

Heisenberg's Uncertainty Principle :

Defination : It states that it is impossible to determine simultaneously the exact position and exact momentum (or velocity) of an electron.

Note : - This is the result of dual behaviour of matter and radiation.

Mathematically, it can be expressed as : -

`Delta x xx Delta p_x >= h/(4 pi)`

or ` Delta x xx Delta (mv_x) >= h/(4 pi)`

or ` Delta x xx Delta v_x >= h/(4 pi m)`

where ` Delta x =` uncertainty in position

` Delta p_x =` uncertainty in momentum

` Delta v_x =` uncertainty in velocity

Q 2682445337

A microscope using suitable photons is employed to locate an electron in an atom within a distance of `0.1 A^0`. What is the uncertainty involved in the measurement of its velocity?

Solution:

`Deltax . Deltap = h/(4pi ) ` or `Deltax . mDeltav = h/(4pi)`

`Deltav = h/(4pi Deltaxm)`

`Deltav = (6.626xx10^(-34) Js)/(4xx3.14xx0.1xx10^(-10)mxx 9.11xx10^(-31)kg)`

` = 0.579xx10^7 ms^(-1) (1J = 1kgm^2 s^(-2))`

` = 5.79xx10^6 ms^(-1)`
Q 2652745634

The uncertainties in the velocities of two particles A and B are `0.05` and `0.02ms^(-1)` respectively. The mass of B is five times to that of mass A. What is the ratio of uncertainties `((Deltax_A)/(Deltax_B))` in their positions?

(A)

`2`

(B)

`0.25`

(C)

`4`

(D)

`1`

Solution:

According to Heisenberg `Deltax xx m xx Deltav = h/(4pi)` ........(1)

Now,

`Deltax_A xx m xx0.05 = h/(4pi)`

`Deltax_B xx 5m xx 0.02 = h/(4pi)` .............(2)


From eq (1) and (2)

`(Deltax_A xx m xx 0.05)/(Deltax_B xx 5m xx 0.02) = 1` or `(Deltax_A)/(Deltax_B) = 2`
Correct Answer is `=>` (A) `2`

Explanation :

(i) For getting accurate results, the instrument used should have smaller units then the actual units of the thing to be measured.

(ii) For electron, the dimension of electron is very small i.e. it is considered as a point charge, so if we radiate electron with a light of having wavelength smaller than the dimension of electron, we can accurately measure the position of electron. The light having smaller wavelength will have higher energy. But as the energy of the electrons would change after collision, so we would be able to know very little about the velocity of the electron.


Significance of Uncertainty Principle :

Important Conclusion -

(i) It rules out existence of definite paths or trajectories of electrons and other similar particles.

(ii) The trajectories of an object is measured by its location and velocity.

If position of a body is known at some instant and also we know its velocity and forces acting on it at the same time we can interpret where the body would be after sometime. Since it is not possible for a sub-atomic particle to measure both position and velocity precisely and accurately, we cannot tell about its trajectory.

(iii) The principle is not applicable to macroscopic objects.
Q 2612545430

A golf ball has a mass of 40g, and a speed of 45 m/s. If the speed can be measured within accuracy of 2%, calculate the uncertainty in the position?

Solution:

The uncertainty in the speed is 2%, i.e.,

`45 xx 2/100 = 0.9ms^(-1)`

Using the equation

`Deltax = h/(4pi m Deltav)`

` = (6.626xx10^(-34) Js)/(4xx3.14xx40gxx10^(-3) kg g^(-1) (0.9 ms^(-1)))`

` = 1.46xx10^(-33) m`

This is nearly `~ 10^(18)` times smaller than the diameter of a typical atomic nucleus. For large particles, the uncertainty principle sets no meaningful limit to the precision of measurements.

Quantum Mechanical Model of Atom :

Classic mechanics describe the motion of macroscopic objects but not of microscopic objects.

Note : This is due to the fact that classical mechanics does not consider the dual behaviour of matter and uncertainty principle.

