Anand Classes provides detailed NEET PYQs Chapterwise Solutions for Periodic Table Properties to help students strengthen their preparation for competitive exams. With step-by-step explanations and accurate answers, these solutions make it easier to understand the concepts and improve problem-solving skills. Whether you are revising important topics or practicing previous year questions, this resource will boost your confidence for NEET Chemistry. Click the print button to download study material and notes.
NEET 2018 – Periodic Table & Classification of Elements
NEET Q.1 : The correct order of atomic radii in group 13 elements is:
(a) B < Ga < Al < Tl < In
(b) B < Al < Ga < In < Tl
(c) B < Al < In < Ga < Tl
(d) B < Ga < Al < In < Tl [NEET 2018]
Answer: (d)
Explanation
The atomic radii as well as ionic radii increase on moving down the group 13 elements because of the successive addition of one extra shell of electrons.
However, there is an anomaly at Gallium (Ga) in case of atomic radii. The atomic radius of Ga is smaller than that of Al.
Gallium (Ga) has the electronic configuration:
$$ \text{Ga}: [Ar] \, 3d^{10} \, 4s^{2} \, 4p^{1} $$
The extra $d$-electrons do not shield the nucleus effectively. Consequently, the valence electrons of Ga experience greater attraction from the nucleus, leading to a smaller atomic radius.
Correct Order of Atomic Radii (Group 13 Elements)
$$ \text{B (85 pm)} < \text{Ga (135 pm)} < \text{Al (143 pm)} < \text{In (167 pm)} < \text{Tl (170 pm)} $$
Thus, the increasing order is:
$$ \text{B} < \text{Ga} < \text{Al} < \text{In} < \text{Tl} $$
✔ Final Answer: (d)
NEET 2016 (Phase I) – Periodic Table & Properties of Classification of Elements
NEET Q.2 : In which of the following options the order of arrangement does not agree with the variation of property indicated against it?
(a) B < C < N < O (increasing first ionisation enthalpy)
(b) I < Br < Cl < F (increasing electron gain enthalpy)
(c) Li < Na < K < Rb (increasing metallic radius)
(d) $ \text{Al}^{3+} < \text{Mg}^{2+} < \text{Na}^{+} < \text{F}^{-} $ (increasing ionic size)
NEET 2016 (Phase I)
Answer: (a), (b)
Explanation
Option (a) – First Ionisation Enthalpy
First ionisation enthalpy is the energy required to remove an electron from the outermost shell.
The correct order is:
$$ \text{B} < \text{C} < \text{O} < \text{N} $$
Hence, the given order B < C < N < O is incorrect.
Option (b) – Electron Gain Enthalpy
Electron gain enthalpy is the energy released when an electron is added to the outermost shell.
The correct order is:
$$ \text{I} < \text{Br} < \text{F} < \text{Cl} $$
Hence, the given order I < Br < Cl < F is incorrect.
Option (c) – Metallic Radius
As we move down the group in alkali metals, the metallic radius increases:
$$ \text{Li} < \text{Na} < \text{K} < \text{Rb} $$
This order is correct.
Option (d) – Ionic Size
For isoelectronic species, as positive charge decreases (or negative charge increases), the ionic size increases.
$$ \text{Al}^{3+} < \text{Mg}^{2+} < \text{Na}^{+} < \text{F}^{-} $$
This order is correct.
✔ Final Answer: (a), (b)
AIPMT 2015 – Periodic Table & Classification of Elements PYQs MCQs
NEET Q.3 : The species Ar, $K^{+}$ and $Ca^{2+}$ contain the same number of electrons. In which order do their radii increase?
(a) Ar < $K^{+}$ < $Ca^{2+}$
(b) $Ca^{2+}$ < Ar < $K^{+}$
(c) $Ca^{2+}$ < $K^{+}$ < Ar
(d) $K^{+}$ < Ar < $Ca^{2+}$ [AIPMT 2015]
Answer: (c)
Explanation
Ar, $K^{+}$ and $Ca^{2+}$ are isoelectronic species because each has 18 electrons. Isoelectronic species differ only in the number of protons present in the nucleus. The greater the nuclear charge, the more strongly the nucleus attracts the electrons, and the smaller the atomic or ionic radius becomes.
$Ca^{2+}$ has 20 protons but only 18 electrons. The strong nuclear pull contracts the electron cloud, giving the smallest radius.
$K^{+}$ has 19 protons and 18 electrons. Its nuclear pull is slightly weaker than $Ca^{2+}$, so its radius is larger than $Ca^{2+}$ but still smaller than Ar.
Ar is a neutral atom with 18 protons and 18 electrons. The pull of the nucleus is balanced by the electron repulsions, so its radius is the largest among the three.
