NEET PYQs Previous Year Questions MCQs on Classification of Elements and Periodicity in Properties help students understand important concepts frequently asked in medical entrance examinations. These questions cover periodic trends, atomic radius, ionization enthalpy, electron gain enthalpy, modern periodic table, electronic configuration, and periodic classification of elements. Practicing previous year questions improves conceptual clarity, accuracy, and exam confidence.
TOPIC 1 : Periodic Table and Classification of Elements
From the following pairs of ions which one is not an iso-electronic pair?
(a) $\ce{O^{2-}}$, $\ce{F-}$
(b) $\ce{Na+}$, $\ce{Mg^{2+}}$
(c) $\ce{Mn^{2+}}$, $\ce{Fe^{3+}}$
(d) $\ce{Fe^{2+}}$, $\ce{Mn^{2+}}$
[NEET 2021]
Ans. (d)
| Ion | Number of electrons |
|---|---|
| $\ce{O^{2-}}$ | 10 |
| $\ce{F-}$ | 10 |
| $\ce{Na+}$ | 10 |
| $\ce{Mg^{2+}}$ | 10 |
| $\ce{Mn^{2+}}$ | 23 |
| $\ce{Fe^{3+}}$ | 23 |
| $\ce{Fe^{2+}}$ | 24 |
| $\ce{Mn^{2+}}$ | 23 |
$\ce{O^{2-}}$ and $\ce{F-}$ are iso-electronic pair.
$\ce{Na+}$ and $\ce{Mg^{2+}}$ are iso-electronic pair.
$\ce{Mn^{2+}}$ and $\ce{Fe^{3+}}$ are iso-electronic pair.
$\ce{Fe^{2+}}$ and $\ce{Mn^{2+}}$ are not iso-electronic pair.
Identify the incorrect match.
| Name | IUPAC official name |
|---|---|
| (A) Unnilunium | (i) Mendelevium |
| (B) Unniltrium | (ii) Lawrencium |
| (C) Unnihexium | (iii) Seaborgium |
| (D) Unununnium | (iv) Darmstadtium |
(a) (B), (ii)
(b) (C), (iii)
(c) (D), (iv)
(d) (A), (i)
[NEET (Sep.) 2020]
Ans. (c)
| Name | Atomic number (Z) | IUPAC official name |
|---|---|---|
| (A) Unnilunium | 101 | Mendelevium (Md) |
| (B) Unniltrium | 103 | Lawrencium (Lr) |
| (C) Unnihexium | 106 | Seaborgium (Sg) |
| (D) Unununnium | 111 | Roentgenium (Rg) [but (iv) Darmstadtium (Ds), Given] [Z = 110] |
So, D-(iv) is the incorrect match.
The element Z = 114 has been discovered recently. It will belong to which of the following family/group and electronic configuration?
(a) Halogen family, $\ce{[Rn] 5f^{14} 6d^{10} 7s^{2} 7p^{5}}$
(b) Carbon family, $\ce{[Rn] 5f^{14} 6d^{10} 7s^{2} 7p^{2}}$
(c) Oxygen family, $\ce{[Rn] 5f^{14} 6d^{10} 7s^{2} 7p^{4}}$
(d) Nitrogen family, $\ce{[Rn] 5f^{14} 6d^{10} 7s^{2} 7p^{6}}$
[NEET 2017]
Ans. (b)
The element with atomic number, Z = 114 is flerovium (Fl). It is a super heavy artificial chemical element. In the periodic table of the elements, it is a transactinide element in the p-block. It is a member of the 7th period and is the heaviest known member of the carbon family.
Electronic configuration for Z = 114 is
$\ce{[Rn]_{86} 5f^{14} 6d^{10} 7s^{2} 7p^{2}}$
Which one of the elements with the following outer orbital configurations may exhibit the largest number of oxidation states?
