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Structure of Atom - Electronic Configuration of Atoms

Grade 11CBSEChemistry

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

🔑Concepts

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The Aufbau Principle states that in the ground state of an atom, orbitals are filled in order of their increasing energies. The order of filling is generally determined by the (n+l)(n + l) rule.

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Pauli Exclusion Principle: No two electrons in an atom can have the same set of four quantum numbers (n,l,ml,msn, l, m_l, m_s). Consequently, an orbital can accommodate a maximum of 22 electrons with opposite spins.

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Hund's Rule of Maximum Multiplicity: Electron pairing in orbitals belonging to the same subshell (pp, dd, or ff) does not take place until each orbital of that subshell is singly occupied with parallel spins.

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Extra Stability of Half-filled and Fully-filled Subshells: Subshells with exactly half-filled (p3,d5,f7p^3, d^5, f^7) or completely filled (p6,d10,f14p^6, d^{10}, f^{14}) configurations possess extra stability due to symmetrical distribution of electrons and maximum exchange energy.

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The electronic configuration is represented using the notation nlxnl^x, where nn is the principal quantum number, ll is the azimuthal quantum number (represented by letters s,p,d,fs, p, d, f), and xx is the number of electrons in that subshell.

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For the (n+l)(n+l) rule: If two orbitals have the same (n+l)(n+l) value, the orbital with the lower value of nn is filled first (e.g., 3d3d has n+l=3+2=5n+l = 3+2=5 and 4p4p has n+l=4+1=5n+l = 4+1=5; 3d3d is filled first).

📐Formulae

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

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

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

Maximum number of electrons in a subshell=2(2l+1)\text{Maximum number of electrons in a subshell} = 2(2l + 1)

Total nodes=n−1\text{Total nodes} = n - 1

Angular nodes=l\text{Angular nodes} = l

Radial nodes=n−l−1\text{Radial nodes} = n - l - 1

💡Examples

Problem 1:

Write the electronic configuration of Chromium (Z=24Z = 24) and Copper (Z=29Z = 29).

Solution:

Cr(Z=24):[Ar]3d54s1Cr (Z=24): [Ar] 3d^5 4s^1 and Cu(Z=29):[Ar]3d104s1Cu (Z=29): [Ar] 3d^{10} 4s^1

Explanation:

Expected configurations were [Ar]3d44s2[Ar] 3d^4 4s^2 and [Ar]3d94s2[Ar] 3d^9 4s^2. However, an electron shifts from the 4s4s to the 3d3d subshell to achieve half-filled (d5d^5) and fully-filled (d10d^{10}) states respectively, which provide extra stability.

Problem 2:

Determine the set of four quantum numbers for the 4th4^{th} electron in a Carbon atom (Z=6Z = 6).

Solution:

n=2,l=0,ml=0,ms=−12n=2, l=0, m_l=0, m_s=-\frac{1}{2}

Explanation:

The electronic configuration of Carbon is 1s22s22p21s^2 2s^2 2p^2. The 4th4^{th} electron enters the 2s2s orbital. For 2s2s: the principal quantum number n=2n=2, the azimuthal quantum number l=0l=0 (for ss), the magnetic quantum number ml=0m_l=0, and since it is the second electron in that orbital, its spin ms=−12m_s = -\frac{1}{2}.

Problem 3:

Which orbital is filled first: 4s4s or 3d3d?

Solution:

4s4s orbital is filled before 3d3d.

Explanation:

Using the (n+l)(n+l) rule: For 4s4s, n=4n=4 and l=0l=0, so n+l=4n+l = 4. For 3d3d, n=3n=3 and l=2l=2, so n+l=5n+l = 5. Since 4s4s has a lower (n+l)(n+l) value, it has lower energy and is filled first.

Electronic Configuration of Atoms Revision - Class 11 Chemistry CBSE