Quantum number is an important concept for understanding the structure of atom. Here get a free quantum number worksheet on Quantum number to excel in the examination. You may also want to look at our post on Quantum Numbers Chart.

## Quantum Numbers Practice Problems Worksheet

**Question 1**. What is the total number of orbitals associated with the principal quantum number n = 3?

## Answer

For n = 3, the possible values of l are 0, 1 and 2. Thus there is one 3s orbital (n = 3, l = 0 and ml = 0); there are three 3p orbitals (n = 3, l = 1 and ml = –1, 0, +1); there are five 3d orbitals (n = 3, l = 2 and ml = –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. 32 = 9.

**Question 2.** 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

(e) n=1 , l-0

(g) n=4, l=3

## Answer

(a) 2p

(b) 4s

(c) 5f

(d) 3d

(e) 1s

(f) 4f

**Question 3.** Which of the following sets of Quantum is not possible

(a) n = 0, l = 0, m = 0, m = 1/2

(b) n = 1, l = 0, m = 0, m = -1/2

(c) n = 1, l = 1, m = 0, m = -1/2

(d) n = 2, l = 1, m = 0, m = -1/2

(e) n = 3, l = 3, m = –3, m = 1/2

(f) n = 3, l = 2, m = 0, m = 1/2

(g) n = 1,l = 1,m = 0,s = +1/2

## Answer

3. (e) (a) (c)

**Question 4.** The quantum numbers of six electrons are given below.

i. n = 4, l = 2, ml = –2 , ms = –1/2

ii. n = 3, l = 2, ml = 1 , ms = +1/2

iii. n = 4, l = 1, ml = 0 , ms = +1/2

iv. n = 3, l = 2, ml = –2 , ms = –1/2

v. n = 3, l = 1, ml = –1 , ms = +1/2

vi. n = 4, l = 1, ml = 0 , ms = +1/2

Answer below questions

(a) which of the electrons are in same energy level

(b) which of the electron is having lowest energy

(c) which of the electron is having highest energy

## Answer

(a) 2 and 4 electron, 3rd and sixth electron

(b) 5th electron

(c) Ist electron

**Question 5.** The total number of electrons that can be accommodated in all the orbitals having principal quantum number 3 azumuthal quantum number 1 is

(a) 2

(b) 4

(c) 6

(d) 8

## Answer

(c)

**Question 6.** The principal quantum number of an atom represents

a. Spin Angular momentum

b. Size of the Orbital

c. Orbital Angular momentum

d. Space orientation of the orbital

## Answer

( b)

**Question 7.** A p-orbital can accommodate

a. 4 electrons

b. 6 electrons

c. 2 electron with opposite spins

d. 2 electron with parallel spins

## Answer

(c)

**Question 8.** What are n,l,m values for electrons in below orbitals

a. $2p_x$

b $4p_y$

c. $3p_z$

## Answer

a. n=2, l=1 ,m=-1

b. n=2,l=1,m=0

c. n=3,l=1,m=1

**Question 9.** An electron is in one of the 3d orbitals. Give the possible values of n, l and ml for

this electron.

## Answer

Principal quantum number (n) = 3

Azimuthal quantum number (l) = 2

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

**Question 10.** Principal,azimuthal and magnetic quantum numbers are respectively related to

a. Size,shape and orientation

b.Shape,size and orientation

c.Size,orientation and shape

d.None of the above

## Answer

(a)

**Question 11.** What is the maximum number of electrons that may be present in all the atomic orbitals with Principal quantum number 3 and azimuthal quantum number 2 ?

(a) 10

(b) 18

(c) 12

(d) 14

## Answer

For n=3 and l=2, we have 5 orbitals with m=-2,-1,0,1,2 .Since an orbital can occupy only 2 electrons,the maximum number of electrons= 5 x 2= 10

hence (a) is correct

**Question 12.** Designate the Orbitals having

(a) n=2 , l=1

(b) n=4 ,l=0

(c) n=3,l=2,$m=\pm 2$

(d) n=3,l=2

## Answer

(a) 2p

(b) 4s

(c) $3d_{x^2 – y^2}$

(d) 3d

**Question 13.** Write down all the Four quantum numbers of the fourth electron of Be(Z=4)

## Answer

Be (Z=4) has E.C as $1s^2 \ 2s^2$

So for fourth electron

n=2,l=0,m=0,s=-1/2

**Question 14.** Write the orbital Notation for the following set of Quantum Numbers

(a) n=1,l=0,m=0

(b) n=3,l=2 ,m=-1

## Answer

(a) 1s

(b) $3d_{yz}$

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