(a) $\ce{3d^{3} 4s^{2}}$
(b) $\ce{3d^{5} 4s^{1}}$
(c) $\ce{3d^{5} 4s^{2}}$
(d) $\ce{3d^{2} 4s^{2}}$
[CBSE AIPMT 2009]
Ans. (c)
The sum of number of electrons (unpaired) in d-orbitals and number of electrons in s-orbital gives the number of oxidation states (os) exhibited by a d-block element. Therefore,
(a) $\ce{3d^{3} 4s^{2}}$ $\Rightarrow$ OS = $3 + 2 = 5$
(b) $\ce{3d^{5} 4s^{1}}$ $\Rightarrow$ OS = $5 + 1 = 6$
(c) $\ce{3d^{5} 4s^{2}}$ $\Rightarrow$ OS = $5 + 2 = 7$
(d) $\ce{3d^{2} 4s^{2}}$ $\Rightarrow$ OS = $2 + 2 = 4$
Hence, element with $\ce{3d^{5} 4s^{2}}$ configuration exhibits largest number of oxidation states.
An atom has electronic configuration $\ce{1s^{2} 2s^{2} 2p^{6} 3s^{2} 3p^{6} 3d^{2} 4s^{2}}$, you will place it in
(a) fifth group
(b) fifteenth group
(c) second group
(d) third group
[CBSE AIPMT 2002]
Ans. (a)
The amount of energy required to remove an electron from the outermost orbit of a gaseous atom is known as ionisation potential. Elements having half-filled or completely filled orbitals are more stable than partially filled orbitals.
In a period from left to right ionisation potential increases as the atomic number increases. The given elements (Be, B, C, N, O) are present in II period and their Ionisation potential increases as Be < B < C < N < O
But in case of Be and B, Be has higher ionisation potential due to stable configuration.
Be (Z = 4) : 1s² 2s² (stable)
B (Z = 5) : 1s² 2s² 2p¹ (unstable)
In the same way in case of N and O, N has higher ionisation potential than O due to stable configuration (half-filled).
N (₇N) = 1s² 2s² 2p³ (stable)
O (₈O) = 1s² 2s² 2p⁴ (unstable)
So, the correct order of increasing ionisation potential will be
B < Be < C < O < N
The element with the atomic number 118, will be
(a) alkali
(b) noble gas
(c) lanthanide
(d) transition element
[CBSE AIPMT 1996]
Ans. (b)
The outermost electronic configuration of element with atomic number 118 is 7s² 7p⁶, so it will be a noble gas.
The electronic configuration of an element is 1s² 2s² 2p⁶ 3s² 3p³. What is the atomic number of the element, which is present just below the above element in the periodic table?
(a) 33
(b) 34
(c) 36
(d) 49
[CBSE AIPMT 1995]
Ans. (a)
The element which is present just below the given element will have outermost electronic configuration as 4s² 4p³, so its full electronic configuration is 1s² 2s² 2p⁶ 3s² 3p⁶ 4s² 3d¹⁰ 4p³ and hence, its atomic number is 33.
If the atomic number of an element is 33, it will be placed in the periodic table in the
(a) first group
(b) third group
(c) fifteen group
(d) seventh group
[CBSE AIPMT 1993]
Ans. (c)
The electronic configuration of element with atomic number 33 is 1s² 2s² 2p⁶ 3s² 3p⁶ 4s² 3d¹⁰ 4p³. As its last shell has five electrons, its group is 10 + 5 = 15ᵗʰ or V A group.
The electronic configuration of four elements are given below. Which element does not belong to the same family as others?
(a) [Xe] 4f¹⁴ 5d¹⁰ 6s²
(b) [Kr] 4d¹⁰ 5s²
(c) [Ne] 3s² 3p⁵
(d) [Ar] 3d¹⁰ 4s²
[CBSE AIPMT 1989]
Ans. (c)
In a family, all elements have same outermost electronic configuration. Since [Ne] 3s² 3p⁵ (chlorine) belongs to halogen family while the remaining three are in the same group i.e. group 12.
Hg (₈₀Hg) = [Xe] 4f¹⁴ 5d¹⁰ 6s²
Cd (₄₈Cd) = [Kr] 4d¹⁰ 5s²
Zn (₃₀Zn) = [Ar] 3d¹⁰ 4s²
TOPIC 2 – Periodic Properties
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]
Ans. (d)
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. Atomic radius of Ga is lesser as compared to Al.
