NEET Chemistry PYQs Solutions | Periodic Table Properties pdf download

Anand Classes provides detailed NEET PYQs MCQs Solutions on Periodic Table Properties with clear explanations and solved answers to help students master important concepts for competitive exams. These step-by-step solutions cover tricky questions on periodic trends such as atomic radius, ionization enthalpy, electronegativity, and the properties of s-block, p-block, d-block, and f-block elements. Designed for quick revision and exam preparation, this study material ensures conceptual clarity for NEET aspirants. Click the print button to download study material and notes.

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NEET 2025 – Periodic Table & Classification of Elements (Topic – Periodic Properties)

NEET Q.1 : Which of the following statements are true?
(A) Unlike Ga that has a very high melting point, Cs has a very low melting point.
(B) On Pauling scale, the electronegativity values of N and Cl are not the same.
(C) Ar, $K^{+}$, $Cl^{-}$, $Ca^{2+}$, and $S^{2-}$ are all isoelectronic species.
(D) The correct order of the first ionization enthalpies of Na, Mg, Al, and Si is $Si > Al > Mg > Na$.
(E) The atomic radius of Cs is greater than that of Li and Rb. [NEET 2025]
Options:
(1) A, B, and E only
(2) C and E only
(3) C and D only
(4) A, C, and E only [NEET 2025]

Answer : (2)

Explanation :

Statement A
Claim: “Unlike Ga that has a very high melting point, Cs has a very low melting point.”
Analysis: Gallium (Ga) actually has an unusually low melting point (~29.8 °C). Cesium (Cs) also has a very low melting point (~28.5 °C). The clause that Ga has a very high melting point is incorrect.
Conclusion: False

Statement B
Claim: “On Pauling scale, the electronegativity values of N and Cl are not the same.”
Analysis: Precise modern Pauling values differ slightly ($\chi_{\text{N}}\approx 3.04$, $\chi_{\text{Cl}}\approx 3.16$), but many exam approximations treat N and Cl as having essentially the same value (≈3.0). For typical exam-level reasoning the statement is treated as False.
Conclusion: False

Statement C
Claim: “Ar, $K^{+}$, $Cl^{-}$, $Ca^{2+}$, and $S^{2-}$ are all isoelectronic species.”
Analysis: Electron counts: Ar = 18, $K^{+}$ = 19−1 = 18, $Cl^{-}$ = 17+1 = 18, $Ca^{2+}$ = 20−2 = 18, $S^{2-}$ = 16+2 = 18. All have 18 electrons → isoelectronic.
Conclusion: True

Statement D
Claim: “The correct order of the first ionization enthalpies of Na, Mg, Al, and Si is $Si > Al > Mg > Na$.”
Analysis: Approximate first ionization energies (kJ mol⁻¹): Na ≈ 496, Mg ≈ 738, Al ≈ 578, Si ≈ 787. Correct order (highest → lowest) is $Si > Mg > Al > Na$. The given order is incorrect.
Conclusion: False

Statement E
Claim: “The atomic radius of Cs is greater than that of Li and Rb.”
Analysis: In Group 1, atomic radius increases down the group: $Li < Rb < Cs$. So Cs has a larger radius than both Li and Rb.
Conclusion: True

Final Answer :

True statements: C and E only.

Correct Option: (2) C and E only

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NEET 2025 – Periodic Table & Classification of Elements (Topic – Main Group Elements)

NEET Q.2 : Which among the following electronic configurations belong to main group elements?
A. [Ne]3s¹
B. [Ar]3d³4s²
C. [Kr]4d¹⁰5s²5p⁵
D. [Ar]3d¹⁰4s¹
E. [Rn]5f⁰6d²7s²
Options:
(1) B and E only
(2) A and C only
(3) D and E only
(4) A, C and D only [NEET 2025]

Answer :

• (A) $[Ne]3s^{1}$ → Sodium (Na), belongs to s-block (main group).
• (B) $[Ar]3d^{3}4s^{2}$ → Vanadium (V), belongs to d-block (transition element).
• (C) $[Kr]4d^{10}5s^{2}5p^{5}$ → Iodine (I), belongs to p-block (main group).
• (D) $[Ar]3d^{10}4s^{1}$ → Copper (Cu), belongs to d-block (transition element).
• (E) $[Rn]5f^{0}6d^{2}7s^{2}$ → Thorium (Th), belongs to f-block (actinide).

