Hydrogen

- Statement I is correct because sodium hexametaphosphate is used to remove permanent hardness of water in the calgon method. - Statement II is incorrect because lithium generally forms covalent hydrides, not interstitial hydrides. Thus, the correct answer is (3).

Step 1: Analyze Reaction (I) - Sodium with Water:
Balanced Equation: .

• Oxidation State of H in : .
• Oxidation State of H in : .
• Change: (Decrease in oxidation number = Gain of electrons).
• Conclusion: Water acts as an oxidizing agent and gets reduced to Hydrogen gas ( ). Sodium is a strong reducing agent.

Step 2: Analyze Reaction (II) - Fluorine with Water:
Balanced Equation: .

• Oxidation State of O in : .
• Oxidation State of O in : .
• Change: (Increase in oxidation number = Loss of electrons).
• Conclusion: Water acts as a reducing agent and gets oxidized to Oxygen gas ( ). Fluorine is a very strong oxidizing agent.

Step 3: Final Answer:
In (I) water is reduced to , and in (II) water is oxidized to . This matches Option (D).

Step 1: Identify the type of water hardness.
The presence of soluble bicarbonates of calcium ( ) and magnesium ( ) causes temporary hardness in water.

Step 2: Write the reaction for the decomposition of Calcium bicarbonate upon boiling.
When water containing is boiled, the soluble bicarbonate decomposes to form insoluble calcium carbonate ( ) precipitate. The precipitate from is .

Step 3: Write the reaction for the decomposition of Magnesium bicarbonate upon boiling.
When water containing is boiled, the soluble bicarbonate decomposes to form magnesium carbonate ( ). However, unlike calcium carbonate, is slightly soluble in hot water and reacts further with water via hydrolysis (or decomposition under high temperature) to form the less soluble magnesium hydroxide ( ) precipitate. The final precipitate from is .

Step 4: Conclude the final precipitates.
The precipitates formed are and .

This question concerns the anomalous expansion of water (H O) and heavy water (D O).
The temperature at which normal water (H O) exhibits its maximum density is approximately 4 C.
The question provides the variable y for this temperature in Kelvin.
y = K.
Heavy water (D O) also shows a similar property, but its maximum density occurs at a different temperature.
The temperature of maximum density for heavy water is approximately 11.2 C.
The variable x represents this temperature in Kelvin.
x = K.
We are asked to find the difference (x - y).
x - y = K.
This calculated value of 7.2 K is approximately equal to 7.0 K.

Step 1: Understanding the Concept:
Hydrides can be broadly classified into ionic, covalent, and metallic hydrides. Covalent hydrides are further classified based on the number of electrons around the central atom relative to a complete octet.
- Electron-deficient hydrides: The central atom has fewer than 8 valence electrons. Typically formed by Group 13 elements.
- Electron-precise hydrides: The central atom has exactly 8 valence electrons (a complete octet). Typically formed by Group 14 elements.
- Electron-rich hydrides: The central atom has a complete octet and one or more lone pairs of electrons. Typically formed by Group 15, 16, and 17 elements.
Step 2: Analyzing the Options:
- (A) B₂H₆, AlH₃: Boron and Aluminum are in Group 13. Their hydrides, diborane and alane, are electron-deficient. B in B₂H₆ does not have a complete octet.
- (B) NaH, MgH₂: Sodium and Magnesium are Group 1 and Group 2 metals. They form ionic or saline hydrides, which consist of metal cations (Na⁺, Mg²⁺) and hydride anions (H⁻). They are not classified under the electron-deficient/precise/rich scheme for covalent hydrides.
- (C) HCl, H₂S: Chlorine (Group 17) and Sulfur (Group 16) are p-block non-metals. In HCl, the Cl atom has 3 lone pairs of electrons in addition to the bonding pair. In H₂S, the S atom has 2 lone pairs in addition to the two bonding pairs. Since the central atoms have excess electrons in the form of lone pairs, these are electron-rich hydrides.
- (D) CH₄, SiH₄: Carbon and Silicon are in Group 14. In methane (CH₄) and silane (SiH₄), the central atom forms four single bonds and has a complete octet with no lone pairs. These are electron-precise hydrides.
Step 3: Final Answer:
HCl and H₂S are the hydrides where the central atom has lone pairs, making them electron-rich. Therefore, option (C) is correct.

1. For CuSO H O: Four H O molecules coordinate with Cu in a square planar arrangement, and one is hydrogen-bonded to SO . Thus, statement I is incorrect as more than one H O is involved in hydrogen bonding.
2. For BaCl H O: Water molecules occupy interstitial positions in the lattice, not coordinated to Ba , making statement II correct.
3. For CrCl H O: Typically [Cr(H O) Cl ]Cl H O, where four H O molecules coordinate with Cr . However, in some forms like [Cr(H O) Cl ] H O, three H O coordinate, making statement III correct in context.
4. Thus, statements II and III are correct, so the answer is (3) II, III only.

• H O decomposes easily.
• Urea acts as a stabilizer by complexing trace metal ions that catalyze decomposition.