NEET PYQs (Periodic Table) Competency Based Questions | Classification of Elements and Periodicity in Properties

Anand Classes provides NEET PYQs (Periodic Table) Competency Based Questions (CBQs) with detailed step-by-step solutions from the chapter Classification of Elements and Periodicity in Properties. These solved past year questions help students understand periodic trends like ionisation energy, electronegativity, atomic size, metallic character, and chemical reactivity. Mastering these concepts is essential for cracking NEET Chemistry with accuracy and speed. Click the print button to download study material and notes.

AIPMT 1996 – Periodic Table & Classification of Elements (Ionic Radii)

NEET Q.1 : Which one of the following ions will be smallest in size?(a) $Na^+$
(b) $Mg^{2+}$
(c) $F^-$
(d) $O^{2-}$
[CBSE AIPMT 1996]

Answer : Correct Option: (b)

Step 1: Concept of ionic size in isoelectronic species

When species are isoelectronic (same number of electrons), the ionic size depends mainly on the nuclear charge (Z):

• Higher nuclear charge → stronger attraction → smaller ionic radius.
• Lower nuclear charge → weaker attraction → larger ionic radius.

Step 2: Isoelectronic nature of given ions

All the given species have 10 electrons:

• $Na^+$ : 11 − 1 = 10 electrons
• $Mg^{2+}$ : 12 − 2 = 10 electrons
• $F^−$ : 9 + 1 = 10 electrons
• $O^{2−}$ : 8 + 2 = 10 electrons

Thus, they are isoelectronic.

Step 3: Nuclear charge comparison

Step 4: Conclusion

Among these, $Mg^{2+}$ has the highest nuclear charge (Z = 12) for the same number of electrons, so it pulls its electron cloud most strongly, giving the smallest size.

Hence, the correct answer is (b) $Mg^{2+}$, a key concept highlighted in NEET PYQs Chapterwise Solutions for Periodic Table Properties.

📝 Concept Takeaway

• Isoelectronic species → size decreases with increasing nuclear charge.
• Cations are always smaller than their parent atoms, while anions are larger.
• Here, $Mg^{2+}$ is the smallest ion due to its maximum effective nuclear charge among the given species.

CBSE AIPMT 1994 -Periodic Table & Classification of Elements (Basic Character trend)

NEET Q.2 : Among the following, the one which is most basic is
(a) ZnO
(b) MgO
(c) Al₂O₃
(d) N₂O₅
[CBSE AIPMT 1994]

Answer : Correct Option: (b)

Step 1: Types of oxides

• Basic oxides: Formed by metals, react with acids to form salts and water.
• Acidic oxides: Formed by non-metals, react with bases to form salts and water.
• Amphoteric oxides: Show both acidic and basic behaviour.

Step 2: Nature of the given oxides

• $ZnO$: Amphoteric (reacts with both acids and bases).
• $Al_2O_3$: Amphoteric.
• $N_2O_5$: Acidic (oxide of non-metal, nitrogen).
• $MgO$: Basic (typical metallic oxide).

Step 3: Comparison

Step 4: Conclusion

Since $MgO$ is a typical metallic oxide and reacts strongly as a base, it is the most basic among the given oxides.

Hence, the correct answer is (b) MgO, a concept emphasized in NEET Chemistry Solutions for Periodic Table Properties.

📝 Concept Takeaway

$MgO$ is the most basic oxide among the given options.

Metallic oxides are generally basic.

Non-metallic oxides are generally acidic.

Some oxides like $Al_2O_3$ and $ZnO$ are amphoteric.

CBSE AIPMT 1993 -Periodic Table & Classification of Elements (Successive Ionization Enthalpy Trend)

NEET Q.3 : Which electronic configuration of an element has abnormally high difference between second and third ionisation energy?
(a) 1s² 2s² 2p⁶ 3s¹
(b) 1s² 2s² 2p⁶ 3s² 3p¹
(c) 1s² 2s² 2p⁶ 3s² 3p²
(d) 1s² 2s² 2p⁶ 2s² 2p⁶
[CBSE AIPMT 1993]

Answer : Correct Option: (d)

Step 1: Understanding ionisation energy

• Ionisation energy is the energy required to remove an electron.
• Successive ionisation energies increase because electrons are removed from an increasingly positive ion.
• A sharp increase occurs when electrons are removed from a stable noble gas configuration.

