Isotopes and Isobars — Long Answer Questions (Class 9 Chemistry)
Medium Level (Application & Explanation)
Q1. What are isotopes? Explain with examples and mention how their chemical and physical properties differ.
Answer:
Isotopes are atoms of the same element that have the same atomic number but different mass numbers. This means they have the same number of protons but different numbers of neutrons.
Examples: hydrogen has three isotopes: Protium (11H), Deuterium (12H) and Tritium (13H). Carbon has 612C and 614C. Chlorine has 1735Cl and 1737Cl.
Their chemical properties are essentially the same because chemical behaviour depends on the electrons and atomic number, which do not change among isotopes. For example, all isotopes of chlorine form the same kinds of compounds.
Their physical properties can be different due to mass differences: melting/boiling points, densities, and rates of diffusion change slightly. Also, some isotopes are radioactive (e.g., tritium), while others are stable.
In short: same chemical behavior, but different physical behavior and some differing nuclear properties (stability, radioactivity).
Q2. How do you calculate the average atomic mass of an element? Use chlorine with isotopes 35 u and 37 u in a 3:1 ratio as an example.
Answer:
The average atomic mass of an element is the weighted mean of the masses of its naturally occurring isotopes, weighted by their relative abundances. The formula is:
Average mass = (mass₁ × abundance₁ + mass₂ × abundance₂ + ...) / (total abundance).
For chlorine with isotopes 35 u and 37 u in a 3:1 ratio, the abundances are 75% for 35 u and 25% for 37 u. Calculate step by step:
Multiply masses by their percent abundances: 35 × 75 = 2625 and 37 × 25 = 925.
Add the results: 2625 + 925 = 3550.
Divide by total percent (100): 3550 / 100 = 35.50 u.
So the average atomic mass of chlorine is 35.5 u. This value appears in the periodic table and reflects the natural mix of isotopes. It helps chemists predict the mass of atoms in samples and perform calculations in chemical equations.
Q3. Describe three important applications of isotopes in daily life, medicine, and environmental science.
Answer:
Medicine (diagnosis and treatment): Radioactive isotopes such as Iodine‑131 and Cobalt‑60 are used in diagnosis (imaging) and treatment (radiotherapy). For example, I‑131 helps detect and treat thyroid problems because the thyroid absorbs iodine. Cobalt‑60 is used to kill cancer cells in radiotherapy. These isotopes allow doctors to locate problems or destroy diseased tissue.
Tracers in biology and industry: Stable or radioactive isotopes (like Carbon‑14 or Deuterium) are used as tracers to follow the path of substances in chemical reactions, metabolic pathways, or industrial processes. Tracers help scientists understand where atoms go during reactions or how pollutants move in the environment.
Environmental science and archaeology:Carbon‑14 dating is used to estimate the age of ancient organic materials (like wood or bones). Isotope ratios in ice cores or tree rings (oxygen and hydrogen isotopes) give information about past climates and temperatures. These uses show how isotopes are powerful tools for research across fields.
Q4. What are isobars? Give examples and explain why isobars show different chemical behaviour despite having the same mass number.
Answer:
Isobars are atoms of different elements that have the same mass number (same total number of nucleons: protons + neutrons) but different atomic numbers (different numbers of protons).
Examples: 2040Ca (calcium) and 1840Ar (argon) are isobars because both have mass number 40 but calcium has atomic number 20 and argon has 18. Another pair is 714N and 614C.
They show different chemical behaviour because chemical properties depend on the number of protons (which sets the identity of the element) and on the electron configuration. Even though isobars have the same total nucleons, their proton numbers differ, so their electrons and electronic arrangements differ. For example, calcium is a metal that tends to lose electrons and form positive ions, while argon is a noble gas that is chemically inert with a full valence shell.
In nuclear processes, isobars are important because mass number conservation is considered, but chemical properties are governed by atomic number.
Q5. Explain how isotopes affect the physical properties of a compound using heavy water (D2O) versus ordinary water (H2O) as an example.
