Hydrogen

When zinc reacts with excess aqueous alkali, such as with a strong base like NaOH, hydrogen gas is evolved, and an anionic complex forms. This reaction can be represented by the chemical equation:

In this reaction:

1. Metallic zinc reacts with sodium hydroxide, a strong alkali.
2. The water present in an aqueous medium helps to facilitate the reaction, resulting in the formation of sodium zincate and hydrogen gas.
3. The complex formed, sodium zincate, contains the anion .

To understand why is formed rather than other options:

• Zn(OH)₂: Zinc hydroxide is typically formed when zinc reacts with limited alkali (e.g., not in excess), resulting in an insoluble hydroxide which does not occur in this scenario.
• ZnO: Zinc oxide would require a different reaction condition such as heating zinc with oxygen or insufficient alkali; it is not produced in this aqueous alkaline scenario.
• [Zn(OH)₄]^2–: Though [Zn(OH)₄]^2– is a tetrahydroxo complex, the correct answer in the context of the reaction with excess alkali typically leads to the formation of [ZnO₂]^2–, the sodium zincate ion.

Therefore, the correct product formed when zinc reacts with excess aqueous alkali is the complex ion , making the answer: [ZnO₂]^2–.

The correct answer is (D) : Atomic mass
¹H, ²D and ³H have same atomic number but their atomic masses are different. Isotopes have same atomic number i.e. same number of protons.

In reaction (3):
hydrogen peroxide ( ) acts as a reducing agent. Here:
donates electrons (is oxidized) to form .
is reduced to .
Oxidation and reduction:


Thus, is a reducing agent in this reaction.

The correct answer is 46

To solve this problem, we can use the balanced chemical equation for the reaction between nitrogen gas (N₂) and hydrogen gas (H₂) to form ammonia gas (NH₃):
N₂ + 3H₂ → 2NH₃
According to the stoichiometry of the balanced equation, 1 mole of N₂ reacts with 3 moles of H₂ to produce 2 moles of NH₃.
Given: Volume of nitrogen gas (N₂) = 56.0 L Volume of ammonia gas (NH₃) produced = 20 L
Using the ideal gas law, we can relate the volume of a gas to the number of moles of that gas:
n = PV / RT
where: n is the number of moles of the gas P is the pressure of the gas V is the volume of the gas R is the gas constant (0.0821 L·atm/(mol·K)) T is the temperature in Kelvin
Let's calculate the number of moles of nitrogen gas (N₂):
n(N₂) = (P(N₂) * V(N₂)) / (R * T)
Since the pressure, volume, and temperature are not specified, we can assume they are constant throughout the reaction. Therefore, we can write:
n(N₂) = n(NH₃) * (1 mole N₂ / 2 moles NH₃)
n(N₂) = (20 L * 1 mole N₂) / (2 moles NH₃)
n(N₂) = 10 L
Now, let's calculate the volume of unused nitrogen gas (N₂):
V(unused N₂) = V(N₂ initial) - V(N₂ used)
V(unused N₂) = 56.0 L - 10 L
V(unused N₂) = 46.0 L
Therefore, the volume of unused nitrogen gas is 46.0 L.

The Correct Option is (C) : B > A > C
The cage like structure of ice means that the solid is actually slightly less dense than the liquid, which explains why ice floats on water rather than sinks. The structure of liquid water is very similar, but in the liquid, the hydrogen bonds are continually broken and formed because of rapid molecular motion.

Step 1: Analyze the Structures
All four structures represent ascorbic acid (vitamin C), but they differ in the position of the double bond within the ring and the configuration of the hydroxyl groups.
Step 2: Consider Resonance Stabilization
The most stable structure will be the one with the greatest resonance stabilization. Structure (1) has the most resonance structures possible because the double bond is conjugated with the carbonyl group, allowing for delocalization of electrons. This delocalization stabilizes the structure more than the other structures. Also, the hydroxyl group on C2 will donate electron density to the carbonyl group at C1, whereas in structure 2 the carbonyl group will pull electrons making it unstable.

Stabilization and Decomposition of Hydrogen Peroxide (H₂O₂):

Hydrogen peroxide (H₂O₂) decomposes readily in the presence of light, heat, and impurities like dust or alkali. Urea acts as a stabilizer for hydrogen peroxide by forming a complex that reduces the rate of decomposition.

Analysis of Options:

1. Dust: Catalyzes the decomposition of H₂O₂.
2. Urea: Stabilizes H₂O₂ and reduces its decomposition.
3. Glass containers: Do not actively stabilize H₂O₂.
4. Alkali: Increases the decomposition rate of H₂O₂.

