JEE Main (PYQs) Previous Year Questions MCQs on Classification of Elements and Periodicity in Properties are extremely important for mastering periodic trends and concepts. This topic forms the backbone of inorganic chemistry and frequently appears in the exam. Practicing topicwise PYQs helps students understand question patterns, strengthen conceptual clarity, and improve accuracy in solving numerical and theoretical problems related to atomic size, ionization enthalpy, electronegativity, and periodic behavior.
TOPIC 1 : Periodic Table and Classification of Elements
01 Identify the element for which electronic configuration in +3 oxidation state is $[ \text{Ar} ] 3d^{5}$.
(a) Ru
(b) Mn
(c) Co
(d) Fe
[2021, 1 Sep Shift-II]
Ans. (d)
Iron (Fe) has electronic configuration $[ \text{Ar} ] 3d^{5}$ in +3 oxidation state.
Electronic configuration of Fe = $[ \text{Ar} ] 3d^{6} \, 4s^{2}$
Electronic configuration of $\text{Fe}^{3+} = [ \text{Ar} ] 3d^{5}$.
02 Outermost electronic configuration of a group-13 element E is $4s^{2} \, 4p^{1}$. The electronic configuration of an element of p-block period-five placed diagonally to element E is
(a) $[ \text{Kr} ] 3d^{10} \, 4s^{2} \, 4p^{2}$
(b) $[ \text{Ar} ] 3d^{10} \, 4s^{2} \, 4p^{2}$
(c) $[ \text{Xe} ] 5d^{10} \, 6s^{2} \, 6p^{2}$
(d) $[ \text{Kr} ] 4d^{10} \, 5s^{2} \, 5p^{2}$
[2021, 20 July Shift-II]
Ans. (d)
| 13th | 14th | 15th | |
|---|---|---|---|
| 2nd period | $2s^{2} \, 2p^{1}$ (B) | $2s^{2} \, 2p^{2}$ (C) | $2s^{2} \, 2p^{3}$ (N) |
| 3rd period | $3s^{2} \, 3p^{1}$ (Al) | $3s^{2} \, 3p^{2}$ (Si) | $3s^{2} \, 3p^{3}$ (P) |
| 4th period | $4s^{2} \, 4p^{1}$ (Ga) | $4s^{2} \, 4p^{2}$ (Ge) | $4s^{2} \, 4p^{3}$ (As) |
| 5th period | $5s^{2} \, 5p^{1}$ (In) | $5s^{2} \, 5p^{2}$ (Sn) | $5s^{2} \, 5p^{3}$ (Sb) |
That element is $^{50}\text{Sn} \implies [ \text{Kr} ] 4d^{10} \, 5s^{2} \, 5p^{2}$.
03 The atomic number of the element unnilennium is
(a) 109
(b) 102
(c) 108
(d) 119
[2020, 3 Sep Shift-I]
Ans. (a)
The atomic number of the element unnilennium is 109.
Atomic numbers and ‘ium’ is added at the end.
| Digit | Name | Abbreviation |
|---|---|---|
| 1 | un | u |
| 0 | nil | n |
| 1 | un | u |
04 The process that is not endothermic in nature is
(a) $\text{O}^{-}(g) + e^{-} \to \text{O}^{2-}(g)$
(b) $\text{Na}(g) \to \text{Na}^{+}(g) + e^{-}$
(c) $\text{H}(g) + e^{-} \to \text{H}^{-}(g)$
(d) $\text{Ar}(g) + e^{-} \to \text{Ar}^{-}(g)$
[2020, 4 Sep Shift-II]
Ans. (c)
(a) $\text{O}^{-}(g) + e^{-} \to \text{O}^{2-}(g)$ (endothermic)
(b) $\text{Na}(g) \to \text{Na}^{+}(g) + e^{-}$ (endothermic)
(c) $\text{H}(g) + e^{-} \to \text{H}^{-}(g)$ (exothermic)
(d) $\text{Ar}(g) + e^{-} \to \text{Ar}^{-}(g)$ (endothermic)
Reaction (c) is not endothermic. It is exothermic. Electron gain enthalpy of H(g) is negative, due to ‘s1’ configuration while that of Ar(g) is positive due to $ns^{2}np^{6}$ configuration.
Second electron gain enthalpy is always positive for an atom.
Ionisation potential of an atom is positive.