Therefore, the correct increasing order of radii is
$$ Ca^{2+} < K^{+} < Ar $$
✔ Final Answer: (c)
AIPMT 2014 – Periodic Table & Classification of Elements
NEET Q.4 : Which of the following orders of ionic radii is correctly represented?
(a) $H^{-} > H^{+} > H$
(b) $Na^{+} > F^{-} > O^{2-}$
(c) $F^{-} > O^{2-} > Na^{+}$
(d) $Al^{3+} > Mg^{2+} > N^{3-}$ [AIPMT 2014]
Answer: (*) No option is correct
Explanation
(a) For hydrogen species:
It is known that the radius of a cation is always smaller than that of the neutral atom due to the decrease in the number of orbits. The radius of an anion is always greater than the neutral atom due to the decrease in effective nuclear charge.
Hence, the correct order is
$$ H^{-} > H > H^{+} $$
(b) For $Na^{+}$, $F^{-}$, and $O^{2-}$:
These species are isoelectronic as they all contain 10 electrons. For isoelectronic species,
$$ \text{Ionic radius} \propto \frac{1}{\text{atomic number}} $$
Ion: $Na^{+},\ F^{-},\ O^{2-}$
Atomic number: 11, 9, 8
Therefore, the correct order is
$$ O^{2-} > F^{-} > Na^{+} $$
(c) Similarly, the correct order is
$$ O^{2-} > F^{-} > Na^{+} $$
(d) For $Al^{3+}$, $Mg^{2+}$, and $N^{3-}$:
Atomic numbers are 13, 12, and 7 respectively. For isoelectronic species, the size decreases with increasing nuclear charge.
Therefore, the correct order is
$$ N^{3-} > Mg^{2+} > Al^{3+} $$
AIPMT 2012 – Periodic Table & Classification of Elements (isoelectronic species)
NEET Q.5 : Identify the wrong statement in the following.
(a) Amongst isoelectronic species, smaller the positive charge on the cation, smaller is the ionic radius
(b) Amongst isoelectronic species, greater the negative charge on the anion, larger is the ionic radius
(c) Atomic radius of the elements increases as one moves down the first group of the periodic table
(d) Atomic radius of the elements decreases as one moves across from left to right in the 2nd period of the periodic table [AIPMT 2012]
Answer: (a)
Explanation
Let us check each statement carefully:
(a) This is wrong. In isoelectronic species, the number of electrons is the same but the nuclear charge differs. A cation with a higher positive charge pulls its electrons more strongly, reducing its radius. Therefore, a smaller positive charge actually means a larger radius, not smaller. For example, compare $Na^{+}$ and $Mg^{2+}$ (both 10 electrons):
$$ Na^{+} > Mg^{2+} $$
Hence, statement (a) is incorrect.
(b) This is correct. For isoelectronic anions, more negative charge means more electron–electron repulsion and weaker nuclear pull, which increases the radius. For example:
$$ O^{2-} > F^{-} $$
(c) This is correct. As we move down the first group, new electron shells are added. The outermost electron is farther from the nucleus, so the atomic radius increases:
$$ Li < Na < K < Rb < Cs $$
(d) This is correct. As we move left to right in a period, protons are added to the nucleus while electrons are added to the same shell. The increased nuclear charge pulls electrons more strongly without much increase in shielding, so atomic radius decreases:
$$ Li > Be > B > C > N > O > F $$
Therefore, the wrong statement is (a).
AIPMT 2010 – Periodic Table & Classification of Elements (isoelectronic species)
NEET Q.6 : The correct order of the decreasing ionic radii among the following isoelectronic species is:
(a) $Ca^{2+} > K^{+} > S^{2-} > Cl^{-}$
(b) $Cl^{-} > S^{2-} > Ca^{2+} > K^{+}$
(c) $S^{2-} > Cl^{-} > K^{+} > Ca^{2+}$
(d) $K^{+} > Ca^{2+} > Cl^{-} > S^{2-}$
Answer: (c)
Explanation
Key Idea:
For isoelectronic species (same number of electrons), the size of the ion depends mainly on the nuclear charge.
- Cations: When a cation forms, electrons are lost from the outer shell. The same number of protons now pulls fewer electrons, so the effective nuclear attraction per electron increases. This pulls the electron cloud inward and decreases the radius. Thus, higher positive charge $\Rightarrow$ smaller radius.
For example:
$$ Ca^{2+} < K^{+} $$
- Anions: When an anion forms, extra electrons are added to the same outer shell. This increases electron–electron repulsion and reduces the effective pull of the nucleus on each electron, increasing the radius. Thus, higher negative charge $\Rightarrow$ larger radius.
