krit.club logo

Structure of Atom - Electronic Configuration of Atoms

Grade 11ICSEChemistry

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

🔑Concepts

•

The Aufbau Principle states that in the ground state of an atom, electrons fill orbitals in order of increasing energy levels: 1s<2s<2p<3s<3p<4s<3d<4p…1s < 2s < 2p < 3s < 3p < 4s < 3d < 4p \dots

•

The (n+l)(n+l) Rule determines orbital energy: Orbitals with a lower (n+l)(n+l) value are filled first. If two orbitals have the same (n+l)(n+l) value, the one with the lower principal quantum number (nn) has lower energy.

•

Pauli Exclusion Principle states that 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.

•

Hund's Rule of Maximum Multiplicity states that for degenerate orbitals (orbitals of the same energy, like the three pp orbitals), electron pairing does not occur until each orbital is singly occupied with an electron of parallel spin.

•

Stability of Half-filled and Fully-filled Subshells: Atoms like Chromium (Z=24Z=24) and Copper (Z=29Z=29) exhibit anomalous configurations because half-filled (d5d^5) and fully-filled (d10d^{10}) subshells provide extra stability due to symmetrical distribution and high exchange energy.

•

The electronic configuration of an atom is represented as nlxnl^x, where nn is the principal quantum number, ll is the orbital type (s,p,d,fs, p, d, f), and xx is the number of electrons in that subshell.

📐Formulae

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

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

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

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

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

💡Examples

Problem 1:

Write the electronic configuration of Chromium (Z=24Z = 24) and explain the reason for any anomaly.

Solution:

The expected configuration is [Ar]3d44s2[Ar] 3d^4 4s^2, but the actual configuration is [Ar]3d54s1[Ar] 3d^5 4s^1.

Explanation:

This occurs because the 3d54s13d^5 4s^1 configuration results in a half-filled dd-subshell and a half-filled ss-subshell. Half-filled subshells are exceptionally stable due to the symmetrical distribution of electrons and the maximum exchange energy released when electrons with the same spin exchange positions.

Problem 2:

Determine the set of four quantum numbers for the valence electron of a Sodium atom (Z=11Z = 11).

Solution:

n=3,l=0,ml=0,ms=+12n = 3, l = 0, m_l = 0, m_s = +\frac{1}{2} (or −12-\frac{1}{2})

Explanation:

The electronic configuration of NaNa is 1s22s22p63s11s^2 2s^2 2p^6 3s^1. The valence electron is in the 3s3s orbital. For 3s3s: the principal quantum number n=3n=3, for ss-orbital l=0l=0, which forces ml=0m_l=0, and the spin msm_s can be +12+\frac{1}{2}.

Problem 3:

Compare the energies of 4s4s and 3d3d orbitals using the (n+l)(n+l) rule.

Solution:

For 4s4s: n=4,l=0⇒n+l=4n=4, l=0 \Rightarrow n+l = 4. For 3d3d: n=3,l=2⇒n+l=5n=3, l=2 \Rightarrow n+l = 5. Since 4<54 < 5, the 4s4s orbital has lower energy than 3d3d.

Explanation:

According to the (n+l)(n+l) rule, the orbital with the lower sum of nn and ll has lower energy and is filled first. Therefore, electrons enter the 4s4s orbital before the 3d3d orbital.

Electronic Configuration of Atoms Revision - Class 11 Chemistry ICSE