Quantum Mechanical Model of Atom - 2 - Free MCQ Practice Test with solutions,

ψ² is known as probability density and is always positive. From the value of ψ² at different points with in an atom it is possible to predict the region around the nucleus where electron will most probably be found.

For electrons present in M shell the value of other quantum numbers are same. But, the value of spin quantum number will be different.

For g orbital, value of l is 4.
(0 = s, 1 = p , 2 = d, 3 = f, 4 = g)
Since l = n - 1, n should be 5.

For nth orbital possible values of azimuthal quantum number (subshell), l are from 0 to (n-1). Total of 'n' values.

In n=4, l=0,1,2,3 thus there are 4 subshells i.e.s,p,d,f respectively.

Magnetic quantum number ml have values from -l to +l and total of 2l+1 values.

For n=4, possible values of l and ml are:

ml=0 for l=0; total ml values =1

ml=−1,0,1 for l=1; total ml values =3

ml=−2,−1,0,1,2 for l=2; total ml values =5

ml=−3,−2,−1,0,1,2,3 for l=3; total ml values =7

Total number of orbitals = total values of ml

for n=4,

∴1+3+5+7=16 orbitals

Each orbital can occupy maximum of two electron

Number of electrons =2×16=32

Therefore in n=4, number of subshells=4, orbitals=16 and number of electrons =32.

For 4f;  n = 4, l = 3,
m_l = 7 or -3 , -2, -1 , 0 , +1, +2, +3

(i) n = 3, l = 2 ⇒ 3 d
(ii) n = 4 , l = 3 ⇒ 4f

Quantum Numbers Description

i) n = 3, l = 2

- For the quantum numbers n = 3 and l = 2, the orbital is in the 3d subshell.
- The 3d orbital has a complex shape with two angular nodes and can hold a maximum of 10 electrons.
- Electrons in the 3d orbital have higher energy compared to those in the s and p orbitals.

ii) n = 4, l = 3

- For the quantum numbers n = 4 and l = 3, the orbital is in the 4f subshell.
- The 4f orbital has a more complex and intricate shape compared to lower energy orbitals.
- The 4f orbital can hold a maximum of 14 electrons and is located further from the nucleus due to higher energy levels.

n = 4
l = 0, 1 , 2 , 3
m_l = 0, (-1, 0, +1), (-2, -1, 0, +1, +2), (-3, -2, -1, 0, +1, +2, +3)

Total number of electrons with m_s = - 1/2 will be 16. 

n = 3, l = 1 represents 3p orbital. Since p has three orientations p_x, p_y and 6 electrons will show same quantum number values of n and l.

When l = 3, magnetic quantum number has 7 values m_l = (2l + 1). These values are represented as -3, -2, -1, 0, +1, +2, +3

For 2s-orbital, n = 2,1 = 0

Spin angular momentum of the electron, a vector quantity can have two orientations relative to the chosen axis. These two orientations take the values of +1/2 or -1/2 and are called two spin states of the electrons.

d_z2 has electron density in all three axes.

Total number of nodes = n - 1
For d-orbital, radial nodes = n - 3 and there are 2 angular nodes.

ns orbital has (n - 1) nodes.

If two orbitals have same value for (n + l), the orbital with lower value of n will have lower energy.

(i) 4p, (ii) 4s, (iii) 3d, (iv) 3p
The order of increasing energy
(iv) < (ii) < (iii) < (i)

The electron will enter in to an orbital with minimum value of n + l.

n = 4, l = 0,1, 2, 3

Total number of orbitals 1s + 3p + 5d + 7f = 16

The Aufbau principle states that in the ground state of an atom, an electron enters the orbitals of lowest energy first and subsequent electrons are filled in the order of increasing energies. In the given orbital diagram, option (c) violates the Aufbau principle as electrons switch to p orbitals without completely filling the s orbital which has comparatively low energy.

3s, 3p, 4s, 3d, 4p, 5s, 4d, 5p is the correct order of orbitals with increasing energy.

O^2- ion contains 2 electrons more than O. Hence its electronic configuration will be 1s² 2s² 2p⁶.

Phosphorus has atomic number 15.
Electronic configuration: 1s² 2s² 2p⁶ 3s² 3p³ or [Ne] 3s² 3p³ 

Z = 9 is the atomic number of fluorine which is the most electronegative element.

The outermost shell configuration of the element shows 8 electrons hence it is a noble gas.

M orbital with 13 electrons and N orbital with 1 electron correspond to chromium.

Atomic number 22 has the following configuration:
₂₂X = 1s² 2s² 2p⁶ 3s² 3p⁶ 4s² 3d²

₁₁Na = 1s² 2s² 2p⁶ 3s¹
For 3s¹, n = 3, l = 0, m_l = 0, s = -1/2

The configuration does not follow Hund’s rule of maximum multiplicity because 3p will be fully filled before the electrons go to 4s.

The outer electronic configuration of chromium atom is 3d⁵ 4s¹.