Quantum Mechnics :

Definition :- The branch of science which considers the dual behaviour of matter is called quantum mechanics.

Notes :- When quantum mechanics is applied to macroscopic objects, the result obtained is same as those obtained from the classical mechanics.

It was develop independently (1926) by Werner Heisenberg and Erwin Schrodinger.

Schrodinger developed the fundamental equation of quantum mechanics. This is expressed as : -

` hat H psi = E psi`

where ` hat H ` is called Hamiltonian operator.

Note :- This expression is applicable to a system whose energy does not change with time.

The total energy of the system considers the kinetic energies of all the sub-atomic particles, attractive potential between electrons and nuclei individually.

Solution of this equation gives `E` and `psi`.


Important Feature of the Quantum Mechanical Model of Atom :

Hydrogen Atom and the Schrodinger equation : -

(a) The solution of Schrodinger equation for `H`-atom gives the possible energy level the electron can occupy and the corresponding wave function (`psi`) of the electron associated with each energy level.

(b) These quantized energy level and corresponding wave function are characterised by quantum number which are as follow :-

(i) Principle Quantum number (`n`)

(ii) Azimuthal Quantum number (`l`)

(iii) Magnetic Quantum number (`m_l`)

(c) The wave function corresponding to a particular energy state gives all information about the electron.

(d) The value of wave function depends upon the coordinates of the electron in the atom.

(e) The wave function `psi` has no physical meaning .

(f) Wave function of the species having only one electron are called atomic orbitals. These wave functions are called one electron systems.

(g) The probability of finding an electron at a point within an atom `prop | psi |^2 ` at that points.

(h) The quantum mechanical results explain all the aspect of `H`-atom spectrum.

Note : - Schrodinger equation cannot be solved exactly for a multi-electron atom. So for multi-electron system we use approximate method.

(i) Orbitals in atoms other than `H`-atom do not differ very much from the hydrogen atom. The principle difference is : -

(i) only nuclear charge increases and as a result orbitals get contracted.

(ii) For `H`-atom, energy of orbital `prop n`

(j) For multi-electron atoms, energy of orbital depends on (`n` and `l`).

Orbitals and Quantum Numbers :-

(i) There are a large number of orbitals possible in an atom.

(ii) These orbitals are distinguised by Quantum Numbers.

(iii) Each orbital is designated by three Quantum Numbers `n`, `l` and `m_l`.

Principle Quantum Numbers (`n`) :

(a) It can have any positive integer values i.e. `n = 1,2,3,...`

(b) It determines size and energy of the orbital.

(c) Size `prop n` and energy `prop l`

(d) It also identifies the ''shell''

(e) The number of allowed orbitals `= n^2`

(f) All Orbitals belonging to a given value of `n` constitute a shell of atom and it is represented as

`n \ \ \ \ \ \ \ \ \ \ \ \ \ \ 1 \ \ \ \ \ \ 2 \ \ \ \ \ 3 \ \ \ \ \ \ 4`
`text(shell) \ \ \ \ \ \ \ \ K \ \ \ \ L \ \ \ \ M \ \ \ \ N`


Q 2632545432

What is the total number of orbitals associated with the principal quantum number n = 3 ?

Solution:

For `n = 3`, the possible values of `l` are `0`, 1 and 2. Thus there is one `3s` orbital (`n = 3, l = 0` and `m_l = 0`);

there are three `3p` orbitals (`n = 3, l = 1` and `m_l = –1, 0, +1`);

there are five `3d` orbitals (`n = 3, l = 2` and `m_l = –2, –1, 0, +1+, +2`).

Therefore, the total number of orbitals is `1+3+5 = 9`

The same value can also be obtained by using the relation; number of orbitals
`= n^2, i.e. 3^2 = 9.`

Azimuthal Quantum Number (`l`) :

(a) It is also known as orbital angular momentum or subsidiary quantum number.

(b) It defines 3D shape of the orbital.