Gallium (Ga) with electronic configuration [Ar] 3d¹⁰ 4s² 4p¹ has extra d-electrons which do not screen the nucleus effectively. Consequently, electrons of Ga are more attracted by the nucleus.
Thus, the increasing order of atomic radii of the group 13 elements is
B (85 pm) < Ga (135 pm) < Al (143 pm) < In (167 pm) < Tl (170 pm).
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) Al³⁺ < Mg²⁺ < Na⁺ < F⁻ (increasing ionic size)
[NEET 2016, Phase I]
Ans. (a, b)
For option (a) : First ionisation energy is the energy required to remove an electron from outermost shell.
Hence, correct order is B < C < O < N.
For option (b) : Electron gain enthalpy is the energy required to gain an electron in the outermost shell.
Hence, the correct order is I < Br < F < Cl.
For option (c) : As we move down the group in alkali metal, metallic radius increases as Li < Na < K < Rb.
For option (d) : In case of isoelectronic species, as positive charge decreases or negative charge increases the ionic size of the species increases and vice-versa i.e.
Al³⁺ < Mg²⁺ < Na⁺ < F⁻
The species Ar, K⁺ and Ca²⁺ contain the same number of electrons. In which order do their radii increase?
(a) Ar < K⁺ < Ca²⁺
(b) Ca²⁺ < Ar < K⁺
(c) Ca²⁺ < K⁺ < Ar
(d) K⁺ < Ar < Ca²⁺
[CBSE AIPMT 2015]
Ans. (c)
Ar, K⁺ and Ca²⁺ are isoelectronic (18 electrons). For isoelectronic species, ionic radii decrease with increase in effective nuclear charge. Also Ar, K and Ca belong to the same period (3rd period). Thus their ionic radii as follows :-
Ca²⁺ < K⁺ < Ar
Which of the following orders of ionic radii is correctly represented?
(a) H⁻ > H > H⁺
(b) Na⁺ > F⁻ > O²⁻
(c) F⁻ > O²⁻ > Na⁺
(d) Al³⁺ > Mg²⁺ > N³⁻
[CBSE AIPMT 2014]
Ans. (a)
(a) H⁻ > H > H⁺
It is known that radius of a cation is always smaller than that of a neutral atom due to decrease in the number of orbits. Whereas, the radius of anion is always greater than a cation due to decrease in effective nuclear charge.
Hence, the correct order is H⁻ > H > H⁺
(b) Na⁺ > F⁻ > O²⁻
The given species are isoelectronic (same number of electrons = 10). For isoelectronic species,
Ionic radii ∝ 1 / atomic number
| Ion | Na⁺ | F⁻ | O²⁻ |
|---|---|---|---|
| Atomic number (Nuclear Charge) | 11 | 9 | 8 |
Hence, the correct order of ionic radii is O²⁻ > F⁻ > Na⁺
(c) Similarly, the correct order is O²⁻ > F⁻ > Na⁺
(d) Ions : Al³⁺, Mg²⁺, N³⁻
The given species are isoelectronic (same number of electrons = 10). For isoelectronic species,
Ionic radii ∝ 1 / atomic number
| Ion | Al³⁺ | Mg²⁺ | N³⁻ |
|---|---|---|---|
| Atomic number (Nuclear Charge) | 13 | 12 | 7 |
Hence, the correct order is N³⁻ > Mg²⁺ > Al³⁺
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.
[CBSE AIPMT 2012]
Ans. (a)
Atomic radius of the elements decreases across a period from left to right due to increase in effective nuclear charge. On moving down a group, since number of shells increases, atomic radius increases.
Amongst isoelectronic species, ionic radius increases with increase in negative charge or decrease in positive charge.
The correct order of the decreasing ionic radii among the following isoelectronic species is
(a) Ca²⁺ > K⁺ > S²⁻ > Cl⁻
(b) Cl⁻ > S²⁻ > Ca²⁺ > K⁺
(c) S²⁻ > Cl⁻ > K⁺ > Ca²⁺
(d) K⁺ > Ca²⁺ > Cl⁻ > S²⁻
[CBSE AIPMT 2010]
Ans. (c)
Ionic radii ∝ negative charge on anion
∝ 1 / positive charge on cation
During the formation of a cation, the electrons are lost from the outer shell and the remaining electrons experience a greater force of attraction by the nucleus i.e. attracted more towards the nucleus. In other words, nucleus hold the remaining electrons more tightly and this results in decreased radii.
