1.Apply molecular orbital theory to oxygen molecule. Or Explain the formation of O2 / 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 O2 molecule there are 16 electrons.
2. The electronic configuration of O2 is:
O2: KK (σ2s)2 (σ*2s)2 (σ2pz)2 (π2px)2 = (π2py)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 O2 molecule is:
4. The bond order of O2 can be calculated as follows:
Nb = 8 and Na = 4
Bond order = Nb - Na / 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 = mc2 (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 = mc2
But,
ν = c / λ
h . c / λ = mc2
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 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
4.
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 (N2) is 14.
2. The electronic configuration of N2:
N2: KK (σ2s)2 (σ*2s)2 (π2px)2 = (π2py)2 (σ2pz)2
Here (σ1s)2 (σ*1s)2 part of the configuration is abbreviated as KK.
3. The molecular orbital energy level diagram of N2 is:
4. The bond order of N2 can be calculated as follows:
Here, Nb = 8 and Na = 2
Bond order = Nb - Na / 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 orbital 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 10-27g. Calculate the uncertainty in its velocity. If the uncertainty in its position were of the order of 10-11m.
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 106 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 m2 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
12. Explain the types of hydrogen bond with example.
ONE MARKS THREE MARKS
10. Discuss the shapes of d orbital
12. Explain the types of hydrogen bond with example.
ONE MARKS THREE MARKS
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