Step 2: Checking the configurations
(a) 1s² 2s² 2p⁶ 3s¹ → after losing 2 electrons, not noble gas.
(b) 1s² 2s² 2p⁶ 3s² 3p¹ → after losing 2 electrons, still not noble gas.
(c) 1s² 2s² 2p⁶ 3s² 3p² → after losing 2 electrons, still not noble gas.
(d) 1s² 2s² 2p⁶ → this is the neon configuration (noble gas).

Step 3: Explanation

• In option (d), after losing two electrons, the element attains the stable noble gas configuration of neon.
• Removing a third electron would mean disturbing this stable closed-shell configuration.
• Hence, the difference between the second and third ionisation energies is abnormally high.

Conclusion
The correct answer is (d) 1s² 2s² 2p⁶, since removal of the third electron requires a very high amount of energy after achieving noble gas stability. This type of question is frequently seen in NEET Previous Year Questions on periodic table properties.

📝 Concept Takeaway

• Successive ionisation energies increase, but the jump is largest when a stable noble gas configuration is reached.
• Elements with configurations close to noble gases show abnormally high differences in ionisation energies.
• In this case, after losing 2 electrons, the atom resembles neon, so the third ionisation energy is much higher.

CBSE AIPMT 1993 -Periodic Table & Classification of Elements (Atomic Volume Trend)

NEET Q.4 : In the periodic table from left to right in a period, the tomic volume
(a) decreases
(b) increases
(c) remains same
(d) first decreases then increases
[CBSE AIPMT 1993]

Answer : Correct Option: (d)

Step 1: Definition of atomic volume
Atomic volume = Molar mass / Density
It is indirectly related to the atomic size (or radius) of an element.

Step 2: Trend across a period

• Moving left to right in a period:
  • Nuclear charge increases.
  • Atomic radius generally decreases.
  • Therefore, atomic volume decreases initially.
• Towards the end of a period (halogens, noble gases):
  • Interatomic distances increase due to electron–electron repulsions.
  • This leads to an increase in atomic volume again.

Step 3: Trend summary

Conclusion
The atomic volume first decreases and then increases from left to right in a period. Hence, the correct answer is (d), an important fact often seen in NEET Important Concepts for periodic properties.

📝 Concept Takeaway

• Atomic volume is closely linked to atomic size.
• Across a period:
  • Decrease due to increasing nuclear charge.
  • Increase again near the end due to larger interatomic distances in non-metals and noble gases.
• Therefore, the trend is: first decreases, then increases.

CBSE AIPMT 1993 -Periodic Table & Classification of Elements (Non-Metals Properties)

NEET Q.5 : One of the characteristic properties of non-metals is that
(a) are reducing agents
(b) form basic oxides
(c) form cations by electron gain
(d) are electronegative
[CBSE AIPMT 1993]

Answer : Correct Option: (d)

Step 1: General properties of non-metals

• Non-metals have high ionisation energies and high electronegativities.
• They tend to gain electrons rather than lose them.
• This makes them strong oxidising agents, not reducing agents.

Step 2: Checking the options
(a) Reducing agents → typical of metals, not non-metals.
(b) Form basic oxides → metals form basic oxides, non-metals form acidic oxides.
(c) Form cations by electron gain → incorrect, electron gain leads to anions, not cations.
(d) Are electronegative → correct, non-metals gain electrons easily due to high electronegativity.

Conclusion
Non-metals are characterised by their tendency to gain electrons and form negative ions. Therefore, they are highly electronegative. The correct answer is (d), a point often stressed in NEET Exam Chapterwise Practice on periodic properties.

📝 Concept Takeaway

High electronegativity is the most defining property of non-metals.

Metals → electropositive, lose electrons, form cations.

Non-metals → electronegative, gain electrons, form anions.