Key Idea

Main group elements = s-block + p-block elements only.

Thus, the main group configurations are A and C only.

Final Answer

Correct Option: (2) A and C only

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NEET 2024 – Periodic Table & Classification of Elements (Topic – Electronegativity)

NEET Q.3 : Arrange the following elements in increasing order of electronegativity:
N, O, F, C, Si
Options:
(1) Si < C < N < O < F
(2) Si < C < O < N < F
(3) O < F < N < C < Si
(4) F < O < N < C < Si [NEET 2024]

Answer : (1)

Electronegativity trend:

• Across a period (left → right): increases due to increasing nuclear charge.
• Down a group: decreases due to increasing atomic size and shielding.
• Silicon (Si) is in Group 14, Period 3 → lowest among the given.
• Carbon (C) is in Group 14, Period 2 → higher than Si.
• Nitrogen (N) is in Group 15, Period 2 → higher than C.
• Oxygen (O) is in Group 16, Period 2 → higher than N.
• Fluorine (F) is in Group 17, Period 2 → the highest.

Thus, the increasing order is:
$$ Si < C < N < O < F $$

Final Answer

Correct Option: (1) Si < C < N < O < F

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NEET 2024 – Periodic Table & Classification of Elements (Topic – Ionization Enthalpy)

NEET Q.4 : Arrange the following elements in increasing order of first ionization enthalpy:
$Li,\; Be,\; B,\; C,\; N$
(1) $Li < Be < B < C < N$
(2) $Li < B < Be < C < N$
(3) $Li < Be < C < B < N$
(4) $Li < Be < N < B < C$ [NEET – 2024]

Answer: (2)

Explanation

Definition of first ionization enthalpy

First ionization enthalpy ($\Delta_{i}H$) is the minimum energy required to remove the outermost electron from a neutral gaseous atom:

$$ X(g) \;\xrightarrow{\Delta_i H}\; X^{+}(g) + e^{-} $$

Higher $\Delta_i H$ → electron is harder to remove.

General periodic trend

• Across a period (left → right): $\Delta_i H$ increases due to increasing effective nuclear charge ($Z_{\text{eff}}$).
• Down a group: $\Delta_i H$ decreases due to increasing distance of the valence electron from the nucleus.

Configurations of the given atoms

• $Li: 1s^2\,2s^1$
• $Be: 1s^2\,2s^2$
• $B: 1s^2\,2s^2\,2p^1$
• $C: 1s^2\,2s^2\,2p^2$
• $N: 1s^2\,2s^2\,2p^3$

Observed values (kJ mol$^{-1}$)

• $Li$: 520
• $B$: 801
• $Be$: 899
• $C$: 1086
• $N$: 1402

Order: $Li < B < Be < C < N$

Explanation of anomalies (stable electronic configurations)

(i) Why $B < Be$?

• In $Be$, the electron to be removed is from a completely filled $2s^2$ subshell. A filled subshell is relatively stable because of symmetry and penetration effects. Removing from $2s$ requires more energy.
• In $B$, the electron is from a single $2p^1$ orbital. The $2p$ orbital is less penetrating, more shielded, and less stable than $2s$. Hence it is easier to remove.

Thus, although $B$ is to the right of $Be$, its ionization enthalpy is lower.

(ii) Why $N$ is the highest?

• $N$ has a half-filled $2p^3$ configuration.
• Half-filled subshells are especially stable due to exchange energy stabilization and symmetric distribution of electrons.
• Removing one electron from $N$ disrupts this stable arrangement, so its ionization enthalpy is much higher.

Final Order

$$ Li < B < Be < C < N $$

✔ Correct Option: (2)

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NEET 2023 – Periodic Table & Classification of Elements (Topic – Isoelectronic Series and Ionic Size)

NEET Q.5 : The element expected to form the largest ion to achieve the nearest noble gas configuration is
(1) O
(2) F
(3) N
(4) Na [NEET 2023]

Answer (3) N

Detailed explanation

When each element attains the nearest noble-gas configuration they form these ions:
$N \to N^{3-},\; O \to O^{2-},\; F \to F^{-},\; Na \to Na^{+}$. All of these ions are isoelectronic with neon (10 electrons).