Answer:
Heavy water (D2O) contains deuterium (12H) instead of the usual protium (11H). Both are isotopes of hydrogen, so chemically both water types behave similarly in forming H‑bonds and participating in reactions. However, because deuterium is heavier, heavy water has noticeably different physical properties.
For example, boiling point and melting point of D2O are higher than H2O. Density of D2O is greater, so objects float slightly differently. The vibrational frequencies of bonds change because heavier nuclei vibrate more slowly; this changes spectroscopic signatures (useful in research).
Biological systems often react more slowly with D2O because of the kinetic isotope effect: reaction rates involving bond breaking to hydrogen differ when hydrogen is replaced by deuterium. This can affect enzymes and metabolic rates if large amounts of heavy water are used.
Thus, isotopic substitution keeps chemical identity but alters physical and kinetic behaviour, which is useful in experiments and has practical consequences.
High Complexity (Analytical & Scenario-Based)
Q6. Analytical: A sample of element X contains three isotopes: Z29X (mass 28.97 u) at 5%, Z30X (mass 29.97 u) at 20%, and Z31X (mass 30.97 u) at 75%. Calculate the average atomic mass and explain how changing the abundance of the heaviest isotope would affect the average mass.
Answer:
First, calculate the weighted average using the given abundances:
Contribution from 28.97 u isotope: 28.97 × 5 = 144.85
Contribution from 29.97 u isotope: 29.97 × 20 = 599.40
Contribution from 30.97 u isotope: 30.97 × 75 = 2322.75
Sum = 144.85 + 599.40 + 2322.75 = 3067.00
Divide by total percentage (100): Average mass = 30.67 u.
Interpretation: The average atomic mass (≈ 30.67 u) reflects the dominant contribution of the most abundant isotope (75% at 30.97 u). If the abundance of the heaviest isotope (30.97 u) increased, the average atomic mass would increase proportionally, because heavier isotopes contribute more to the weighted mean. Conversely, reducing the heaviest isotope’s abundance in favour of lighter isotopes would decrease the average mass. This shows how natural or artificial changes in isotopic composition (isotopic enrichment or depletion) directly change the average atomic mass value reported in tables and measured by instruments.
Q7. Scenario: During a laboratory demonstration of beta decay, you observe that carbon‑14 (614C) transforms to nitrogen‑14 (714N). Explain why the two nuclei are isobars, and discuss the conservation laws and changes in subatomic particles occurring in this decay.
Answer:
In beta decay of carbon‑14, a neutron transforms into a proton, an electron (beta particle), and an antineutrino. The nuclear equation is: 614C → 714N + e⁻ + ν̄.
Both 614C and 714N have the same mass number 14 (14 nucleons), so they are isobars. The atomic number changes from 6 to 7 because one neutron became a proton. This changes the element from carbon to nitrogen, hence different chemical identity.
Conservation laws:
Mass number (A) is conserved because a neutron becomes a proton — total nucleons remain 14.
Charge (Z) is conserved overall because when a neutron becomes a proton the nucleus charge increases by +1 but an electron (beta particle) is emitted carrying negative charge, keeping total charge balanced.
Energy and lepton number are conserved with emission of the electron and an antineutrino.
This example shows how nuclear transformations can produce isobars and why isobars are important in nuclear physics: the element changes, mass number remains the same, and nuclear stability considerations determine whether such decay occurs.
Q8. Scenario & Analysis: You are asked to separate a small sample containing isotopes of an element in a school project. Explain two feasible methods (one theoretical and one practical for a high school lab), and discuss limitations and safety concerns.
Answer:
Theoretical method (accurate): Mass spectrometry separates isotopes by their mass-to-charge ratio. Ions are accelerated and bent by a magnetic field; lighter ions curve more and hit different positions on a detector, allowing precise separation and measurement. This method is very accurate and widely used in research and industry.
Limitation: Mass spectrometers are expensive and require trained operators, vacuum systems, and safety protocols — not usually available in a school lab.
Practical school-lab method (demonstration-level): Fractional distillation or electrolysis can separate isotopes in...