Conclusion: The correct answer is option (2).

Properties of Alkenes:

• Melting Point: The melting point of trans alkene is greater than that of cis alkene due to the higher symmetry of trans isomers, which allows better packing in the solid state.
• Boiling Point: The boiling point of cis alkene is greater than that of trans alkene due to stronger intermolecular forces, as cis isomers are more polar.
• Bromine Addition: Bromine (Br₂) adds more easily to cis alkene than trans alkene because of the closer proximity of substituents in cis isomers.
• Isomers: A is cis–but–2–ene, and B is trans–but–2–ene.
• Solubility: A (cis–but–2–ene) is more soluble due to its polarity, as cis isomers have a net dipole moment.

The correct answer is option (C): B, C & D only

Deuterated hydrogen peroxide ( ) can be conveniently prepared in the laboratory by reacting with deuterated sulfuric acid ( ) in (heavy water). This method allows for the direct incorporation of deuterium into the hydrogen peroxide molecule.

Solution:
Match the terms in Column-I with their description in Column-II and choose the correct option:

Column 1(Type of hydride) Column II
A. Electron deficient 1. MgH₂
B. Electron precise 2. HF
C. Electron rich 3. CH₄
D. Saline Hydride 4. B₂H₆

Matching:

A. Electron deficient hydrides are those which have less than 8 electrons in their valence shell. B₂H₆ is an example of an electron deficient hydride.
Therefore, A matches with 4.

B. Electron precise hydrides are those which have exactly 8 electrons in their valence shell. CH₄ is an example of an electron precise hydride.
Therefore, B matches with 3.

C. Electron rich hydrides are those which have more than 8 electrons in their valence shell due to the presence of lone pairs. HF has lone pairs on fluorine, making it an electron rich hydride.
Therefore, C matches with 2.

D. Saline hydrides are ionic hydrides formed by alkali and alkaline earth metals. MgH₂ is an example of a saline hydride.
Therefore, D matches with 1.

Correct Matching:
A - 4 (B₂H₆)
B - 3 (CH₄)
C - 2 (HF)
D - 1 (MgH₂)

Analysis of Assertion (A):

• The hydrogen ion (H⁺) is essentially a proton.
• Since a proton is a subatomic particle with an extremely small size, its radius is approximately 1.5 × 10^–3 pm.
• Thus, the Assertion (A) is correct.

Analysis of Reason (R):

• H⁺ is a bare proton without any electrons. Due to its high charge density and extremely small size, it cannot exist independently in nature.
• Instead, it immediately associates with other atoms or molecules, typically forming complexes like H₃O⁺ in aqueous solutions.
• Thus, the Reason (R) is also correct.

Conclusion:

Both the Assertion (A) and Reason (R) are correct. Furthermore, the Reason (R) correctly explains the Assertion (A). Therefore, the correct answer is:

Option A: Both A and R are true, and R is the correct explanation of A.

The correct answer is option (C) BeH₂

BeH₂ is the least stable as it has a significant covalent character and is an electron-deficient hydride.

- Hydrogen storage materials should exhibit properties such as high absorption capacity and easy reversibility.
- is a hydride-forming alloy that can efficiently store and release hydrogen under suitable conditions.
- Other options, such as and di-isobutylaluminium hydride, do not possess such characteristics suitable for hydrogen storage.

1. Energy Levels and Wavelengths: The energy of a transition in a hydrogen-like atom is given by:

where is the atomic number and is the principal quantum number. The wavelength is related to the energy difference between levels:

where is the Rydberg constant, and and are the lower and higher energy levels, respectively.
2. Lyman Series (Hydrogen): The shortest wavelength in the Lyman series corresponds to a transition from to . For hydrogen ( ):

3. Balmer Series (He ): For He ( ), the longest wavelength in the Balmer series corresponds to a transition from to . The wavelength is given by:

Simplify:

Thus:

Final Answer: .

The energy ordering of orbitals for hydrogen-like atoms is governed by the principle that the energy increases as the principal quantum number (n) increases, but within the same shell, orbitals with higher angular momentum (l) have higher energy.
- (A) is correct as it correctly orders the orbitals: .
- (B) is incorrect as , and .
- (C) is correct as it follows the correct ordering of orbitals for hydrogen.
- (D) is incorrect because has lower energy than , so this ordering is wrong.
Therefore, the correct answers are (A) and (C).

- For a cyclic process, the internal energy change ( ) is always zero as the system returns to its initial state.
- Since is a state function, it depends only on the initial and final states, which are the same for all three cases.