Classification of Elements and Periodicity in PropertiesNEET Chemistry · Class 11 · NCERT Chapter 3

As you move across a period from left to right, the atomic number increases. More protons in the nucleus pull the electrons with greater force. The electrons are added to the same energy level (same shell), so shielding increases only slightly. The greater nuclear pull contracts the electron cloud, reducing the atomic radius.

Nitrogen has the electron configuration 2s² 2p³. All three 2p orbitals are half-filled with one electron each (Hund's rule). This gives the configuration extra stability due to symmetric electron distribution and maximum exchange energy. Oxygen (2s² 2p⁴) has one paired electron in a 2p orbital. The repulsion between paired electrons makes it slightly easier to remove one, so IE₁(O) < IE₁(N).

Fluorine is smaller than chlorine. The 2p orbitals of fluorine are compact and already hold 5 electrons, creating strong electron-electron repulsion. When a new electron is added to F, the repulsion in the crowded 2p subshell partially offsets the energy released. In chlorine, the 3p orbitals are larger with less repulsion, so more energy is released when an electron is added. Hence: EGA(Cl) = −349 kJ/mol is more negative than EGA(F) = −328 kJ/mol.

The effective nuclear charge (Z_eff) is the actual nuclear charge experienced by a valence electron after accounting for the shielding effect of inner electrons. Z_eff = Z − σ, where σ is the shielding constant. Across a period, Z increases but the number of inner shielding electrons barely changes, so Z_eff increases. This explains why atomic radius decreases and IE increases across a period.

For s and p-block elements: the period number equals the highest principal quantum number (n) in the configuration. For the group, count the total number of valence electrons (s + p electrons in the outermost shell). For d-block elements: the period equals the highest n; the group number = (electrons in outermost s) + (electrons in the penultimate d). For example, [Ar] 3d⁵ 4s² has n=4 (Period 4), and 5+2=7 valence d+s electrons, so it is in Group 7 (Mn).

Isoelectronic species have the same number of electrons but different nuclear charges (different elements). Example: O²⁻, F⁻, Na⁺, Mg²⁺, Al³⁺ all have 10 electrons. Within this series, higher nuclear charge pulls the 10 electrons more tightly, giving a smaller radius. So ionic radius order: Al³⁺ < Mg²⁺ < Na⁺ < F⁻ < O²⁻.