In case of anion formation, addition of electron(s) takes place in the same outer shell, thus the hold of nucleus on the electrons decreases and this results in increased ionic radii.
Thus, the correct order of ionic radii is
S²⁻ > Cl⁻ > K⁺ > Ca²⁺
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]
Ans. (b)
Electron gain enthalpy generally increases in a period from left to right and decreases in a group on moving downwards. However, members of third period have some what higher electron gain enthalpy as compared to the corresponding members of second period because of their larger size.
O and S belong to group 16, Cl and F belong to group 17. Thus, electron gain enthalpy of Cl and F is higher than O and S.
Between Cl and F, Cl has higher electron gain enthalpy then the F, since the incoming electron experiences a greater force of repulsion because of small size of F-atom. Thus electron gain enthalpy order is Cl > F
Similar is true in case of O and S, i.e. the electron gain enthalpy of S is higher as compared to O due to its small size. Thus electron gain enthalpy order is S > O.
Thus, the correct order of electron gain enthalpy of given elements is
O < S < F < Cl
Amongst the elements with following electronic configurations, which one may have the highest ionisation energy?
(a) [Ne] 3s² 3p³
(b) [Ne] 3s² 3p²
(c) [Ar] 3d¹⁰ 4s² 4p³
(d) [Ne] 3s² 3p¹
[CBSE AIPMT 2009]
Ans. (a)
Across a period, increasing nuclear charge outweighs the shielding, hence the outermost electrons are held more and more tightly and ionisation energy increases across a period while as we move down a group increase in shielding outweighs the increasing nuclear charge and the removal of the outermost electron required less energy down a group.
| Electronic configuration | Group |
|---|---|
| [Ne] 3s² 3p³ | 15 |
| [Ne] 3s² 3p² | 14 |
| [Ar] 3d¹⁰ 4s² 4p³ | 15 |
| [Ne] 3s² 3p¹ | 13 |
Since, ionisation energy increases in a period and decreases in a group, [Ne] 3s² 3p³ (half-filled, stable) configuration has the highest ionisation energy among the
given elements.
Which of the following oxides is not expected to react with sodium hydroxide?
(a) B₂O₃
(b) CaO
(c) SiO₂
(d) BeO
[CBSE AIPMT 2009]
Ans. (b)
Sodium hydroxide, NaOH, being a strong alkali never react with a basic oxide (compound). Among the given options, B2O3 and BeO are amphoteric oxides, SiO2 is an acidic oxide and CaO is a basic oxide. Therefore, NaOH does not react with CaO.
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]
Ans. (a)
Second ionisation enthalpy is the energy required to remove an electron from unipositive ion. In Ti, V, Cr and Mn, generally second ionisation energy increases with increase in atomic number but second ionisation potential of Cr is greater than that of Mn due to the presence of exactly half-filled d-subshell in Cr+. Cr has very high second IE due to stable d⁵ configuration after removal of first electron from 4s. Thus, the order of second ionisation enthalpy is
Cr > Mn > V > Ti
Which of the following electronic configuration of an atom has the lowest ionisation enthalpy?
(a) 1s² 2s² 2p⁵
(b) 1s² 2s² 2p³
(c) 1s² 2s² 2p⁶ 3s¹
(d) 1s² 2s² 2p⁶
[CBSE AIPMT 2007]
Ans. (c)
The electronic configuration 1s² 2s² 2p⁶ 3s¹ shows the lowest ionisation energy because this configuration belongs to third period and all others are belong to second period of periodic table. Also 1s² 2s² 2p⁶ 3s¹ belongs to element sodium, which is a alkali metal. Alkali metals are largest in size in the periodic table. Hence, less energy is required to remove the electron.
Identify the correct order of the size of the following.
