SENAPREP
Laws of Chemical Combination (excluding problems) Questions
Chapter: Atomic Structure & Laws | Total Available: 35 Real Questions
Which one of the following is the correctmolecular formula of ammonium carbonateif the valency of ammonium ion is (+1) andcarbonate anion is (–2)?
Correct Answer
📖 Governing Law / Core Concept
The molecular formula of a compound is determined by the valence (or charge) of its constituent ions. For ammonium (NH₄⁺) and carbonate (CO₃²⁻), balancing their charges yields the formula.
🧪 Step-by-Step Breakdown
Given:
- Valency of Ammonium ion: +1 (NH₄⁺)
- Valency of Carbonate ion: -2 (CO₃²⁻)
The charges must be balanced in the compound:
Let x = number of ammonium ions and y = number of carbonate ions:
This means:
The final molecular formula is:
🔍 Option Analysis
- Option A: (NH₄)₂CO₃ - Correct formula with balanced charges.
- Option B: NH₄(CO₃)₂ - Indicates incorrect composition, suggesting two carbonate ions.
- Option C: (NH₃)₂CO₂ - Does not represent ammonium; incorrect.
- Option D: NH₄CO₃ - Incorrect ratio; lacks charge balance.
⚡ Mnemonic / Speed-Run
Remember: "Ammonium and Carbonate, pair them to get rates!" For charge balancing, always cross-multiply the valencies.
📊 Visual Suggestion
6. Visual Suggestion
A diagram illustrating the balance of charges between the ammonium and carbonate ions, showing one carbonate ion attracting two ammonium ions to form the compound.
📖 Factual Verification & Reference
Verified against NCERT Class XI Chemistry, Chapter on States of Matter.
Which one of the following particles in the nucleus of an atom was discovered by J.Chadwick?
1. Correct Answer
Option D is correct. Neutron was discovered by J. Chadwick.
📖 Short Explanation
James Chadwick discovered the neutron, a neutrally charged particle within the atomic nucleus, through his experiments in 1932.
💡 3. Why Correct Answer is Correct
❌ 4. Why Other Options are Wrong
- •Option A: Electrons are not constituents of the nucleus; they orbit around the nucleus in different energy levels and were discovered by J.J. Thomson.
- •Option B: Protons are positively charged particles within the nucleus discovered by Ernest Rutherford.
- •Option C: Positrons are the antimatter counterpart of electrons and are not found in the nucleus. They were discovered by Carl Anderson.
NDA Speed-Run / Mnemonic Shortcut
Remember: N for Neutron and N for Neutral (no charge), both discovered by Nobel Prize laureate J. Chadwick.
5. Key Learning Point
The neutron is a neutral particle within an atom's nucleus, fundamental to atomic structure and nuclear physics, discovered by J. Chadwick in 1932.
6. Visual Suggestion
Include a schematic depiction of an atom showing the nucleus with protons and neutrons in the center and electrons orbiting, highlighting how neutrons lack charge but contribute to atomic mass.
Which one of the following findings is NOT observed in Rutherford’s Nuclear Model of Atom?
1. Correct Answer
Option A is correct. Rutherford’s Nuclear Model of Atom did not propose the presence of a neutral center in the atom, such as a nucleus.
📖 Short Explanation
Rutherford’s Nuclear Model revolutionized atomic theory by introducing the existence of a small, dense nucleus, but it did not suggest the nucleus was neutral.
💡 3. Why Correct Answer is Correct
❌ 4. Why Other Options are Wrong
- •Option B: Correct as Rutherford’s model does indicate that nearly all the mass of the atom resides in the nucleus.
- •Option C: Correct because the model depicts electrons revolving around the nucleus in circular paths.
- •Option D: Correct as it reflects the fact that the size of the nucleus is very small compared to the rest of the atom.
NDA Speed-Run / Mnemonic Shortcut
Remember that Rutherford's atom model introduced the nucleus and its positive charge without detailing the presence of neutrons which are neutral.
5. Key Learning Point
Rutherford's experiment identified a nucleus filled with protons; it did not account for neutrons.
6. Visual Suggestion
A diagram showing a compact nucleus at the center with electrons orbiting in elliptical paths and differential sizes between the nucleus and the atom as a whole.
Element X forms a chloride with the formula XCl2, which is a solid with high melting poin t.X would most likely be in the same group ofperiodic table as
1. Correct Answer
Option C is correct. Element X is most likely magnesium (Mg), which is in the same group of the periodic table as magnesium.
📖 Short Explanation
The high melting point of the solid XCl2 indicates that element X forms ionic bonds. Such characteristics are typical of alkaline earth metals like magnesium (Mg), which has a valency of 2, matching the formula XCl2.
💡 3. Why Correct Answer is Correct
❌ 4. Why Other Options are Wrong
- •Option A: Sodium (Na) forms NaCl, not a compound like NaCl2, because it has a valency of +1.