Non-metals generally form acidic oxides.

CBSE AIPMT 1993 -Periodic Table & Classification of Elements (Isoelectronic Species and Ionic Sizes Trend)

NEET Q.6 : Na⁺, Mg²⁺, Al³⁺ and Si⁴⁺ are isoelectronic. The order of their ionic size is
(a) Na⁺ > Mg²⁺ < Al³⁺ < Si⁴⁺
(b) Na⁺ < Mg²⁺ > Al³⁺ > Si⁴⁺
(c) Na⁺ > Mg²⁺ > Al³⁺ > Si⁴⁺
(d) Na⁺ < Mg²⁺ > Al³⁺ < Si⁴⁺
[CBSE AIPMT 1993]

Answer : Correct Option: (c)

Step 1: Concept of isoelectronic species

• Isoelectronic species have the same number of electrons but different nuclear charges.
• Greater nuclear charge (Z) → stronger attraction on electrons → smaller ionic radius.

Step 2: Nuclear charges of given ions

• Na⁺ → Z = 11, electrons = 10
• Mg²⁺ → Z = 12, electrons = 10
• Al³⁺ → Z = 13, electrons = 10
• Si⁴⁺ → Z = 14, electrons = 10

Step 3: Trend
As Z increases across these isoelectronic ions, the nuclear pull on electrons increases, causing ionic radius to shrink.

Step 4: Comparison

Conclusion
Thus, the ionic size order is:
Na⁺ > Mg²⁺ > Al³⁺ > Si⁴⁺

Hence, the correct answer is (c), a frequently asked concept in NEET Question Bank Solutions on periodic properties.

📝 Concept Takeaway

Order here: Na⁺ (largest) > Mg²⁺ > Al³⁺ > Si⁴⁺ (smallest).

For isoelectronic species: greater nuclear charge → smaller ionic size.

Cations shrink in size as charge increases.

CBSE AIPMT 1993 -Periodic Table & Classification of Elements (Hydration Energy Concept)

NEET Q.7 : One would expect proton to have very large
(a) charge
(b) ionisation potential
(c) hydration energy
(d) radius
[CBSE AIPMT 1993]

Answer : Correct Option: (c)

Step 1: Key concept of hydration energy

• Hydration energy is the energy released when one mole of ions is surrounded by water molecules.
• It depends mainly on the size and charge of the ion.
• Smaller the size → higher charge density → stronger attraction for water molecules → greater hydration energy.

Step 2: Special case of proton

• Proton (H⁺) is the smallest possible cation with practically negligible radius.
• This extremely small size gives it very high charge density.
• Thus, the hydration energy of H⁺ is the maximum among all ions.

Step 3: Comparison of options

• (a) Charge → Proton has +1 charge, but not “very large.”
• (b) Ionisation potential → Refers to removing an electron, but proton already has no electron.
• (c) Hydration energy → Very large because of tiny radius. ✅
• (d) Radius → Proton has the smallest radius, not large.

Conclusion
Therefore, the proton has very large hydration energy, making option (c) correct.

📝 Concept Takeaway

Proton (H⁺), being the smallest cation, shows the highest hydration energy.

Hydration energy ∝ 1 / ionic size.

CBSE AIPMT 1993 -Periodic Table & Classification of Elements (Anions Formation)

NEET Q.8 : Which of the following sets has strongest tendency to form anions?
(a) Ga, In, Tl
(b) Na, Mg, Al
(c) N, O, F
(d) V, Cr, Mn
[CBSE AIPMT 1993]

Answer : Correct Option: (c)

Step 1: Recall definition of anions

• Anions are formed when atoms gain electrons.
• The tendency to gain electrons depends on electronegativity and electron affinity.

Step 2: Compare the given sets

• (a) Ga, In, Tl → Metals, generally form cations, not anions.
• (b) Na, Mg, Al → Also metals, prefer to lose electrons (cation formation).
• (c) N, O, F → Highly electronegative non-metals, with strong tendency to gain electrons → strongest tendency to form anions. ✅
• (d) V, Cr, Mn → Transition metals, prefer cation formation.