(a) Ca²⁺ < K⁺ < Ar < S²⁻ < Cl⁻
(b) Ca²⁺ < K⁺ < Ar < Cl⁻ < S²⁻
(c) Ar < Ca²⁺ < K⁺ < Cl⁻ < S²⁻
(d) Ca²⁺ < Ar < K⁺ < Cl⁻ < S²⁻
[CBSE AIPMT 2007]
Ans. (b)
A cation has always lesser ionic size than the parent metal atom due to loss of electrons. An anion has always greater size than the parent atom due to gain of electrons. The given species are isoelectronic species as they contain same number of electrons (= 18).
For isoelectronic species : Ionic radii ∝ 1/atomic number
| Ion | Ca²⁺ | K⁺ | Ar | Cl⁻ | S²⁻ |
|---|---|---|---|---|---|
| Atomic number | 20 | 19 | 18 | 17 | 16 |
So, the correct order of size is :
Ca²⁺ < K⁺ < Ar < Cl⁻ < S²⁻
Ionic radii are
(a) inversely proportional to effective nuclear charge
(b) inversely proportional to square of effective nuclear charge
(c) directly proportional to effective nuclear charge
(d) directly proportional to square of effective nuclear charge
[CBSE AIPMT 2004]
Ans. (a)
Ionic radii ∝ 1/Zeff
where Zeff = effective nuclear charge = Z − σ (σ = screening constant).
This value of screening constant is based upon the number of electrons in valence shell as well as in penultimate shells.
The ions O²⁻, F⁻, Na⁺, Mg²⁺ and Al³⁺ are isoelectronic. Their ionic radii show
(a) an increase from O²⁻ to F⁻ and then decrease from Na⁺ to Al³⁺
(b) a decrease from O²⁻ to F⁻ and then increase from Na⁺ to Al³⁺
(c) a significant increase from O²⁻ to Al³⁺
(d) a significant decrease from O²⁻ to Al³⁺
[CBSE AIPMT 2003]
Ans. (d)
On increasing atomic number of isoelectronic species, ionic radii decrease due to increasing effective nuclear charge (Zeff).
Radius ∝ 1/atomic number ∝ 1/Zeff
So, as the negative charge increases ionic radii increases while on increasing positive charge ionic radii decreases. Anions having higher ionic radii than the cation. Hence, order of radii :
O²⁻ > F⁻ > Na⁺ > Mg²⁺ > Al³⁺
Which of the following order is wrong?
(a) NH₃ < PH₃ < AsH₃ – Acidic
(b) Li < Be < B < C – 1st Ionisation potential
(c) Al₂O₃ < MgO < Na₂O < K₂O – Basic
(d) Li⁺ < Na⁺ < K⁺ < Cs⁺ – Ionic radius
[CBSE AIPMT 2002]
Ans. (b)
Li, Be, B and C are present in the 2nd period. In a period, ionisation potential generally increases from left to right. But in case of Be and B, Be has higher ionisation potential than B due to stable fully-filled 2s² configuration of Be. So, the correct order of ionisation potential of given elements is :
Li < B < Be < C
Correct order of 1st ionisation potential (IP) among following elements Be, B, C, N, O is
(a) B < Be < C < O < N
(b) B < Be < C < N < O
(c) Be < B < C < N < O
(d) Be < B < C < O < N
[CBSE AIPMT 2001]
Ans. (a)
Ionisation potential generally increases from left to right across a period due to increasing effective nuclear charge. However, there are two important exceptions in the 2nd period :
Be has electronic configuration 1s² 2s² which is stable fully filled 2s orbital configuration. While B has 1s² 2s² 2p¹ configuration. It is easier to remove the 2p¹ electron from B than a 2s electron from Be. Hence Be has higher ionisation potential than B.
N has electronic configuration 1s² 2s² 2p³ which is extra stable half-filled p subshell configuration. While O has 1s² 2s² 2p⁴ configuration, which is not stable. Removing one electron from O gives half-filled 2p³ configuration, so it is easier to remove an electron from O than from N. Hence N has higher ionisation potential than O. So the correct order is :
B < Be < C < O < N
The first ionisation potential (in eV) of Be and B, respectively are
(a) 8.29, 9.32
(b) 9.32, 9.32
(c) 8.29, 8.29
(d) 9.32, 8.29
[CBSE AIPMT 1998]
Ans. (d)
First ionisation potential of Be (9.32 eV) is greater than B (8.29 eV) due to stable 2s² configuration in Be.