(c) For a given value of `n`, `l` can have `n` values ranging from `0` to (`n - 1`) i.e. `l = 0, 1, 2, 3 ......... ( n - 1)`

e.g when `n = 1 , l = 0`

when `n = 2 , l = 0, 1`

(d) There can be one or more sub-shell or sub levels for a given shell.

(e) the number of sub-shell = n

e.g for `n =1`, number of sub-shell = 1
for `n = 2`, number of sub-shell = 2

(f) Sub-shells with different values of `l` are represented as : -

`text(Value for) \ \ l : \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ 0 \ \ \ \ 1 \ \ \ \ 2 \ \ \ \ \ 3 \ \ \ \ 4 \ \ \ \ 5`
`text(Notation for sub shell ) : \ \ s \ \ \ \ p \ \ \ \ d \ \ \ \ \ f \ \ \ \ g \ \ \ \ h`

See table 2.4

Q 2652545434

Using s, p, d, f notations, describe the orbital with the following quantum numbers

(a) `n = 2, l = 1,` (b) `n = 4, l = 0,` (c)` n = 5, l = 3,` (d) `n = 3, l = 2`




Solution:

`\ \ \ \ \ \ \ n \ \ l \ \ \ text(orbital)`
`(a) \ \ \ \ \ \ \ \ \ \ \ \ \ 2 \ \ \ 1 \ \ \ \ 2p`
`(b) \ \ \ \ \ \ \ \ \ \ \ \ \ 4 \ \ \ \0 \ \ \ \ 4s`
`(c) \ \ \ \ \ \ \ \ \ \ \ \ \ 5 \ \ \ \ 3 \ \ \ \ 5f`
`(d) \ \ \ \ \ \ \ \ \ \ \ \ \ 3 \ \ \ \ 2 \ \ \ \ 3d`

Magnetic Orbital Quantum Number (`m_l`) :

(a) It gives details about the spatial orientation of the orbital in the space.

(b) for any value of sub shell i.e. `l` , these can be (`2l + 1`) value of `m_l`. These values are given as :

`m_l = - l , - (l - 1) , - ( l - 2) ....... 0,1,2, ....... ( l- 2),(l - 1) , l`

e.g for `l = 0, m_l = ( 2 xx 0 + 1) = 1` (one s-orbital)
for `l = 1, m_l = ( 2 xx 1 + 1) = 3` (three p-orbital)
and `m_l = -1, 0, 1`

(c) The relation between the sub-shell and the number of orbitals associated with it is given in the Table.
Q 1935801762

An electron is in one of the `3d` orbitals. Give the possible values of `n, l` and `m_l` for this electron
Class 11 Exercise 2 Q.No. 25
Solution:

For 3d orbital, `n=3,l=2.` For `l=2,m =-2-1,`
`0, +1+2.`

Spin Quantum Number (`m_s`) :

The above three quantum numbers were unable to explain the line spectra observed in the case of multi-electron atoms. This suggests the presences of a few more energy levels.

George Uhlenbeck and Samuel Goudsmit proposed the presence of the fourth quantum number known as the electron spin quantum numbers (`m_s`).

An electron spins around its own axis. So, it has intrinsic spin angular quantum number. Since spin angular momentum is vector quantity it can have two orientations are distinguished relative to its axis and these two orientations are distintinguised by spin quantum number (`m_s`). `m_s` can have values of ` + 1//2` or ` - 1//2`. These are called the two spin states of the electron and represented as :-

`uparrow` ( spin up) and `downarrow`

An orbital can not hold more then two electrons and these two electron have opposite spins.
Q 1965312265

Explain, giving reasons, which of the following
sets of quantum numbers are not possible.
(a) `n=0, l=0, m_1=0, m_s=+ 1/2`

(b) `n=1, l=0, m_1=0, m_s= -1/2`

(c) `n=1, l=1, m_1=0, m_s=+1/2`

(d) `n =2, l = 1, m_1=0, m_s=-1/2`

(e) ` n=3, l=3, m_1=-3, m_5 =+1/2`

(f) `n=3, l= 1, m_1=0, m_s=+ 1/2`

Class 11 Exercise 2 Q.No. 30
Solution:

(a) Not possible because `n` can never be equal to, zero.
(b) Possible
(c) Not possible
(d) Possible
(e) Not possible because when `n = 3, l ne 3` as
value of `l` varies from `0` to `n- 1`.
(f) Possible

Shapes of Atomic Orbitals :

`psi` has no physical meaning. It is simply a mathematical function of the coordinates of the electron. Fig 2.12(a)

According to Max Born, the sqare of wave function(i.e, ` psi^2`) at a point gives the probability density of the electron at that point. Fig 2.12(b).