- •Option B: Aluminum (Al) forms AlCl3 because it has a valency of +3, which does not match XCl2.
- •Option D: Potassium (K) forms KCl, with a valency of +1, and does not align with the formula XCl2.
NDA Speed-Run / Mnemonic Shortcut
Remember that elements in the same group share similar valency. Group 1 forms MCl, Group 2 forms MCl2, and so forth. This pattern quickly identifies which group the element is likely in.
5. Key Learning Point
Compounds like XCl2 indicate a typical Group 2 element behavior, associating with alkaline earth metals such as magnesium (Mg).
6. Visual Suggestion
A diagram of the periodic table highlighting alkaline earth metals with visual indicators for typical compounds like XCl2 would elucidate their common valency in teaching settings.
Which one of the following statements is NOT correct ?
1. Correct Answer
Option D is correct. 4 grams of hydrogen gas contains 12.044 × 10²³ molecules, not 6.022 × 10²³.
📖 Short Explanation
The problem addresses the application of the concept of moles and Avogadro's number. Specifically, it examines whether the given quantities of different gases align with the expected number of molecules and volumes at standard temperature and pressure (STP).
💡 3. Why Correct Answer is Correct
According to Avogadro's Law, one mole of any gas at STP occupies a volume of 22.4 liters and contains Avogadro's number of molecules, which is 6.022 × 10²³. For hydrogen gas (H₂), which has a molar mass of 2 grams:
Thus, 4 grams of H₂ would amount to:
Therefore, the statement in option D is incorrect, hence it is the correct choice for the question.
❌ 4. Why Other Options are Wrong
- •Option A: A half mole of nitrogen gas (N₂) indeed occupies 11.2 liters at STP, as 1 mole occupies 22.4 liters.
- •Option B: 17 grams of ammonia (NH₃) indeed contains 6.022 × 10²³ molecules at STP, since it's equivalent to 1 mole.
- •Option C: 22.4 liters of CO₂ gas at STP indeed corresponds to 44 grams, which is 1 mole.
NDA Speed-Run / Mnemonic Shortcut
Remember the universal constants at STP: 1 mole = 22.4 liters = 6.022 × 10²³ molecules. Double-check these values for any quick mole-to-molecule calculations.
5. Key Learning Point
Always convert given mass to number of moles, and recall Avogadro's number for the number of molecules/net calculations at STP conditions.
6. Visual Suggestion
Consider constructing a simple table summarizing various gases, their molar masses, volumes at STP, and number of molecules to visually aid students in remembering key relationships.
Which of the following sets of elements has the same valency?
Correct Answer
📖 Governing Law / Core Concept
Valency is defined as the combining capacity of an element, often determined by the number of electrons in the outermost shell (valence shell).
🧪 Step-by-Step Breakdown
The elements in question and their common valency are:
- Na (Sodium) - Valency: +1
- Mg (Magnesium) - Valency: +2
- Ca (Calcium) - Valency: +2
- Al (Aluminum) - Valency: +3
- K (Potassium) - Valency: +1
- Ba (Barium) - Valency: +2
Elements Mg, Ca, and Ba are from Group 2 of the periodic table (alkaline earth metals) which consistently have a valency of +2 in their common ionic compounds:
This equation illustrates magnesium forming a +2 ion with nitrate ions, resulting in magnesium nitrate.
🔍 Option Analysis
All other options contain a mix of elements with differing valencies:
- Option A (Na, Mg, Ca): Na is +1
- Option B (Na, Mg, Al): Na is +1, Al is +3
- Option C (Mg, Ca, K): K is +1
⚡ Mnemonic / Speed-Run
Remember "Mighty Calcium Battles" to recall that Mg, Ca, and Ba all share a +2 valency.
6. Visual Suggestion
6. Visual Suggestion
Consider a periodic table highlighting alkaline earth metals (Be, Mg, Ca, Sr, Ba, Ra) and their +2 valency, with their common compounds illustrated.
📖 Factual Verification & Reference
Verified against NCERT Class 10 Science, Chapter on the Periodic Table and Element Properties.
Which one of the following conclusions could not be derived from Rutherfords -particle scattering experiment?
Correct Answer
📖 Governing Law / Core Concept
The foundational concept from Rutherford's experiment illustrates that atoms consist primarily of empty space, with most mass concentrated in a dense nucleus.
🧪 Step-by-Step Breakdown
Rutherford's gold foil experiment revealed several key points:
- Most of the atom's volume is empty.
- The nucleus is significantly smaller than the atom (radius ratio: ~105).
- Rutherford did not propose a model for electron orbits, which was later introduced by Bohr.
Thus, the statement regarding fixed energy orbits for electrons could not be derived from his findings.
🔍 Option Analysis
- A: Correctly states the atom's empty space.
- B: Accurate ratio derived from the nucleus's size.
- D: Correctly identifies that most atomic mass is nuclear.
- C: Incorrect inference as stationary orbits were not a part of Rutherford's conclusions.