Be has electronic configuration 1s² 2s² (stable), while B has 1s² 2s² 2p¹ configuration. Order of attraction of electrons towards nucleus is 2s > 2p, so more amount of energy is required to remove the electron from 2s orbital in comparison to 2p orbital. So, ionisation potential of Be is 9.32 eV and B is 8.29 eV.
In crystals of which of the following ionic compounds would you expect maximum distance between centres of cations and anions?
(a) LiF
(b) CsF
(c) CsI
(d) LiI
[CBSE AIPMT 1998]
Ans. (c)
On moving from top to bottom in a group of the periodic table, the distance between ions in ionic compounds increases. Hence, it is maximum in CsI.
Which one of the following ions will be smallest in size?
(a) Na⁺
(b) Mg²⁺
(c) F⁻
d) O²⁻
[CBSE AIPMT 1996]
Ans. (b)
Na⁺, Mg²⁺, O²⁻ and F⁻ are isoelectronic, but Mg²⁺ has 12 protons in its nucleus.
For isoelectronic species : Ionic radii ∝ 1/atomic number
So the attraction force on the valence shell is maximum for Mg²⁺ and hence it has the smallest size.
Among the following, the one which is most basic is
(a) ZnO
(b) MgO
(c) Al₂O₃
(d) N₂O₅
[CBSE AIPMT 1994]
Ans. (b)
ZnO and Al₂O₃ are amphoteric oxides, N₂O₅ is an oxide of a non-metal so it is acidic. Hence, MgO is the most basic among the given oxides.
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 3s²
[CBSE AIPMT 1993]
Ans. (d)
The element with configuration 1s² 2s² 2p⁶ 3s² after losing two electrons acquires the stable noble gas (Ne) configuration. Removing the third electron requires a large amount of energy. Hence, there is a large difference between second and third ionisation energies.
In the periodic table from left to right in a period, the atomic volume
(a) decreases
(b) increases
(c) remains same
(d) first decreases then increases
[CBSE AIPMT 1993]
Ans. (d)
In the periodic table the atomic size first decreases from left to right in period, so the atomic volume first decreases and then increases because atomic size in last of any period (noble gases) increases (van der Waals radius).
One of the characteristic properties of non-metals is that they
(a) are reducing agents
(b) form basic oxides
(c) form cations by electron gain
(d) are electronegative
[CBSE AIPMT 1993]
Ans. (d)
Non-metals easily gain electrons to form negative ions, so they are electronegative in nature.
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]
Ans. (c)
In isoelectronic species the number of electrons are same but nuclear charge is different. As the nuclear charge increase, the attraction force on last electron increases, so the size decreases or in other words
Ionic Size ∝ 1/Charge on cation
Hence correct order is Na⁺ > Mg²⁺ > Al³⁺ > Si⁴⁺
One would expect proton to have very large
(a) charge
(b) ionisation potential
(c) hydration energy
(d) radius
[CBSE AIPMT 1993]
Ans. (c)
Proton (H⁺) has very small size, so it has large hydration energy. The degree of hydration depends upon the size of the cation. Smaller the size of a cation greater the hydration energy.
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]
Ans. (c)
N, O, F are highly electronegative elements, so they have the strongest tendency to accept electrons and form anions.
The ionisation of hydrogen atom would give rise to
(a) hydride ion
(b) hydronium ion
(c) proton
(d) hydroxyl ion
[CBSE AIPMT 1990]
Ans. (c)
Hydrogen have one proton and one electron, when it ionise, i.e. it lose one electron, then only proton is left in the nucleus, so H+ ion is formed during ionisation which is also called proton.
H → H⁺ + e⁻
(H⁺ is also called a proton).
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]
Ans. (a)
In periodic table, the metallic character increases down the group because the ionisation enthalpy decreases down the group and metallic character decreases from left to right because the ionisation enthalpy increases from left to right.
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]
Ans. (a)
Pauling’s electronegativity values are useful in determination of polarity of the bond in molecules. If electronegativity difference is zero, then the molecule is non-polar otherwise it is polar.
