Exams › JEE Advanced › Chemistry

The successive ionization energies of an atom are: IE1 = 7.5 eV, IE2 = 25.6 eV, IE3 = 48.6 eV, IE4 = 170.6 eV. The ground-state electronic configuration of this atom is:

1. 1s² 2s² 2p⁶ 3s¹
2. 1s² 2s² 2p⁶ 3s² 3p¹
3. 1s² 2s² 2p⁶ 3s² 3p³
4. 1s² 2s² 2p⁶ 3s²

Correct answer: 1s² 2s² 2p⁶ 3s² 3p¹

Solution

The large jump between IE3 (48.6 eV) and IE4 (170.6 eV) indicates that 3 electrons are in the outermost shell (valence electrons), and the 4th electron must come from an inner shell. An element with 3 valence electrons and total electrons in the 3rd shell: configuration is 1s² 2s² 2p⁶ 3s² 3p¹ (aluminum-like, Z=13). This is consistent with IE1 being moderate (3p electron, partially shielded) and the big jump after 3 electrons are removed.

Related JEE Advanced Chemistry questions

• When the kinetic energy of an electron is quadrupled, how does the wavelength of its associated de-Broglie wave change?
• An electron labeled e₁ is in the fifth energy level, while another electron labeled e₂ is in the fourth energy level. The orbital radius of e₁ is five times that of e₂. What is the ratio of the speed of e₁ (v₁) to the speed of e₂ (v₂)?
• If the minimum wavelength of the spectral line in the Lyman series for a hydrogen atom is X, what is the maximum wavelength of the spectral line in the Balmer series for a Li²⁺ ion?
• For a sub-energy level with an azimuthal quantum number of 4, what are the highest and lowest possible spin multiplicities?
• The wave number is given by the formula ν = 1/λ = kZ² [1/n₁² - 1/n₂²]. For the Lyman series, the shortest wavelength (λₘₐₓ) occurs when n₁ = 1, n₂ = ∞, and Z = 1 (for a hydrogen atom). This simplifies to 1/λ = kZ² [1/1² - 1/∞²] ⇒ λ = 1/x. Which of the following statements is correct?
• In Bohr's theory, the electron with the greatest kinetic energy is found in:

⚔️ Practice JEE Advanced Chemistry free + battle 1v1 →