**1.**

**Apply molecular orbital theory to oxygen molecule.**

**Or Explain the formation of O**

**2**

**/ oxygen molecule using molecular orbital theory.**

1. The

**electronic configuration of oxygen**atom (Z = 8) in the ground state is 1s22s22p4. Each oxygen atom has 8 electrons, hence, in**O****2****molecule there are 16 electrons**.
2. The

**electronic configuration of O****2**is:**O**

**2**

**: KK (σ**

**2s**

**)**

**2**

**(σ***

**2s**

**)**

**2**

**(σ**

**2pz**

**)**

**2**

**(π**

**2px**

**)**

**2**

**=**

**(π**

**2py**

**)**

**2**

**(π***

**2px**

**)**

**1**

**=**

**(π***

**2py**

**)**

**1**

Here (σ1s)2 (σ*1s)2 part of the configuration is abbreviated as KK.

3. The

**molecular orbital energy level diagram of O****2****molecule**is:
4. The

**bond order of O****2**can be calculated as follows:
N

_{b}= 8 and N_{a}= 4**Bond order**

**= N**

_{b}

**- N**

_{a}**/ 2**

= 8 - 4 / 2 =

**2**
5.

**Nature of bond:***A bond order of 2 means that***a double bond is present**in a molecule of oxygen.
6.

**Paramagnetic nature:***Since***two electrons in oxygen**[(π*2px) & (π*2py)]**are unpaired,**it is**paramagnetic**in nature.**2.**

**Derive / Defne de Broglie's equation.**

In case of a photon, if it is assumed to have wave character, its energy is given by

**E = hν**(according to the

**Planck’s quantum theory**) ... 1

Where,

**ν**is the frequency of the wave and**h**is Planck’s constant.

If the photon is supposed to have particle character, its energy is given by

**E = mc**

**2**(according to

**Einstein equation**) ... 2

Where,

**m**is the mass of photon and**c**is the velocity of light.

From equations 1 and 2, we get

**h ν = mc**

**2**

But,

**ν = c / λ**

**h . c / λ = mc**

**2**

or

**λ = h / mc**

For any material particle like electron, we may write

**λ = h / mv**

or

**λ = h / p**

Where

**mv = p**is the momentum of the particle.
The above equation is called

**de Broglie equation**and ‘**λ**’ is called de Broglie wavelength.**3.**

**Discuss Davisson and Germer’s experiment.**

1. A

**beam of electrons**obtained**from**a heated**tungsten filament is****accelerated**by using a high positive potential.
2. When this fine beam of accelerated electron is

**allowed to fall on a large single crystal of nickel**, the electrons are**scattered**from the crystal in different directions.
3. The

4.**diffraction pattern**so**obtained is****similar to the diffraction pattern obtained by Bragg’s experiment**on diffraction of X-rays from a target in the same way
5. Since

**X-rays have wave character**, therefore, the**electrons must also have wave character**associated with them.
6. Moreover, the

**wave length of the electrons as determined by the diffraction experiments were found to be in****agreement with the values calculated from de-Broglie equation**.
7. From the above discussion, it is clear that an

**electron behaves both as a particle and as a wave i.e., it has dual character**.
1. The

**electronic configuration of nitrogen**atom (Z=7) in the ground state is 1s22s22px12py12pz1. Therefore, the**total number of electrons present in nitrogen molecule (N****2****) is 14**.
2. The

**electronic configuration of N****2**:**N**

**2**

**: KK (σ**

**2s**

**)**

**2**

**(σ***

**2s**

**)**

**2**

**(π**

**2px**

**)**

**2**

**=**

**(π**

**2py**

**)**

**2**

**(σ**

**2pz**

**)**

**2**

Here (σ1s)2 (σ*1s)2 part of the configuration is abbreviated as KK.