Nodal surface or Node :

The region where the probability density function reaches to zero.

For `ns`-orbital, there are (`n - 1`) nodes,

Representation of probability density curve : -

The probability density variation can be imagined in terms of charge could diagrams [fig 2.13 (a)]



Boundary Surface Diagram :

This gives fairly good representation of the shapes of the orbitals.

In this, a boundary surface or contour surface is drawn in space for an orbital on which the value `|psi|^2 ` is constant.

Definition : For a given orbital, only that boundary surface diagram of constant probability density is taken which encloses a region or volume in which the probability of finding the electron is `90%`. fig [2.13 (b)].

Note : - We do not draw boundary surface diagram of `100%` probability because ` psi^2` has always some value at any finite distance from the nuclues, so it is not possible to draw a boundary surface diagram of a rigid size of `100%` probability.

For `s`-orbital, boundary surface diagram is spherical in shape. i.e the probability of finding the electron in all direction is equal.


Size of `s`-orbital `prop 'n'`

For `p`-orbital (Fig 2.14), the boundary surface diagram is not spherical. But it has two lobes which are on either side of the plane passing through the nucleus. At this plane, probability density function is zero.

The shape, size and energy of the three orbitals are identical. The lobes lie along the x, y and z axis and they are given the designations `2p_x`, `2p_y` and `2p_z`.

For `p`-orbital, the energy of `p`-orbitals `prop n`

For `d`-orbitals, `l = 2`. And there are five `d`-orbitals, designated as `d_(xy), d_(yz), d_(zx), d_(x^2-y^2)` and `d_(z^2)`. See fig 2.15.

All the five `d`-orbitals have same energy.

Angular Nodes :

The probability density functions at the plane passing through the nuecleus is zero. This is called angular node.

No. of angular nodes = `l`

Total no. of nodes = `n-1`

No. of radial nodes ` = (n-l-1)`
Q 2612045839

The nodes present in `3p` - orbitals are


(A)

two spherical

(B)

two planar

(C)

one planar

(D)

one planar one spherical

Solution:

The nodes present in `3p`-orbitals are one spherical and one planar.
Correct Answer is `=>` (D) one planar one spherical

Energies of Orbitals :

Energy of an electron in a `H`-atom is determined by `n`.

The order of energy follows as

`1s < 2s = 2p < 3s = 3p = 3d < 4s = 4p = 4d = 4f < ................`

See fig.2.16

Note : The `2s` and `2p` - oribitals have same energy even though their shape is different.

Degenerate Orbitals :

Orbitals having same energy are called degenerate orbitals.

The `1s`-orbital corresponds to the most stable condition and is called the ground state.

The electron is this orbital is most strongly held by the nucleus.

Electron of `H`-atom present in `2s, 2p` or higher orbitals is said to be in excited state.

The energy of an electron in a multi-electron atom depends on both `n` and `l`.



Reason :

The energy of an electron in a multi-electron atom depends on both `n` and `l` the mutual repulsion among the electrons.

The reason of stability of multi-electron system is

`text(total attractive interaction > total repulsive interactions)`

Repulsive interaction is among the electrons of outer shell and inner shell.

Attractive interaction is between the electrons and nucleus.

As the positive charge `(Ze)` on the nucleus increases, the attractive interaction increases.

Due to the presence of inner electrons, outer electrons will not experiene the full positive charge of the nucleus `(Ze)`.