05 The correct electronic configuration and spin-only magnetic moment (BM) of Gd³⁺ (Z = 64), respectively, are
(a) $[ \text{Xe} ] 4f^{7}$ and 7.9
(b) $[ \text{Xe} ] 5f^{7}$ and 7.9
(c) $[ \text{Xe} ] 5f^{7}$ and 8.9
(d) $[ \text{Xe} ] 4f^{7}$ and 8.9
[2020, 5 Sep Shift-I]
Ans. (a)
Electronic configuration of Gd³⁺ is
$^{64}\text{Gd}^{3+} = [ \text{Xe} ] 4f^{7}$
$$ \text{Gd}^{3+} \quad \begin{array}{c} \uparrow\uparrow\uparrow\uparrow\uparrow\uparrow\uparrow \end{array} $$
Gd³⁺ having 7 unpaired electrons.
Magnetic moment $(\mu) = \sqrt{n(n+2)}$ BM
$\Rightarrow \mu = \sqrt{7(7+2)}$ BM $= 7.9$ BM
where, $n =$ number of unpaired electrons.
06 In the sixth period, the orbitals that are filled are
(a) 6s, 5f, 6d, 6p
(b) 6s, 4f, 5d, 6p
(c) 6s, 5d, 5f, 6p
(d) 6s, 6p, 6d, 6f
[2020, 5 Sep Shift-I]
Ans. (b)
Filling of electrons in orbitals in any period takes place as :
$s \to ns$ $(n-2)f$ $(n-1)d$ $np$
(if possible) (if possible)
∴ A per $(n+1)$ rule for sixth period $n = 6$, orbitals that are filled are
6s, 4f, 5d and 6p.
07 The atomic number of unnilunium is …… .
[2020, 6 Sep Shift-II]
Ans. (107)
Unnilunium
IUPAC symbol = Unu
Atomic number (Z) = 101
Because, the roots are strung together in the order of digits which makes up the atomic numbers and ‘ium’ is added at the end.
| Digit | Name | Abbreviation |
|---|---|---|
| 1 | un | u |
| 0 | nil | n |
| 1 | un | u |
08 The electronic configuration of bivalent europium and trivalent cerium are (atomic number : Xe = 54, Ce = 58, Eu = 63)
(a) $[ \text{Xe} ] 4f^{7}$ and $[ \text{Xe} ] 4f^{1}$
(b) $[ \text{Xe} ] 4f^{7} \, 6s^{2}$ and $[ \text{Xe} ] 4f^{2} \, 6s^{2}$
(c) $[ \text{Xe} ] 4f^{2}$ and $[ \text{Xe} ] 4f^{7}$
(d) $[ \text{Xe} ] 4f^{4}$ and $[ \text{Xe} ] 4f^{8}$
[2020, 9 Jan Shift-I]
Ans. (a)
$\text{Eu} = [ \text{Xe} ] 4f^{7} \, 5d^{0} \, 6s^{2} \implies \text{Eu}^{2+} = [ \text{Xe} ] 4f^{7}$
$\text{Ce} = [ \text{Xe} ] 4f^{1} \, 5d^{1} \, 6s^{2} \implies \text{Ce}^{3+} = [ \text{Xe} ] 4f^{1}$
09 The size of the iso-electronic species $\text{Cl}^{-}$, Ar and $\text{Ca}^{2+}$ is affected by
(a) azimuthal quantum number of valence shell
(b) electron-electron interaction in the outer orbitals
(c) principal quantum number of valence shell
(d) nuclear charge
[2019, 8 April Shift-I]
Ans. (d)
The radius of isoelectronic species is inversely proportional to their nuclear charge or atomic number (Z). Thus, greater the value of Z, lesser the radii of isoelectronic species.
10 The IUPAC symbol for the element with atomic number 119 would be
(a) unh
(b) uue
(c) uun
(d) une
[2019, 8 April Shift-II]
Ans. (b)
Atomic number (119) = 1 1 9 [un un en]
So, symbol of the element = uue
Name of the element = ununennium
It is expected to be s-block element, an alkali metal and the first element in eighth period. It is the lightest element that has not yet been synthesised.