⚡ Mnemonic / Speed-Run
Remember: "Empty Nucleus Relates (E.N.R.)" to recall key findings of Rutherford's model (empty space, nucleus mass concentration).
6. Visual Suggestion
6. Visual Suggestion
A diagram illustrating the Rutherford experiment setup, indicating the alpha particle beam, gold foil, and deflection angles, would clarify the conclusions drawn from the experiment.
📖 Factual Verification & Reference
Verified against NCERT Class XI Chemistry, Chapter 2 (Structure of Atom).
Rutherfords alpha-particle scattering experiment was responsible for the discovery of which one of the following?
Correct Answer
📖 Governing Law / Core Concept
The atomic nucleus is the central component of an atom that contains protons and neutrons, comprising most of an atom's mass.
🧪 Step-by-Step Breakdown
Rutherford's alpha-particle scattering experiment involved directing alpha particles onto a thin gold foil. The key observations were:
- A majority of the alpha particles passed through the foil, indicating that atoms are mostly empty space.
- A small fraction were deflected at large angles, suggesting the presence of a dense, positively charged nucleus.
🔍 Option Analysis
A: Electron - Discovered by J.J. Thomson in 1897, these are negatively charged subatomic particles.
B: Proton - Identified by Ernest Rutherford in 1919, protons are positively charged particles found in the nucleus.
D: Neutron - Discovered by James Chadwick in 1932, neutrons are neutral particles also located in the nucleus.
⚡ Mnemonic / Speed-Run
Remember "Rutherford Rebuilt Atom" to recall that Rutherford's experiment redefined atomic structure with the nucleus at its core.
6. Visual Suggestion
6. Visual Suggestion
A diagram of Rutherford's gold foil experiment showing the paths of alpha particles, illustrating how most pass through while a few are deflected at sharp angles, highlighting the nucleus's presence.
📖 Factual Verification & Reference
Verified against NCERT Class XI Chemistry, Chapter 2 (Structure of Atom).
What is the maximum number of electrons in the M-Shell?
Correct Answer
📖 Governing Law / Core Concept
The maximum number of electrons in a shell is given by the formula \(2n^2\), where \(n\) is the principal quantum number.
🧪 Step-by-Step Breakdown
In this case, for the M-Shell:
Substituting into the formula:
Thus, the M-Shell can hold a maximum of 18 electrons.
🔍 Option Analysis
- A: 6 - Incorrect; this number corresponds to the maximum capacity of the L-Shell (n=2).
- B: 8 - Incorrect; this is the maximum capacity for the K-Shell (n=1), and the L-Shell (n=2) can hold 8 electrons.
- C: 18 - Correct; as derived above, represents M-Shell capacity.
- D: 32 - Incorrect; refers to the maximum capacity of the N-Shell (n=4).
⚡ Mnemonic / Speed-Run
Remember the formula \(2n^2\) and the capacities:
- K-Shell (n=1): 2
- L-Shell (n=2): 8
- M-Shell (n=3): 18
- N-Shell (n=4): 32
📖 Factual Verification & Reference
Verified against NCERT Class 11 Chemistry, Chapter: Structure of Atom.
Chlorine occurs in nature in two isotopic forms of masses 35 u and 37 u in the ratio of 3: 1 respectively. What is the average atomic mass of the Chlorine atom?
Correct Answer
📖 Governing Law / Core Concept
The average atomic mass is calculated using the weighted average of the masses of isotopes based on their natural abundance: \[ \text{Average Atomic Mass} = \frac{(mass_1 \cdot abundance_1) + (mass_2 \cdot abundance_2)}{abundance_1 + abundance_2} \]🧪 Step-by-Step Breakdown
Given: - Mass of \( ^{35}Cl = 35 \, u \) - Mass of \( ^{37}Cl = 37 \, u \) - Ratio of isotopes = 3:1 Calculate abundance: \[ \text{Abundance of } ^{35}Cl = 3 \quad \text{and} \quad \text{Abundance of } ^{37}Cl = 1 \] Therefore, the formula becomes: \[ \text{Average Atomic Mass} = \frac{(35 \cdot 3) + (37 \cdot 1)}{3 + 1} \] Calculating: \[ \text{Average Atomic Mass} = \frac{105 + 37}{4} = \frac{142}{4} = 35.5 \, u \]🔍 Option Analysis
- Option A (36.1 u): Overestimates due to wrong ratio. - Option C (36.5 u): Should not exceed isotopic masses. - Option D (35.1 u): Underestimates average mass.⚡ Mnemonic / Speed-Run
Remember: "Chlorine 3:1, means 35.5 not 36".- Atomic mass = ratio-based calculation
- Average mass formula
A pie chart illustrating the isotopic distribution of Chlorine indicating 75% for 35u and 25% for 37u, enhancing understanding of the average calculation.
Verified against NCERT Class XI Chemistry, Chapter 4 (Structure of Atom).