This is known as the shielding of the outer shell electrons from (the nucleus) by the inner shell electrons, and the net positive charge which outer electrons experience is known as the effective nuclear charge `(Z_text(eff) e)`.

Both attractive and repulsive interaction depends upon the shell and shape of the orbital in which the electron is present.

`s`-orbitals have the most effective shielding power.

The order of shielding power is :

`s > p > d > f`

Within a shell, `s`-orbital electron spends more time closer to the nucleus compared to other orbitals electrons. The order of this is given by

`s > p > d > f`

The more time a orbital electron spends near the nucleus, the more tightly it is bound to the nucleus.

The energy of the `s`-orbital will be lower i.e. more stable than `p`-orbital.

Due to different shielding effect of different orbitals, it leads to the difference in energy of orbitals in same shell.

The lower the value of (`n+l`) for an orbital, the lower is its energy. See Table 2.5.

If two orbitals have same value of `(n+l)`, then the orbital having lower `n` value will have lower energy.

Energies of the orbitals in the same subshell decreases with increase in the atomic number..

e.g. `E_(2s) (H) > E_(2s) (Li) > E_(2s) (Na) > E_(2s) (K)`
Q 1946078873

Among the following pairs of orbitals which
orbital will, experience the larger effective
nuclear charge?

(i) 2s and 3s
(ii) 4d and 4f
(iii) 3d and 3p
Class 11 Exercise 2 Q.No. 64
Solution:

(i) 2s is closer to the nucleus than 3s. Hence
2s will experience larger effective nuclear
charge.
(ii) 4d
(iii) 3p (for same reason)

Filling of Orbitals in Atom :

There are three rules according to which electrons are filled into the orbitals of an atom.

Aufbau Principle :

Aufbau => means building up (in German )

Here, building up means => Filling up of orbitals with electrons.

Statement : In the ground state of the atoms, the orbitals are filled in order of the their increasing energies.

The order of filling of electrons into orbitals is as follows :

`1s < 2s < 2p < 3s < 3p < 4 s < 3d < 4p < 5 s < 4d < 5p < ..............`

Trick : See fig.2.17

Q 2622145931

The orbital diagram(s) which follow the Aufbau principle is/are Choose the correct option

(A)

Only I and II

(B)

Only II and Ill

(C)

Only II. Ill and IV

(D)

All I , II , III and IV

Solution:

In (i) 1s orbital should be filled first before filling of p orbitals. In (iv) p orbitals should be half filled first before pairing.
Correct Answer is `=>` (B) Only II and Ill

Pauli Exclusion Principle :

This principle was given by Walfgang Pauli (1926).

Statement : No two electrons in an atom can have the same set of four quantum numbers.

or Any two electrons can exist in the same orbital with opposite spin.

This rule helps in calculating the number of electrons a subshell can occupy.

So, the maximum number of electrons in a shell = `2n^2` and the maximum capacity of a sub-shell = `2(2l + 1)`
Q 1916180070

(a) How many sub-shells are associated with
`n = 4`?

(b) How many electrons will be present in the ,
sub-shells having `m_s` value of `-1 // 2` for
`n=4` ?
Class 11 Exercise 2 Q.No. 67
Solution:

(a) `n = 4, l = 0, 1, 2, 3` (`4` subshells, viz., `s, p, d`
and `f`)

(b) No. of electrons in 4th shell .
`=2n^2=2xx4^2 =32`

Each orbital has one electron with

`m_s =-1//2`

Hence there will be `16` electrons with

`m_s =-1//2`.

Hund's Rule of Maximum Multiplicity :

This rule is basically about the filling of electrons into the orbitals belonging to the same subshell.

Statement : Pairing of electrons in a orbital of a specific subshell (`s`, `p`, `d`, `f`) does not take place until each orbital of that subshell is singly occupied.

Electronic Configuration of Atom :

Definition : The distribution of electrons into orbitals of an atom is called its electronic configuration.

The electronic configuration of an atom is written according to the three principle studied above.