Chapter Important Interlinks
This section on NEET PYQs Previous Year Questions MCQs: Classification of Elements and Periodicity in Properties provides a comprehensive collection of important previous year questions with detailed solutions and explanations. It covers concepts such as atomic radius, ionic radius, ionization enthalpy, electron gain enthalpy, shielding effect, effective nuclear charge, periodic trends, modern periodic table, and electronic configuration. Students can practice objective questions, assertion-reason problems, conceptual MCQs, and numerical questions designed according to the latest NEET syllabus. These practice sets are highly useful for strengthening concepts, improving speed and accuracy, and preparing effectively for NEET and other competitive examinations.
CHEMISTRY ANAND CLASSES : Complete Chemistry Learning Hub “Explore, Learn, And Excel In Chemistry”
Author : Er Neeraj Anand
Atomic Radii, Ionic Radii And Isoelectronic Ions Trends Conceptual Questions And Answers Of Periodic Table
Radius Of Cation Is Less And Anion Is More Than Its Parent Atom, Size Variation In Isoelectronic Series
Division Of Periodic Table Into s,p,d,f Blocks : Prediction Of Period, Group And Block Of Elements
Screening Effect (Shielding Effect) : Calculation Of Effective Or Reduced Nuclear Charge (Slater’s Rules)
Mendeleev’s Periodic Law And Table : Historical Development Of Periodic Table
Atomic Radius Trends : Variation Of Atomic Radius In A Period And Down The Group
Electron Gain Enthalpy : Definition, Units, Factors, Trends, Successive ΔegH
Ionization Enthalpy Trends Along A Period And Down A Group
Atomic Radius And Its Types : Covalent, Van Der Waal’s, Metallic, Ionic Radii
What Is Ionization Enthalpy? Definition, Units, Factors And Successive IE
Causes Of Periodicity : Why Do Elements Show Periodicity?
Modern Periodic Table Solved Examples, Conceptual Questions Answers, Practice Problems
Classification Of Elements And Periodicity In Properties Class 11 Chemistry Detail Theory And Notes
Modern Periodic Law And Table (Groups, Periods, Blocks) : Elements With Atomic Numbers More Than 100
Class 11 Chemistry Complete Study Material And Notes By Chemistry Anand Classes
FAQs – NEET PYQs Previous Year Questions MCQs: Classification of Elements and Periodicity in Properties
What are NEET PYQs in Classification of Elements and Periodicity in Properties?
NEET PYQs are Previous Year Questions asked in the NEET examination from the chapter Classification of Elements and Periodicity in Properties. These questions help students understand important concepts and exam patterns.
Why should I practice NEET PYQs for Classification of Elements and Periodicity in Properties?
Practicing NEET PYQs improves conceptual clarity, speed, accuracy, and understanding of frequently asked topics such as atomic radius, ionization enthalpy, electron gain enthalpy, and periodic trends.
Are detailed solutions provided for the MCQs?
Yes, detailed and step-by-step solutions are provided for all important NEET previous year questions to help students understand the correct approach and concepts.
Which topics are covered in these NEET PYQs and MCQs?
The questions cover modern periodic table, periodic trends, ionization enthalpy, electron gain enthalpy, shielding effect, effective nuclear charge, atomic and ionic radii, electronic configuration, and classification of elements.
Are these questions useful for other competitive exams besides NEET?
Yes, these MCQs and PYQs are also useful for JEE Main, IMU CET, Merchant Navy sponsorship exams, board examinations, and other entrance tests based on Class 11 Chemistry.
Who is the author of these NEET PYQs and MCQs notes?
These study materials and practice questions are authored by Er Neeraj Anand.
Under which publication are these chemistry notes published?
These chemistry notes and question collections are published under Anand Technical Publishers.
Under which brand are these chemistry classes and notes available?
These chemistry study materials, MCQs, and PYQs are available under the brand Anand Classes.
Are conceptual and assertion-reason questions included?
Yes, the section includes conceptual MCQs, assertion-reason questions, match-the-following problems, and previous year NEET numerical questions for complete exam preparation.
How can these PYQs help in NEET preparation?
Regular practice of previous year questions helps students identify important topics, improve problem-solving techniques, manage time effectively, and gain confidence for the NEET examination.