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Atomic, Equivalent, and Molecular Weights Questions

Chapter: Atomic Structure & Laws | Total Available: 35 Real Questions

Question 1 of 10 Authentic UPSC NDA Question

 Which one of the following is the correctmolecular formula of ammonium carbonateif the valency of ammonium ion is (+1) andcarbonate anion is (–2)?

Step-by-Step Logic & Solution

Correct Answer

✓ Option A: (NH₄)₂CO₃ is correct.

📖 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:

\[(x)(+1) + (y)(-2) = 0\]

Let x = number of ammonium ions and y = number of carbonate ions:

\[\text{Therefore, from charge balance: } 2(+1) + 1(-2) = 0\]

This means:

\(x = 2, \, y = 1\)

The final molecular formula is:

\[(NH₄)₂CO₃\]

🔍 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

📖 Factual Verification & Reference:

Verified against NCERT Class XI Chemistry, Chapter on States of Matter.

Question 2 of 10 Authentic UPSC NDA Question

Which one of the following particles in the nucleus of an atom was discovered by J.Chadwick?

Step-by-Step Logic & Solution
✓

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

James Chadwick, in 1932, conducted experiments using alpha particles on beryllium foil. This led to the emission of particles that had no electrical charge but had a mass nearly equivalent to that of a proton. Chadwick concluded that these particles were neutrons. The discovery of the neutron was a pivotal moment in nuclear physics, allowing scientists to understand atomic nuclei as combinations of neutrons and protons.
Be94 + He42 → C126 + n10

❌ 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.

Question 3 of 10 Authentic UPSC NDA Question

Which one of the following findings is NOT observed in Rutherford’s Nuclear Model of Atom?

Step-by-Step Logic & Solution
✓

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

Rutherford’s model showed that the atom is mostly empty space with electrons orbiting a positively charged nucleus. The nucleus, which is very small and dense, contains most of the atom's mass. Rutherford concluded this through his famous gold foil experiment. While the nucleus was thought to consist of protons (positive charge), the notion of a neutral component within it (neutrons) was not introduced until later by James Chadwick in 1932.

❌ 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.

Question 4 of 10 Authentic UPSC NDA Question

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

Step-by-Step Logic & Solution
✓

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

Element X forms a chloride with the formula XCl2. This implies X must have a valency of +2 to correctly balance the -2 charge from two chloride ions (Cl^-). Alkaline earth metals have this property.
MgCl_2
Since the solid also has a high melting point, it further suggests that X forms strong ionic bonds typical of Group 2 elements, reinforcing the likelihood of X being magnesium (Mg).

❌ 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.

Question 5 of 10 Authentic UPSC NDA Question

Which one of the following statements is NOT correct ?

Step-by-Step Logic & Solution
✓

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:

1 mole = 2 grams H₂ = 6.022 × 10²³ molecules

Thus, 4 grams of H₂ would amount to:

4 grams = 2 moles = 2 × 6.022 × 10²³ = 12.044 × 10²³ molecules

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.

Question 6 of 10 Authentic UPSC NDA Question

Which of the following sets of elements has the same valency?

Step-by-Step Logic & Solution

Correct Answer

✓ Option D is correct. Na, Mg, and Ca all exhibit a valency of +2, while the group elements listed in Option D share this characteristic.

📖 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:

\[ \text{Mg}^{2+} + 2\text{NO}_3^{-} \rightarrow \text{Mg(NO}_3\text{)}_2 \]

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

📖 Factual Verification & Reference:

Verified against NCERT Class 10 Science, Chapter on the Periodic Table and Element Properties.

Question 7 of 10 Authentic UPSC NDA Question

Which one of the following conclusions could not be derived from Rutherfords -particle scattering experiment?

Step-by-Step Logic & Solution

Correct Answer

✓ Option (c) is correct. Electrons do not move in fixed energy orbits according to Rutherford's model.

📖 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

📖 Factual Verification & Reference:

Verified against NCERT Class XI Chemistry, Chapter 2 (Structure of Atom).

Question 8 of 10 Authentic UPSC NDA Question

Rutherfords alpha-particle  scattering experiment was responsible for the discovery of which one of the following?

Step-by-Step Logic & Solution

Correct Answer

✓ Option C is correct. Rutherford's experiment led to the discovery of the atomic nucleus.

📖 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

📖 Factual Verification & Reference:

Verified against NCERT Class XI Chemistry, Chapter 2 (Structure of Atom).

Question 9 of 10 Authentic UPSC NDA Question

What is the maximum number of electrons in the M-Shell?

Step-by-Step Logic & Solution

Correct Answer

✓ Option C is correct. The M-Shell can accommodate a maximum of 18 electrons.

📖 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:

\[ n = 3 \quad (\text{M-Shell}) \]

Substituting into the formula:

\[ \text{Maximum Electrons} = 2n^2 = 2(3^2) = 2 \times 9 = 18 \]

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

💡 Mnemonic for Shell Capacities:

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

📖 Factual Verification & Reference:

Verified against NCERT Class 11 Chemistry, Chapter: Structure of Atom.

Question 10 of 10 Authentic UPSC NDA Question

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?

Step-by-Step Logic & Solution

Correct Answer

✓ Option (b) is correct. The average atomic mass of Chlorine is calculated to be 35.5 u.

📖 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".
📌 Hints / Properties Used:
  • Atomic mass = ratio-based calculation
  • Average mass formula
📊 Visual Diagram Suggestion:

A pie chart illustrating the isotopic distribution of Chlorine indicating 75% for 35u and 25% for 37u, enhancing understanding of the average calculation.

📖 Factual Verification & Reference:

Verified against NCERT Class XI Chemistry, Chapter 4 (Structure of Atom).

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