Electronic configuration of different atoms can be represented as :

(i) `s^a , p^b , d^c .............` notation (`a, b, c` represent the no. of electrons in respective subshell)

(ii) orbital diagram (See Image)

In this, each box represent the orbital of a subshell.

The electron is represented by up arrow (`↑`) positive spin or a down arrow (`↓`) a negative spin.
Q 2642145933

The electronic configuration of an element with atomic number 24 is

(A)

`1s^2 , 2s^2 , 2p^6 , 3s^2 , 3p^6 , 3d^4 , 4s^2`

(B)

`1s^2, 2s^2, 2p^6, 3s^2, 3p^6, 3d^(10)` ,

(C)

`1s^2, 2s^2, 2p^6, 3s^2, 3p^6, 3d^6, `

(D)

`1s^2,2s^2,2p^6,3s^2,3p^6,3d^5,4s^1`

Solution:

The electronic configruation of element with atomic number 24 is

`1s^2,2s^2,2p^6,3s^2,3p^6,3d^5,4s^1`
Correct Answer is `=>` (D) `1s^2,2s^2,2p^6,3s^2,3p^6,3d^5,4s^1`
Q 2662145935

What is the electronic configuration of `Mn^(2+) ?`

(A)

`[Ne] 3d^5 , 4s^0`

(B)

`[Ar] 3d^5 , 4s^2`

(C)

`[Ar] 3d^5 , 4s^0`

(D)

`[Ne] 3d^5 , 4s^2`

Solution:

Electronic configuration of `Mn (25)` is

`1s^2 , 2s^2 , 2p^6 , 3s^2 , 3p^6 , 4s^2 , 3d^5`

`therefore` Electronic configuration of `Mn^(2+)` is

`1s^2 , 2s^2 , 2p^6 , 3s^2 , 3p^6 , 3d^5`

`because Mn^(2+) = [Ar] 3d^5 , 4s^0`
Correct Answer is `=>` (C) `[Ar] 3d^5 , 4s^0`

Advantage of Orbital Diagram :

It gives information about all the four quantum numbers.

Core electrons => These are electrons in the completely filled shells.

Valence electrons => These are the electrons in the shell with the highest principle quantum number.

Importance of Electronic Configuration :

Electronic configuration is useful in understanding and explaining chemical behaviour.

Questions like => (a) why two or more atoms combine to form molecules?

(b) Why some elements are metals and some are non metals?

(c) Why `He` and `Ar` are not reactive but halogens are reactive?

Answer to all these questions are given based on electronic configuration of the element.

Stability of completely filled and half-filled subshells :

The ground state electronic configuration has the lowest energy.

The two main exception of Aufbau principle are :

`Cr_(24) -> 1s^2 , 2s^2 2p^6 , 3s^2 3p^6 3d^4 , 4s^2` (wrong)

`1s^2 , 2s^2 2p^6 , 3s^2 3p^6 3d^5 , 4s^1` (right)

`Cu_(29) -> 1s^2 , 2s^2 2p^6, 3s^2 3p^6 3d^9 , 4s^2` (wrong)

`1s^2 , 2s^2 2p^6 , 3s^2 3p^6 3d^(10) , 4s^1` (right)

This can be explained on the basis of the stability of half-filled and fully- filled sub shells.

Reason of the extra stability of half-filled and fully-filled subshells :

(i) Symmetrical distribution of electrons : These type of electronic configuration have symmetrical distribution of electron in them and therefore more stable. Electrons in the same subshell have equal energy but different spatial distribution and hence less shielding from one another and therefore more strongly attracted by the nucleus.

(ii) Exchange Energy : The electrons in the degenerate orbitals of a subshell with same spin tend to exchange their positions and the energy released due to this is called exchange energy.

When the subshell is either half-filled or completely filled, the no. of exchange is maximum. (Fig 2.18)

So, higher the exchange energy, more is the stability.

Note : (a) Exchange energy is the base of Hund's Rule.

(b) In these configurations, coulmbic repulsion energy is also smaller.

 
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