Electron counts (atomic number $Z$ ± charge):
$N^{3-}: \; 7 + 3 = 10$
$O^{2-}: \; 8 + 2 = 10$
$F^{-}: \; 9 + 1 = 10$
$Na^{+}: \; 11 – 1 = 10$

For an isoelectronic series (same number of electrons), the ionic radius depends mainly on the nuclear charge (number of protons). A larger positive nuclear charge pulls the electron cloud more strongly, producing a smaller ion. Conversely, a smaller nuclear charge gives a larger ion. Thus, ionic size for this set varies inversely with atomic number:

$$\text{ionic size} \propto \frac{1}{Z_{\text{effective}}}$$

Compare the nuclear charges (atomic numbers): $N(7) < O(8) < F(9) < Na(11)$. Therefore the sizes follow:

$$N^{3-} > O^{2-} > F^{-} > Na^{+}$$

So $N^{3-}$ is the largest ion among them. Physically this happens because $N^{3-}$ has the smallest nuclear pull on the same 10 electrons and also has strong electron–electron repulsion (three extra electrons were added), which expands the electron cloud.

Note: $N^{3-}$ is predicted to be the largest, but in practice free $N^{3-}$ ions are highly reactive and rarely observed in isolation; they appear stabilized in ionic lattices (salts). This does not change the relative size argument for the isoelectronic series.

Final answer : (3) N

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NEET 2022 – Periodic Table & Classification of Elements (Topic – Ionization Enthalpy)

NEET Q.6 : The correct order of first ionization enthalpy for the given four elements is:
(a) C < F < N < O
(b) C < N < F < O
(c) C < N < O < F
(d) C < O < N < F [NEET 2022 Phase-2]

Answer : (d) C < O < N < F

Detailed Explanation

Generally, moving left → right across a period the first ionization enthalpy (IE) increases because effective nuclear charge ($Z_{\text{eff}}$) increases while electrons are added to the same shell.

Electronic configurations (valence part):

• C: $2s^{2}2p^{2}$
• N: $2s^{2}2p^{3}$ (half-filled $2p$ subshell — extra stability)
• O: $2s^{2}2p^{4}$
• F: $2s^{2}2p^{5}$

Two important points cause the observed order:

1. Across the period trend: $IE$ increases from C → N → O → F in general because of increasing $Z_{\text{eff}}$.
2. Half-filled stability of N: A half-filled $2p^{3}$ configuration in N is unusually stable (exchange energy and symmetric occupancy). Removing one electron from N disrupts that stable arrangement, so N’s first ionization enthalpy is higher than O’s even though O is to the right of N.

Experimental approximate first ionization enthalpies (kJ mol⁻¹):

• C ≈ 1086
• O ≈ 1314
• N ≈ 1402
• F ≈ 1681

Hence, the increasing order is:
$$ \text{C} < \text{O} < \text{N} < \text{F} $$

✔ Correct option: (d) C < O < N < F

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NEET 2022 – Periodic Table & Classification of Elements (Topic – Poor Shielding Effect of f-block elements)

NEET Q.7 : Decrease in size from left to right in the actinoid series is greater and gradual than that in the lanthanoid series due to:
(a) 5f orbitals have greater shielding effect
(b) 4f orbitals are penultimate
(c) 4f orbitals have greater shielding effect
(d) 5f orbitals have poor shielding effect

Answer : (d) 5f orbitals have poor shielding effect

Detailed Explanation

1. General Trend of Atomic Radius Across a Period

• As we move left to right across a period, the nuclear charge (Z) increases.
• Since additional electrons enter the same shell (or nearly the same energy level), shielding is not very effective.
• Hence, effective nuclear charge ($Z_{\text{eff}}$) increases → electrons are pulled closer → atomic radius decreases.

2. Lanthanoid Series (4f orbitals)

• In lanthanoids, electrons are added to 4f orbitals.
• 4f orbitals are relatively more contracted and lie closer to the nucleus.
• They provide moderate shielding of nuclear charge.
• As a result, there is a steady lanthanoid contraction across the series.