3. The

**molecular orbital energy level diagram of N****2**is:
4. The

**bond order of N****2**can be calculated as follows:
Here, Nb = 8 and Na = 2

**Bond order = N**

_{b}- N_{a}/ 2
= 8 - 2 / 2 =

**3**
5.

**Nature of bond:***A bond order of 3 means that***a triple bond is present**in a molecule of nitrogen.
6.

**Diamagnetic nature:***Since***all the electrons in nitrogen are paired,**it is**diamagnetic**in nature.**5.**

**Give any five postulates of molecular orbital theory.**

1. In a molecule, electrons are present in

**new orbitals**called**molecular orbitals**.
2. Molecular orbitals are formed by

**combination of atomic orbitals of equal energies**(in case of homonuclear molecules)**or of comparable energies**(in case of heteronuclear molecules).
3. The

**number of molecular orbitals formed**is equal to the number of atomic orbitals undergoing combination.
4. Two atomic orbitals can combine to form two molecular orbitals. One of these two molecular orbitals one has a lower energy and the other has a higher energy. The molecular orbital with lower energy is called

**bonding molecular orbita**l and the other with higher energy is called**anti bonding molecular orbital.**
5. The

**shapes of molecular orbitals**depend upon the shapes of combining atomic orbitals.
6. The bonding molecular orbitals are

**represented by**σ (sigma), π (pi), δ (delta) and the antibonding molecular orbitals are represented by σ*, π*, δ*.
7. The molecular orbitals are filled in the increasing order of their energies, starting with orbital of least energy.

**(Aufbau principle)**
8. A molecular orbital can accommodate only two electrons and these two electrons must have opposite spins.

**(Paul’s exclusion principle)**
9. While filling molecular orbitals of equal energy, pairing of electrons does not take place until all such molecular orbitals are singly filled with electrons having parallel spins.

**(Hund’s rule)****6.**

**The approximate mass of an electron is 10**

**-27**

**g. Calculate the uncertainty in its velocity. If the uncertainty in its position were of the order of 10**

**-11**

**m.**

Mass of an electron, m = 10-27g = 10-27 x 10-3 kg = 10-30 kg

Uncertainty in its position, Δx = 10-11m

Uncertainty in its velocity, Δv = ?

**Δx . m.Δv = h / 4π**

Δv = h / 4π x Δx . m

Δv = 6.626 x 10–34 / 4 x 3.14 x10-11 x 10-30

**Δv = 5.275 x 10**

**6**

**m sec**

**-1**

**7.**

**The uncertainty in the position of a moving bullet of mass 10 g is 10**

**-5**

**m. Calculate the uncertainty in its velocity.**

Mass of an electron, m = 10g = 10 x 10-3 kg = 10-2 kg

Uncertainty in its position, Δx = 10-5m

Uncertainty in its velocity, Δv = ?

**Δx . m.Δv = h / 4π**

Δv = h / 4π x Δx.m

Δv = 6.626 x 10–34 / 4 x 3.14 x 10-5 x 10-2

**Δv = 5.275 x 10**

**-28**

**m sec**

**-1**

**8.**

**The wavelength of a moving body of mass 0.1 mg is 3.31 x 10**

**-29**

**m. Calculate its kinetic energy (h = 6.626 x 10**

**-34**

**Js).**

Wavelength, λ = 3.31 x 10-29 m

Mass, m = 0.1 mg = 10-1 x10-3 x 10-3 kg= 10-7 kg

Planck’s constant, h = 6.626 x 10-34 Js

Kinetic energy = ?

According to de-Broglie equation,

Or v = h / m λ

= 6.626 x 10-34 Js / 10-7 kg x 3.31 x 10-29m

v = 2 x 10-2

Kinetic energy = 1 / 2 mv2

= 1 / 2 x 10-7 x (2 x 10-2)2

**Kinetic energy = 2 × 10**

**-3**

**J**

**9.**

**A moving electron has 4.55 × 10**

**-25**

**joules of kinetic energy. Calculate its wavelength (mass = 9.1 × 10**

**-31**

**kg and h = 6.626 × 10**

**-34**

**kg m**

**2**

**s**

**-1**

**).**

Here we are given

Kinetic energy i.e. 1 / 2 mv2 = 4.55 × 10-25 J

Mass, m = 9.1 × 10-31 kg

Planck’s constant, h = 6.626 × 10-34 kg m2 s-1

Wavelength, λ = ?

1 / 2 x (9.1× 10-31) v2 = 4.55 x 10-25

Or v2 = 4.55 x 10-25 x 2 / 9.1× 10-31

Or v =103m sec−1

λ = 6.626 × 10-34 kg m2 s-1/ 9.1 × 10-31 kg x 103 m sec−1

**λ = 7.2 x 10**

**-7**

**m**

**10.**

**Discuss the shapes of d orbital**

**11.**

**Discuss the shapes of p orbital**

**ONE MARKS THREE MARKS**

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