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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 ml for this electron.

Answer:

For the 3d orbital:
Principal quantum number (n) = 3
Azimuthal quantum number (l) = 2
Magnetic quantum number (ml) = – 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 ml ?
(ii) List the quantum numbers (ml 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, ml 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, mlcan 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 

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