11 The isoelectronic set of ions is
(a) $\text{F}^{-}$, $\text{Li}^{+}$, $\text{Na}^{+}$ and $\text{Mg}^{2+}$
(b) $\text{N}^{3-}$, $\text{Li}^{+}$, $\text{Mg}^{2+}$ and $\text{O}^{2-}$
(c) $\text{Li}^{+}$, $\text{Na}^{+}$, $\text{O}^{2-}$ and $\text{F}^{-}$
(d) $\text{N}^{3-}$, $\text{O}^{2-}$, $\text{F}^{-}$ and $\text{Na}^{+}$
[2019, 10 April Shift-I]
Ans. (d)
Key Idea Isoelectronic species contains same number of electrons.
| Species (ions) | At. no. (Z) | No. of electrons |
|---|---|---|
| $\text{N}^{3-}$ | 7 | 10 |
| $\text{O}^{2-}$ | 8 | 10 |
| $\text{F}^{-}$ | 9 | 10 |
| $\text{Na}^{+}$ | 11 | 10 |
| $\text{Li}^{+}$ | 3 | 2 |
| $\text{Mg}^{2+}$ | 12 | 10 |
Thus, option (d) contains isoelectronic set of ions.
12 The group number, number of valence electrons and valency of an element with atomic number 15, respectively, are
(a) 16, 5 and 2
(b) 15, 5 and 3
(c) 16, 6 and 3
(d) 15, 6 and 2
[2019, 12 April Shift-II]
Ans. (b)
Electronic configuration of element having atomic number 15 = $1s^{2} \, 2s^{2} \, 2p^{6} \, 3s^{2} \, 3p^{3}$ (Valence electrons).
As five electrons are present in valence shell, its group number is 15. Valency of element having atomic number 15 is +3 ($8 – 5 = 3$).
13 The element with Z = 120 (not yet discovered) will be an/a
(a) transition metal
(b) inner-transition metal
(c) alkaline earth metal
(d) alkali metal
[2019, 12 Jan Shift-I]
Ans. (c)
The element with Z = 120 will be an alkaline earth metal. Recently, oganesson (Og) with atomic number 118 is named by IUPAC is a noble gas and placed just two places before 120. So, the general electronic configuration is represented as [noble gas] $ns^{2}$ and element with Z = 120 exists as an alkaline earth metal.
14 The group having isoelectronic species is
(a) $\text{O}^{2-}$, $\text{F}^{-}$, $\text{Na}^{+}$, $\text{Mg}^{2+}$
(b) $\text{O}^{-}$, $\text{F}^{-}$, $\text{Na}$, $\text{Mg}^{+}$
(c) $\text{O}^{2-}$, $\text{F}^{-}$, $\text{Na}$, $\text{Mg}^{2+}$
(d) $\text{O}^{-}$, $\text{F}^{-}$, $\text{Na}^{+}$, $\text{Mg}^{2+}$
[JEE Main 2017]
Ans. (a)
| Species | Atomic number | Number of electrons |
|---|---|---|
| $\text{O}^{2-}$ | 8 | 10 |
| $\text{F}^{-}$ | 9 | 10 |
| $\text{Na}^{+}$ | 11 | 10 |
| $\text{Mg}^{2+}$ | 12 | 10 |
∴ Option (a) is correct which contains isoelectronic species $\text{O}^{2-}$, $\text{F}^{-}$, $\text{Na}^{+}$, $\text{Mg}^{2+}$.
15 Which one of the following sets of ions represents a collection of isoelectronic species?
(a) $\text{K}^{+}$, $\text{Cl}^{-}$, $\text{Ca}^{2+}$, $\text{Sc}^{3+}$
(b) $\text{Ba}^{2+}$, $\text{Sr}^{2+}$, $\text{K}^{+}$, $\text{S}^{2-}$
(c) $\text{N}^{3-}$, $\text{O}^{2-}$, $\text{F}^{-}$, $\text{S}^{2-}$
(d) $\text{Li}^{+}$, $\text{Na}^{+}$, $\text{Mg}^{2+}$, $\text{Ca}^{2+}$
[AIEEE 2006]
Ans. (a)
Isoelectronic means having same number of electrons. $\text{K}^{+}$, $\text{Cl}^{-}$, $\text{Ca}^{2+}$, $\text{Sc}^{3+}$ (all are having 18 electrons).
16 According to the periodic law of elements, the variation in properties of elements is related to their
(a) atomic masses
(b) nuclear masses
(c) atomic numbers
(d) nuclear neutron-proton number ratios
[AIEEE 2003]
Ans. (c)
All physical and chemical properties of elements are periodic function of their atomic number — Modern Periodic Law.