For example:
$$ S^{2-} > Cl^{-} $$
Therefore, the correct decreasing order of ionic radii is:
$$ S^{2-} > Cl^{-} > K^{+} > Ca^{2+} $$
✔ Final Answer: (c)
AIPMT 2010 – Periodic Table & Classification of Elements (isoelectronic species)
NEET Q.7 : Which of the following represents the correct order of increasing electron gain enthalpy with negative sign for the elements $O$, $S$, $F$ and $Cl$?
(a) $Cl < F < O < S$
(b) $O < S < F < Cl$
(c) $F < S < O < Cl$
(d) $S < O < Cl < F$
[CBSE AIPMT 2010]
Answer. (b)
Key Idea
Electron gain enthalpy generally increases in a period from left to right and decreases in a group on moving downwards. However, members of the III period have somewhat higher electron gain enthalpy as compared to the corresponding members of the second period, because of their small size.
Explanation
$O$ and $S$ belong to VI A (16) group and $Cl$ and $F$ belong to VII A (17) group. Thus, the electron gain enthalpy of $Cl$ and $F$ is higher as compared to $O$ and $S$.
$$ Cl, F > O, S $$
Between $Cl$ and $F$, $Cl$ has higher electron gain enthalpy than $F$, since the incoming electron experiences a greater force of repulsion because of the small size of $F$-atom.
Similarly, between $O$ and $S$, the electron gain enthalpy of $S$ is higher as compared to $O$ due to its small size.
Final Order
Thus, the correct order of electron gain enthalpy of the given elements is
$$ O < S < F < Cl $$
This concept of electron gain enthalpy trend in the periodic table, including group 16 and group 17 elements, is an important topic for class 11 chemistry and competitive exams like CBSE AIPMT and NEET. Students preparing for exams can also download pdf notes of electron gain enthalpy and related periodic trends for quick revision.
AIPMT 2009 – Periodic Table & Classification of Elements (Ionisation Energy Question)
NEET Q.8 : Amongst the elements with the following electronic configurations, which one may have the highest ionisation energy?
(a) $[Ne] , 3s^2 3p^3$
(b) $[Ne] , 3s^2 3p^2$
(c) $[Ar] , 3d^{10} 4s^2 4p^3$
(d) $[Ne] , 3s^2 3p^1$
[CBSE AIPMT 2009]
Answer: (a)
Key Idea
- Across a period, increasing nuclear charge outweighs the shielding effect.
Hence, the outermost electrons are held more tightly, and ionisation energy increases across a period. - Down a group, the increase in shielding outweighs the increasing nuclear charge.
Thus, the removal of the outermost electron requires less energy, so ionisation energy decreases down a group.
Electronic Configurations and Groups
- $[Ne] , 3s^2 3p^3$ → Group V
- $[Ne] , 3s^2 3p^2$ → Group IV
- $[Ar] , 3d^{10} 4s^2 4p^3$ → Group V
- $[Ne] , 3s^2 3p^1$ → Group III
Final Explanation
Since ionisation energy increases across a period and decreases down a group, the configuration
$[Ne] \, 3s^2 3p^3$
has the highest ionisation energy among the given elements. ✅
AIPMT 2009 – Periodic Table & Classification of Elements (Oxides and Their Reaction with NaOH)
NEET Q.9 : Which of the following oxides is not expected to react with sodium hydroxide?
(a) $B_2O_3$
(b) $CaO$
(c) $SiO_2$
(d) $BaO$
[CBSE AIPMT 2009]
Answer: (b)
Explanation
Sodium hydroxide ($NaOH$), being a strong alkali, never reacts with a basic oxide.
- $B_2O_3$ and BeO are amphoteric oxides.
- $SiO_2$ is an acidic oxide.
- $CaO$ is a basic oxide.
Therefore, $NaOH$ does not react with $CaO$.
Oxides and NaOH reaction notes PDF download for NEET study material
AIPMT 2008 – Periodic Table & Classification of Elements (Second Ionisation Enthalpy of Transition Elements)
NEET Q.10 : The correct order of decreasing second ionisation enthalpy of $Ti(22)$, $V(23)$, $Cr(24)$ and $Mn(25)$ is
(a) $Cr > Mn > V > Ti$
(b) $V > Mn > Cr > Ti$
(c) $Mn > Cr > Ti > V$
(d) $Ti > V > Cr > Mn$
[CBSE AIPMT 2008]
Answer: (a)
Explanation
The amount of energy required to remove an electron from a unipositive ion is referred to as second ionisation enthalpy (or potential).
For $Ti, V, Cr,$ and $Mn$:
- Generally, the second ionisation energy increases with an increase in atomic number.
- However, the second ionisation potential of $Cr$ is greater than Mn due to the presence of a half-filled d-subshell in $Cr$.
Thus, the correct order of second ionisation enthalpy is:
$$Cr > Mn > V > Ti$$
Second ionisation enthalpy notes PDF download for NEET chemistry study material
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