numerical on quantum no

Day 5 

9.5.20

Good Morning Boys,

Today we will continue solving the numerical from structure of atoms.

Google Meet Scheduled today during class hours.

Learning Outcomes,

students will be able to 

•  Apply Formula to questions
• Apply Concept To Questions.
• Identify the nature of questions.

Question 1

What is the lowest value of n that allows g orbitals to exist?

Answer:

For g-orbitals, l = 4.

As for any value ‘n’ of principal quantum number, the Azimuthal quantum number (l) can have a value from zero to (n – 1).

∴ For l = 4, minimum value of n = 5

Question 2.

An electron is in one of the 3d orbitals. Give the possible values of n, l and m_l for this electron.

Answer:

For the 3d orbital:

Principal quantum number (n) = 3

Azimuthal quantum number (l) = 2

Magnetic quantum number (m_l) = – 2, – 1, 0, 1, 2

Question 3

An atom of an element contains 29 electrons and 35 neutrons. Deduce (i) the number of protons and (ii) the electronic configuration of the element.

Answer:

(i) For an atom to be neutral, the number of protons is equal to the number of electrons.

∴ Number of protons in the atom of the given element = 29

(ii) The electronic configuration of the atom is ( 29 ) (Cu ) 

1s² 2s² 2p⁶ 3s² 4p⁶ 4s¹ 3d¹⁰

Question 4

Give the number of electrons in the species, H₂ and 

Answer:

:

Number of electrons present in hydrogen molecule (H₂) = 1 + 1 = 2

∴ Number of electrons in = 2 – 1 = 1

H₂:

Number of electrons in H₂ = 1 + 1 = 2

:

Number of electrons present in oxygen molecule (O₂) = 8 + 8 = 16

∴ Number of electrons in = 16 – 1 = 15

Question 5

(i) An atomic orbital has n = 3. What are the possible values of l and m_l ?

(ii) List the quantum numbers (m_l and l) of electrons for 3d orbital.

(iii) Which of the following orbitals are possible?

1p, 2s, 2p and 3f

Answer:

(i) n = 3 (Given)

For a given value of n, l can have values from 0 to (n – 1).

∴ For n = 3

l = 0, 1, 2

For a given value of l, m_l can have (2l + 1) values.

For l = 0, m = 0

l = 1, m = – 1, 0, 1

l = 2, m = – 2, – 1, 0, 1, 2

∴ For n = 3

l = 0, 1, 2

m₀ = 0

m₁ = – 1, 0, 1

m₂ = – 2, – 1, 0, 1, 2

(ii) For 3d orbital, l = 2.

For a given value of l, m_lcan have (2l + 1) values i.e., 5 values.

∴ For l = 2

m₂ = – 2, – 1, 0, 1, 2

(iii) Among the given orbitals only 2s and 2p are possible. 1p and 3f cannot exist.

For p-orbital, l = 1.

For a given value of n, l can have values from zero to (n – 1).

∴ For l is equal to 1, the minimum value of n is 2.

Similarly,

For f-orbital, l = 4.

For l = 4, the minimum value of n is 5.

Hence, 1p and 3f do not exist.

Question 6

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

(a) n = 1, l = 0; (b) n = 3; l =1 (c) n = 4; l = 2; (d) n = 4; l =3.

Answer:

(a) n = 1, l = 0 (Given)

The orbital is 1s.

(b) For n = 3 and l = 1

The orbital is 3p.

(c) For n = 4 and l = 2

The orbital is 4d.

(d) For n = 4 and l = 3

The orbital is 4f.

Question 7

How many electrons in an atom may have the following quantum numbers?(a) n = 4,  (b) n = 3, l = 0

Answer:

(a) Total number of electrons in an atom for a value of n = 2n²

∴ For n = 4,

Total number of electrons = 2 (4)²

= 32

The given element has a fully filled orbital as

1s² 2s² 2p⁶ 3s² 3p⁶ 4s² 3d¹⁰.

Hence, all the electrons are paired.

∴ Number of electrons (having n = 4 and ) = 16

(b) n = 3, l = 0 indicates that the electrons are present in the 3s orbital. Therefore, the number of electrons having n = 3 and l = 0 is 2.

Question 8

What transition in the hydrogen spectrum would have the same wavelength as the Balmer transition n = 4 to n = 2 of He⁺ spectrum?

Answer:

For He⁺ ion, the wave number associated with the Balmer transition, n = 4 to n = 2 is given by:

Where,

n₁ = 2

n₂ = 4

Z = atomic number of helium

According to the question, the desired transition for hydrogen will have the same wavelength as that of He⁺.

By hit and trail method, the equality given by equation (1) is true only when

n₁ = 1and n₂ = 2.

∴ The transition for n₂ = 2 to n = 1 in hydrogen spectrum would have the same wavelength as Balmer transition n = 4 to n = 2 of He⁺ spectrum.

Question 9

Calculate the energy required for the process

The ionization energy for the H atom in the ground state is 2.18 ×10^–18 J atom^–1

Answer:

Energy associated with hydrogen-like species is given by,

For ground state of hydrogen atom,

For the given process,

An electron is removed from n = 1 to n = ∞.

∴ The energy required for the process