Test: MCQs (One or More Correct Option): The s-Block Elements | JEE Advanced - JEE MCQ

BaO₂.8H₂O + H₂SO₄ → BaSO₄ + H₂O₂ + 8H₂O

Ca is obtained by electrolysis of molten mixture of CaCl₂ mixed with CaF₂.

The free ammoniated electrons make the solution of Na in liquid NH₃ a very powerful reducing agent.
NOTE : The ammonical solution of an alkali metal is rather favoured as a reducing agent than its aqueous solution because in aqueous solution the alkali metal being highly electropositive evolves hydrogen from water (thus H₂O acts as an oxidisng agent) while its solution in ammonia is quite stable, provided no catalyst (transition metal) is present.

Heavy water is D₂O, deuterium oxide.

Na⁺ ion has larger size than Mg²⁺ ion and hence hydration energy of Mg²⁺ is larger than that of Na⁺.

Na₂O₂ + H₂SO₄ (20% ice cold) → Na₂SO₄ + H₂O₂

Glauber ’s salt is Na₂SO₄.10 H₂O.

NOTE : The more electropositive metal will not be reduced by hydrogen.
Among given choices only Al is more electropositive than hydrogen.
∴ It will not be reduced by hydrogen.

NOTE : Acidic and basic salts cannot exist together.
Since NaHCO₃ is an acid salt of H₂CO₃, it reacts with NaOH to form Na₂CO₃ and H₂O.
NaHCO₃ + NaOH → Na₂CO₃ + H₂O

Oscillation of loose electrons.

Volume strength = Normality  × 5.6 = 1.5 × 5.6 = 8.4 L

The increasing thermal stability is

NOTE : Increasing size of cation decreases its polarization ability towards carbonate, making the compound more stable.

In going from top to bottom in a group, the first ionization potential decreases, thus
Be > Mg > Ca

KIO₄ + H₂O₂ → KIO₃ + H₂O + O₂
Thus H₂O₂ is acting as a reducing agent 2NH₂OH + H₂O₂ → N₂ + 4H₂O
Here H₂O₂ is acting as an oxidising agent

Only H₂O₂ (hydrogen peroxide) and BaO2 (Barium peroxide) contain peroxide ions. So (a) and (c) are the correct choices.

A small piece of sodium is cut to expose a fresh surface. The sodium is dropped into liquid ammonia at a temperature of approximately -33 degrees Celsius. Some of the sodium dissolves, forming sodium cations surrounded by ammonia molecules and electrons surrounded by ammonia molecules. The solvated electrons give the blue color to the solution.
Because of the mobility of the electrons, the solution is a good electrical conductor. Bubbles of hydrogen gas are formed by a second reaction that also produces sodium amide. More concentrated solutions appear bronze-colored and have a conductivity similar to metals.
2Na+2NH₃ ​→ 2NaNH₂+H₂​.
Hence options A,B,C & D are correct

Temporary hardness is due to bicarbonates of calcium and magnesium. Temporary hardness can be removed by Clark’s process, which involves the addition of slaked lime, Ca(OH)₂. Washing soda (Na₂CO₃) removes both the temporary and permanent hardness by converting soluble calcium and magnesium compounds into insoluble carbonates.