Classification of Elements and Periodicity in Properties — Practice Quiz

Al₂O₃

Step 1: Oxides are classified based on the position of the element in the periodic table. Elements on the far left (metals) form basic oxides; elements on the far right (non-metals) form acidic oxides. Step 2: Elements near the border between metals and non-metals (metalloids or elements with intermediate character) form amphoteric oxides — they can react with both acids and bases. Step 3: Aluminium (Al) is a metal that lies near the boundary. Al₂O₃ reacts with acids: Al₂O₃ + 6HCl → 2AlCl₃ + 3H₂O AND with bases: Al₂O₃ + 2NaOH → 2NaAlO₂ + H₂O. Step 4: MgO is a basic oxide (Group 2 metal). SO₃ i

The 3p electron in Al is more shielded and at higher energy than the 3s electron in Mg, making it easier to remove.

Step 1: Mg has the configuration [Ne] 3s². The electron removed is a 3s electron. Step 2: Al has the configuration [Ne] 3s² 3p¹. The electron removed is the 3p electron. Step 3: 3p orbitals are higher in energy than 3s orbitals. The 3p electron also has poorer penetration to the nucleus and is more effectively shielded by the 3s² electrons of Al itself. Step 4: Therefore, less energy is needed to remove the 3p electron from Al compared to the 3s electron from Mg — making Al's first ionization enthalpy lower. Step 5: Option A is wrong — Mg is actually larger than Al, but that alone doesn't expl

Mendeleev's table could predict the existence and properties of undiscovered elements.

Step 1: One of the greatest achievements of Mendeleev's Periodic Table was its PREDICTIVE POWER. Step 2: He identified gaps in the table based on patterns in properties and boldly predicted the existence, atomic mass, density, and oxide/chloride formulas of missing elements. Step 3: He named them Eka-Boron (Scandium), Eka-Aluminium (Gallium), and Eka-Silicon (Germanium). The predicted and observed properties matched remarkably well. Step 4: Option A is wrong — Mendeleev himself had to make exceptions, like placing tellurium before iodine ignoring atomic mass order. Step 5: Option C is wrong —

The effective nuclear charge increases across a period as electrons are added to the same shell without significantly increasing shielding.

Step 1: As we move across a period, atomic number (and hence nuclear charge) increases by 1 each step. Step 2: Each new electron is added to the SAME principal energy level (e.g., n=2 for Period 2). Step 3: Electrons in the same shell do not shield each other very effectively from the nucleus — so the shielding does not increase proportionally. Step 4: Result: The effective nuclear charge (= actual nuclear charge − shielding) increases steadily across the period. Step 5: This increased attraction pulls all electrons, including valence electrons, closer to the nucleus → atomic radius decreases.