Chemistry - Hydrogen and Its compound Question with Solution | TestHub

A. Na₆P₆O₁₈ is used in Calgon's method to remove hardness of water due to formation of soluble complex with Ca²⁺ and Mg²⁺ ions.

This statement is CORRECT. Sodium hexametaphosphate (Na₆P₆O₁₈), commonly known as Calgon, is used to remove permanent hardness of water. It forms soluble complexes with Ca²⁺ and Mg²⁺ ions, preventing them from precipitating as insoluble salts. The reaction can be represented as:

$$2{\mathrm{Ca}}^{2+}+{\mathrm{Na}}_2[{\mathrm{Na}}_4({\mathrm{PO}}_3{)}_6]\to {\mathrm{Na}}_2[{\mathrm{Ca}}_2({\mathrm{PO}}_3{)}_6]+4{\mathrm{Na}}^{+}$$

The complex ion, $[{\mathrm{Ca}}_2({\mathrm{PO}}_3{)}_6{]}^{2-}$, is soluble in water, thus effectively removing the hardness-causing ions.

B. In water gas shift reaction CO₂ is removed by using sodium arsenide solution.

This statement is INCORRECT. The water-gas shift reaction is:

$$\mathrm{CO}(g)+{\mathrm{H}}_2\mathrm{O}(g)\rightleftharpoons {\mathrm{CO}}_2(g)+{\mathrm{H}}_2(g)$$

In industrial processes, CO₂ is typically removed from the product stream using various methods, such as absorption with amine solutions (e.g., monoethanolamine, MEA), physical absorption with solvents like Selexol or Rectisol, or pressure swing adsorption (PSA). Sodium arsenide solution is not a standard method for CO₂ removal in this context.

C. Industrially H₂O₂ is prepared by auto-oxidation of 2-alkylanthraquinone.

This statement is INCORRECT. Industrially, hydrogen peroxide (H₂O₂) is prepared by the auto-oxidation of 2-ethylanthraquinol (the reduced form of 2-ethylanthraquinone), not 2-alkylanthraquinone itself. The process involves the catalytic hydrogenation of 2-ethylanthraquinone to 2-ethylanthraquinol, followed by oxidation with air to regenerate 2-ethylanthraquinone and produce H₂O₂.

Step 1: Hydrogenation

$${\mathrm{C}}_{16}{\mathrm{H}}_{12}{\mathrm{O}}_2(2-ethylanthraquinone)+{\mathrm{H}}_2\stackrel{catalyst}{\to }{\mathrm{C}}_{16}{\mathrm{H}}_{14}{\mathrm{O}}_2(2-ethylanthraquinol)$$

Step 2: Auto-oxidation

$${\mathrm{C}}_{16}{\mathrm{H}}_{14}{\mathrm{O}}_2(2-ethylanthraquinol)+{\mathrm{O}}_2\to {\mathrm{C}}_{16}{\mathrm{H}}_{12}{\mathrm{O}}_2(2-ethylanthraquinone)+{\mathrm{H}}_2{\mathrm{O}}_2$$

D. All metals of group 6, 7, 8, 9 do not form metallic hydride.

This statement is INCORRECT. The metals of groups 7, 8, and 9 (Mn, Fe, Co, Ni, etc.) generally do not form stable stoichiometric hydrides under normal conditions. This region of the periodic table is known as the "hydride gap." However, metals in Group 6, specifically Cr, Mo, and W, can form hydrides, although they are often non-stoichiometric or require specific conditions. For example, chromium hydride (CrH) is known. Therefore, the blanket statement that "all metals of group 6, 7, 8, 9 do not form metallic hydride" is false because Group 6 metals can form hydrides.

The only correct statement is A.