Free s-Block and p-Block Elements MCQs with Answers

• A. Mg(g)→ Mg+ (g) + e-
• B. Mg+(g)→ Mg2+(g) + e-
• C. Na(g) → Na+ (g)+ e-
• D. Na+(g) → Na+2(g) + e-

Explanation: Na+ has stable electronic configuration so removal of electron form Na+ to form Na+2 ion requires maximum energy.

• A. Generally reducing character of elements increases in period.
• B. Generally oxidising character of elements increases in period.
• C. Generally, basic character of oxides decreases in a group.
• D. All are correct.

Explanation: Oxidising character of elements increases in a period. Therefore option B is correct.

• A. It forms an basic oxide.
• B. It belongs to II A group.
• C. It belongs to IV period.
• D. It forms an acidic oxide.

Explanation: The element is calcium and hence its oxide (CaO) is basic in nature. So option D is correct.

• A. d5,d3,d1,d4 increasing magnetic moment.
• B. MO, M2O , MO2, M2O5 - decreasing basic strength.
• C. Sc, V, Cr, Mn - increasing number of oxidation states.
• D. Co2+,Fe3+,Cr3+,Sc3+ - increasing stability.

Explanation: d1 ,d3 ,d4 ,d5 - increasing magnetic moment.

• A. 1s2s22sp63s23p1
• B. 1s22s22p63s2
• C. 1s22s2sp63s1
• D. 1s22s2sp63s23p2

Explanation: Sodium has electronic configuration as 1s2 2s2 2p6 3s1. The last electron or the valence electron is in s orbital.

• A. Ionization enthalpy increases for each successive electron.
• B. The greatest increase in ionization enthalpy is experienced on removal of electron from core of noble gas configuration.
• C. End of valence electrons is marked by a big jump in ionization enthalpy.
• D. Removal of electron from orbitals bearing lower n value is easier than from orbital having higher n value.

Explanation: Removal of Electron from orbital bearing lower n value is difficult than from orbitals having higher n values. Option D is correct.

• A. 3, 11, 19, 37.
• B. 5, 13, 21, 39.
• C. 7, 15, 31, 49.
• D. 5, 13, 37, 49.

Explanation: Option D shows elements of group IIIA (5-Boron,13-Aluminum, 31-Gallium, 49-Indium. Hence option D is correct.

• A. IA.
• B. III A.
• C. III B.
• D. Zero Group.

Explanation: Group III B includes lanthanides and actinides and the total number of elements in this group = 32. Hence correct option is C.

• A. Small size
• B. High electronegativity
• C. Low ionization energy
• D. Large atomic radius

Explanation: Lithium and beryllium's small size and high charge density cause strong polarizing effects, leading to covalent bond formation and their…

• A. The properties of elements are periodic functions of their atomic numbers.
• B. Non-metallic elements are fewer in number than metallic elements.
• C. For transition elements, the 3d-orbitals are filled with electrons after 3p-orbitals and before 4s-orbitals.
• D. The first ionisation enthalpies of elements generally increase with an increase in atomic number as we move along a period.

Explanation: Option C is the correct answer because it incorrectly states the order of filling for transition elements.

• A. Increases
• B. Decreases
• C. Constant
• D. None

Explanation: Across a period, added electrons enter the same principal shell, so the inner-shell electrons causing most shielding remain almost…

• A. Al
• B. B
• C. Si
• D. C

Explanation: Borax is the sodium salt of tetraboric acid. It is the most important of all borates .Borax occurs as a natural deposit called tincal in…

• A. Acidic oxide
• B. Basic oxide
• C. Amphoteric oxide
• D. None of these

Explanation: Oxides of relatively less electropositive elements, such as BeO, Al2O3 Bi2O3 and ZnO are amphoteric and behave as acids towards strong…

• A. As antiseptic in medicine
• B. For washing eyes
• C. In soda bottles
• D. For enamels and glazes

Explanation: Uses of Boric Acid1. Boric acid is used in medicines as an antiseptic, e.g.

• A. N2O4
• B. NO
• C. NO2
• D. None of these

Explanation: As Ferrous sulphate reacts with Nitrogen oxide, a compound called Nitroso ferrous sulphate is formed which is responsible for brown ring…

• A. Their oxides are basic
• B. Their oxides and hydroxides are water soluble
• C. both a and b
• D. They are found in earth

Explanation: Alkali metals form basic oxides (e.g., Li₂O, Na₂O) which readily react with water to form strong alkaline solutions.Their hydroxides are…

• A. Small size
• B. High charge density
• C. Lower electropositivity
• D. All of the above

Explanation: Lithium's small size, high charge density, and lower electropositivity compared to its heavier alkali metal family members lead to…

• A. Francium
• B. Cesium
• C. Radium
• D. Rubidium

Explanation: RadiumHere's why:Francium, Caesium, and Rubidium are all part of Group 1 (also known as Group IA) in the periodic table, which corresponds…

• A. Calcium (Ca)
• B. Chromium (Cr)
• C. Both of the above
• D. Barium sulphate

Explanation: Among the given options, Barium sulphate (BaSO4) has the lowest solubility in water.

• A. sodium bicarbonate
• B. sodium carbonate
• C. sodium metal
• D. sodium hydroxide Down's cell is specifically designed for the electrolysis of molten sodium chloride (NaCl) to produce sodium metal (Na) and chlorine gas (Cl2).

Explanation: Down's cell is specifically designed for the electrolysis of molten sodium chloride (NaCl) to produce sodium metal (Na) and chlorine gas…