Step 3: Explanation

• Electronegativity trend in a period increases left to right.
• N, O, and F belong to the second period and have very high electronegativity values (especially fluorine, the most electronegative element).
• This makes them highly effective in accepting electrons to form anions.

Conclusion
Thus, the set N, O, F shows the strongest tendency to form anions.

📝 Concept Takeaway

Metals usually form cations, not anions.

Non-metals with high electronegativity (especially halogens and chalcogens) readily form anions.

CBSE AIPMT 1990 -Periodic Table & Classification of Elements (Ionisation of hydrogen atom)

NEET Q.9 : The ionisation of hydrogen atom would give rise to
(a) hydride ion
(b) hydronium ion
(c) proton
(d) hydroxyl ion
[CBSE AIPMT 1990]

Answer : Correct Option: (c)

Step 1: Recall hydrogen atom structure

• Hydrogen atom has 1 proton and 1 electron.
• Its atomic number is 1, so only one electron orbits the nucleus.

Step 2: What happens on ionisation?

• Ionisation means removal of one electron.
• If hydrogen loses its only electron, the particle left is just the proton in the nucleus.

Equation:

H → H⁺ + e⁻

Step 3: Identify the ion formed

• The H⁺ ion is simply a proton, as no other particle remains.
• It is not a hydride ion (H⁻), not hydronium (H₃O⁺), and not hydroxyl (OH⁻).

Conclusion
Therefore, the ionisation of hydrogen atom gives rise to a proton (H⁺).

📝 Concept Takeaway

This property makes hydrogen behave differently from all other elements in the periodic table.

Hydrogen is unique because its cation (H⁺) is just a proton.

CBSE AIPMT 1989 -Periodic Table & Classification of Elements (Periodic Table Properties)

NEET Q.10 : In the periodic table, with the increase in atomic number, the metallic character of an element
(a) decreases in a period and increases in a group
(b) increases in a period and decreases in a group
(c) increases in a period as well as in the group
(d) decreases in a period and also in the group
[CBSE AIPMT 1989]

Answer : Correct Option: (a)

Step 1: Trend in a group

• As we move down a group, atomic size increases.
• Ionisation enthalpy decreases, so it is easier to lose electrons.
• Metallic character therefore increases.

Step 2: Trend in a period

• As we move across a period (left → right), effective nuclear charge increases.
• Ionisation enthalpy increases, so it is harder to lose electrons.
• Metallic character therefore decreases.

Comparison Table:

Conclusion
Thus, metallic character decreases in a period and increases in a group.

📝 Concept Takeaway

It increases down a group due to lower ionisation enthalpy.

Metallic character ∝ tendency to lose electrons.

It decreases across a period due to high effective nuclear charge.

CBSE AIPMT 1989 -Periodic Table & Classification of Elements (Pauling’s electronegativity values)

NEET Q.11 : Pauling’s electronegativity values for elements are useful in predicting
(a) polarity of the molecules
(b) position in the emf series
(c) coordination numbers
(d) dipole moments
[CBSE AIPMT 1989]

Answer : Correct Option: (a)

Step 1: Role of Pauling’s electronegativity

• Electronegativity is the ability of an atom to attract shared electrons in a bond.
• Pauling assigned numerical values to electronegativities for comparison.

Step 2: Prediction of polarity

• If the electronegativity difference between two bonded atoms is zero, the bond is non-polar.
• If there is a significant difference, the bond is polar, and the molecule develops partial charges.

Formula used:

Δx = |x_A – x_B|

Relation with bond energy:
ΔE = (actual bond energy) – √(E_AA × E_BB)

Step 3: Why other options are wrong

• EMF series is based on electrode potentials, not electronegativity.
• Coordination numbers depend on structure and size, not electronegativity.
• Dipole moment is a vector property; it depends on both polarity and geometry.

Conclusion
Pauling’s electronegativity values are most useful for predicting the polarity of molecules.

📝 Concept Takeaway

Δx > 0 → polar bond.

Greater the difference in electronegativity → greater bond polarity.

Δx = 0 → non-polar bond.

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