3. Actinoid Series (5f orbitals)

• In actinoids, electrons are added to 5f orbitals.
• 5f orbitals are more diffuse and extend further into space.
• They experience stronger penetration from outer electrons and thus provide very poor shielding.
• Because of this poor shielding, the increasing nuclear charge across the series is felt more strongly by the electrons.
• This leads to a much stronger contraction in atomic size across the actinoid series compared to the lanthanoid series.

4. Why the Contraction is Greater in Actinoids

• Poor shielding by 5f orbitals → effective nuclear charge increases significantly.
• Stronger nuclear pull → greater contraction in radius.
• This is why the size decrease in actinoids is larger and more gradual.

✔ Final Point: The key reason is that 5f orbitals have poor shielding effect, so the nuclear charge contracts the orbitals more effectively.

Correct Option: (d) 5f orbitals have poor shielding effect

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NEET 2022 – Periodic Table & Classification of Elements (Topic – First and Second Ionization Enthalpy)

NEET Q.8 : If first ionization enthalpies of elements X and Y are 419 kJ/mol and 590 kJ/mol, respectively, and second ionization enthalpies of X and Y are 3069 kJ/mol and 1145 kJ/mol, respectively, then which statement is correct?
(A) Both X and Y are alkaline earth metals
(B) X is an alkali metal and Y is an alkaline earth metal
(C) X is an alkaline earth metal and Y is an alkali metal
(D) Both X and Y are alkali metals [NEET 2022 Phase-2]

Answer: (B) X is an alkali metal and Y is an alkaline earth metal

Detailed explanation (step-by-step)

1. What ionization enthalpies tell us
   • $IE_1$ = energy to remove the first (most weakly bound) electron.
   • $IE_2$ = energy to remove the second electron from the resulting cation.
   • A very large jump in ionization energy between $IE_1$ and $IE_2$ means that after removing the first electron you have reached a stable noble-gas core (i.e. the next electron to remove comes from a much more tightly bound inner shell). That is characteristic of group-1 (alkali) elements, which have one valence electron.
   • If $IE_1$ and $IE_2$ are both moderate (no huge jump until after $IE_2$), the atom likely had two valence electrons (characteristic of group-2, alkaline earth metals).
2. Compute the jumps
   • For X: $$\Delta = IE_2 – IE_1 = 3069 – 419 = 2650\ \text{kJ mol}^{-1}$$
     This is an enormous jump — indicates that after removing one electron we have a noble-gas configuration and removing the next electron requires pulling from a core shell. This is the classic signature of an alkali metal (one valence electron).
   • For Y: $$\Delta = IE_2 – IE_1 = 1145 – 590 = 555\ \text{kJ mol}^{-1}$$
     This is a moderate increase. Both $IE_1$ and $IE_2$ are of the same order of magnitude, showing that the second electron removed is still a valence electron (i.e. the atom originally had two valence electrons). That fits an alkaline earth metal.
3. Compare to typical/known values (for intuition)
   • Potassium (K, a group-1 element) has $IE_1\approx 419\ \text{kJ mol}^{-1}$ and $IE_2\approx 3052\ \text{kJ mol}^{-1}$ — nearly identical to X. This reinforces that X behaves like an alkali metal (loses one electron easily; second removal is from the noble-gas core).
   • Calcium (Ca, a group-2 element) has $IE_1\approx 590\ \text{kJ mol}^{-1}$ and $IE_2\approx 1145\ \text{kJ mol}^{-1}$ — essentially identical to Y. This matches the behaviour of an alkaline earth metal (two valence electrons removed with moderate energies; a large jump occurs on removing the third).
4. Electronic configuration viewpoint
   • Alkali metal (group-1): general valence configuration $ns^1$. After losing one $ns^1$ electron it attains a noble-gas core; removing a second electron would require breaking into that core (very high $IE_2$).
   • Alkaline earth (group-2): general valence $ns^2$. Removing one electron gives $ns^1$, removing the second still removes a valence electron; the big jump appears after $IE_2$ (to remove a core electron).
5. Conclusion
   • X shows the pattern of an alkali metal (very low $IE_1$, extremely large $IE_2$ jump).
   • Y shows the pattern of an alkaline earth metal (moderate $IE_1$ and $IE_2$, no huge jump until after $IE_2$).

