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Structure of Atom - Quantum Numbers

Grade 11CBSEChemistry

Review the key concepts, formulae, and examples before starting your quiz.

🔑Concepts

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Quantum numbers are a set of four numbers used to completely describe the position and energy of an electron in an atom.

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The Principal Quantum Number (nn) defines the main energy shell and determines the size and energy of the orbital. It can have positive integer values: n=1,2,3,…n = 1, 2, 3, \dots

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The Azimuthal Quantum Number (ll), also known as the orbital angular momentum quantum number, defines the three-dimensional shape of the orbital. For a given nn, ll can range from 00 to (n−1)(n-1). Values 0,1,2,30, 1, 2, 3 correspond to s,p,d,fs, p, d, f orbitals respectively.

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The Magnetic Quantum Number (mlm_l) describes the spatial orientation of the orbital with respect to a standard set of coordinate axes. For a given ll, there are (2l+1)(2l+1) values of mlm_l, ranging from −l-l to +l+l.

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The Spin Quantum Number (msm_s) describes the direction of electron spin. An electron can have two spin states, represented by +1/2+1/2 (spin up) and −1/2-1/2 (spin down).

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Pauli's Exclusion Principle states that no two electrons in an atom can have the same set of four quantum numbers.

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The n+ln + l rule (Aufbau Principle) determines the order of filling of orbitals; the orbital with the lower (n+l)(n + l) value is filled first. If (n+l)(n + l) is the same for two orbitals, the one with the lower nn value is filled first.

📐Formulae

l=0,1,2,…,(n−1)l = 0, 1, 2, \dots, (n-1) (Range of Azimuthal Quantum Number)

ml=−l,…,0,…,+lm_l = -l, \dots, 0, \dots, +l (Range of Magnetic Quantum Number)

Total number of orbitals in a shell=n2\text{Total number of orbitals in a shell} = n^2

Maximum number of electrons in a shell=2n2\text{Maximum number of electrons in a shell} = 2n^2

Number of orbitals in a subshell=2l+1\text{Number of orbitals in a subshell} = 2l + 1

Orbital angular momentum=l(l+1)h2π\text{Orbital angular momentum} = \sqrt{l(l+1)} \frac{h}{2\pi}

💡Examples

Problem 1:

Determine the possible values of mlm_l for an electron in a 3d3d orbital.

Solution:

ml=−2,−1,0,+1,+2m_l = -2, -1, 0, +1, +2

Explanation:

For a 3d3d orbital, the principal quantum number n=3n = 3. For a dd subshell, the azimuthal quantum number l=2l = 2. The magnetic quantum number mlm_l ranges from −l-l to +l+l. Therefore, mlm_l can take (2×2+1)=5(2 \times 2 + 1) = 5 values: −2,−1,0,1,2-2, -1, 0, 1, 2.

Problem 2:

Which orbital has a higher energy: 4s4s or 3d3d?

Solution:

3d3d orbital

Explanation:

Using the n+ln + l rule: For 4s4s, n=4,l=0n = 4, l = 0, so n+l=4+0=4n + l = 4 + 0 = 4. For 3d3d, n=3,l=2n = 3, l = 2, so n+l=3+2=5n + l = 3 + 2 = 5. Since 5>45 > 4, the 3d3d orbital has higher energy than the 4s4s orbital.

Problem 3:

Calculate the orbital angular momentum for an electron in a 2p2p orbital.

Solution:

2h2π\sqrt{2} \frac{h}{2\pi}

Explanation:

For a 2p2p orbital, the value of ll is 11. The formula for orbital angular momentum is l(l+1)h2π\sqrt{l(l+1)} \frac{h}{2\pi}. Substituting l=1l = 1, we get 1(1+1)h2π=2h2π\sqrt{1(1+1)} \frac{h}{2\pi} = \sqrt{2} \frac{h}{2\pi}.

Quantum Numbers - Revision Notes & Key Formulas | CBSE Class 11 Chemistry