Therefore choice (B) X is an alkali metal and Y is an alkaline earth metal is correct.

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NEET 2022 – Periodic Table & Classification of Elements (Topic – IUPAC Names of Elements of Atomic Number > 100)

NEET Q.9 : The IUPAC name of an element with atomic number 119 is:
(A) Ununennium
(B) Unnilennium
(C) Unununnium
(D) Ununoctium

Answer : (A) Ununennium

Explanation

IUPAC gives temporary systematic names to elements beyond those discovered, based on their atomic numbers. The rules are:

• Each digit of the atomic number is replaced with a Latin/Greek prefix:
• 1 → un
• 2 → bi
• 3 → tri
• 4 → quad
• 5 → pent
• 6 → hex
• 7 → sept
• 8 → oct
• 9 → enn
• 0 → nil
• The element name is formed by combining these prefixes, followed by -ium.
• Redundant vowels are omitted for simplicity (for example, “ennn” is shortened to “enn”).

For Z = 119:

• Digits: 1 – 1 – 9 → un – un – enn → ununennium
• Simplified official name: Ununennium (Uue)

Thus, the correct answer is Ununennium (option A).

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NEET 2022 – Periodic Table & Classification of Elements (Topic – Successive Ionization Enthalpy)

NEET Q.10 : Gadolinium has a low value of third ionisation enthalpy because of:
A. small size
B. high exchange enthalpy
C. high electronegativity
D. high basic character [NEET 2022 Pahse-1]

Answer : (B) high exchange enthalpy — more precisely: large exchange stabilization of the half-filled $4f^7$ configuration.

Detailed explanation

1) Electronic configuration and the three ionisations

• Neutral Gd:
  $$\text{Gd}:\; [\text{Xe}]\,4f^{7}\,5d^{1}\,6s^{2}$$
• Remove first two electrons (these are the $6s$ electrons):
  $$\text{Gd}^{2+}:\; [\text{Xe}]\,4f^{7}\,5d^{1}$$
• Remove the third electron (this removes the $5d^{1}$ electron) and you get:
  $$\text{Gd}^{3+}:\; [\text{Xe}]\,4f^{7}$$

So the third ionisation produces a half-filled $4f^{7}$ subshell.

2) What is exchange energy (exchange stabilization)?

• Exchange energy is a stabilization that arises when electrons occupy degenerate orbitals with parallel spins (Hund’s rule).
• The more parallel-spin electrons you have in degenerate orbitals, the larger the exchange stabilization.
• A half-filled subshell (like $4f^{7}$) gives a maximum exchange stabilization for the $f$ set.

3) Why this makes IE₃ unusually low

• Ionisation enthalpy $IE_3$ is the energy required to go from Gd²⁺ → Gd³⁺ + e⁻.
• The product Gd³⁺ has the specially stable half-filled $4f^7$ configuration. Because the final ion is strongly stabilized (lower in energy) by exchange effects, the energy gap that must be supplied to remove the third electron is smaller than expected.
• In simple terms: removing the $5d$ electron creates a particularly stable ion (Gd³⁺), so the required energy for that removal (IE₃) is reduced.

(Analogy: consider O vs N — O (2p⁴) has a lower first IE than N (2p³) because removing an electron from O yields the stable half-filled 2p³. The same “stabilized product lowers the required IE” idea applies.)

4) Why the other options are incorrect

• A (small size): Smaller size normally increases ionisation enthalpy (electrons are held more tightly), so small size would not explain a decrease in IE₃.
• C (high electronegativity): Higher electronegativity means stronger attraction for electrons → higher ionisation enthalpy, not lower.
• D (high basic character): Basic character describes oxide/basicity trends, not directly the cause of an unusually low third IE here.

5) Extra points (useful context)

• This is a classic f-block anomaly: half-filled (or completely filled) $f$ subshells impart extra stability via exchange energy, producing irregularities in successive ionisation energies.
• In the periodic table, many lanthanoids/actinoids show similar small/large jumps at ionizations that create/destroy half-filled $f$ subshells.

Conclusion

Gd has a low third ionisation enthalpy because formation of Gd³⁺ yields a half-filled $4f^{7}$ subshell which is strongly stabilized by exchange energy.

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