TOPIC 2 : Periodic Properties
17 The correct order of ionic radii for the ions, $\text{P}^{3-}$, $\text{S}^{2-}$, $\text{Ca}^{2+}$, $\text{K}^{+}$, $\text{Cl}^{-}$ is
(a) $\text{P}^{3-} > \text{S}^{2-} > \text{Cl}^{-} > \text{K}^{+} > \text{Ca}^{2+}$
(b) $\text{Cl}^{-} > \text{S}^{2-} > \text{P}^{3-} > \text{Ca}^{2+} > \text{K}^{+}$
(c) $\text{P}^{3-} > \text{S}^{2-} > \text{Cl}^{-} > \text{Ca}^{2+} > \text{K}^{+}$
(d) $\text{K}^{+} > \text{Ca}^{2+} > \text{P}^{3-} > \text{S}^{2-} > \text{Cl}^{-}$
[2021, 27 Aug Shift-II]
Ans. (a)
The electronic configuration of given ions are as follows:
$\text{K}{[19]} = 2,8,8,1 \quad \Rightarrow \quad \text{K}^{+} = 2,8,8$
$\text{Ca}{[20]} = 2,8,8,2 \quad \Rightarrow \quad \text{Ca}^{2+} = 2,8,8$
$\text{P}{[15]} = 2,8,5 \quad \Rightarrow \quad \text{P}^{3-} = 2,8,8$
$\text{S}{[16]} = 2,8,6 \quad \Rightarrow \quad \text{S}^{2-} = 2,8,8$
$\text{Cl}{[17]} = 2,8,7 \quad \Rightarrow \quad \text{Cl}^{-} = 2,8,8$
As the effective nuclear charge increases from P to Ca, the atomic size decreases in the same order.
Order of ionic radii
$\text{P}^{3-} > \text{S}^{2-} > \text{Cl}^{-} > \text{K}^{+} > \text{Ca}^{2+}$
18 Given below are two statements : one is labelled as Assertion (A) and the other is labelled as Reason (R).
Assertion (A) Metallic character decreases and non-metallic character increases on moving from left to right in a period.
Reason (R) It is due to increase in ionisation enthalpy and decrease in electron gain enthalpy, when one moves from left to right in a period.
In the light of the above statements, choose the most appropriate answer from the options given below.
(a) (A) is false but (R) is true.
(b) (A) is true but (R) is false.
(c) Both (A) and (R) are correct and (R) is the correct explanation of (A).
(d) Both (A) and (R) are correct but (R) is not the correct explanation of (A).
[2021, 31 Aug Shift-I]
Ans. (b)
On moving from left to right in periodic table, metallic character decreases while non-metallic character increases. This is due to decrease in ionisation enthalpy and increase in electron gain enthalpy when we move from left to right in periodic table. Hence, (A) is true while (R) is false.
19 The correct order of first ionisation enthalpy is
(a) Mg < S < Al < P
(b) Mg < Al < S < P
(c) Al < Mg < S < P
(d) Mg < Al < P < S
[2021, 27 July Shift-II]
Ans. (c)
The correct order of first ionisation enthalpy is Al < Mg < S < P.
Generally, it increases on moving from left to right in a period. But due to the presence of half-filled ($3s^{2} \, 3p^{3}$) orbital in P and fully filled ($3s^{2}$) in Mg, results in the high ionisation energy.
20 The ionic radii of $\text{F}^{-}$ and $\text{O}^{2-}$ respectively are 1.33 Å and 1.4 Å while the covalent radius of N is 0.74 Å. The correct statement for the ionic radius of $\text{N}^{3-}$ from the following is
(a) it is smaller than $\text{F}^{-}$ and N
(b) it is bigger than $\text{O}^{2-}$ and $\text{F}^{-}$
(c) it is bigger than $\text{F}^{-}$ and N, but smaller than $\text{O}^{2-}$
(d) it is smaller than $\text{O}^{2-}$ and $\text{F}^{-}$, but bigger than of N
[2021, 25 July Shift-II]
Ans. (b)
The electronic configuration of $\text{F}^{-}$, $\text{O}^{2-}$ and $\text{N}^{3-}$ are given below.
$ \text{F}^{-} = 1s^{2} \, 2s^{2} \, 2p^{6} \ $
$ \text{O}^{2-} = 1s^{2} \, 2s^{2} \, 2p^{6} \ $
$ \text{N}^{3-} = 1s^{2} \, 2s^{2} \, 2p^{6} \ $
All the above ins are iso electronic species because they contain same number of electrons (10).
Among all these, $\text{N}^{3-}$ has least number of protons due to which its size increases as least nuclear attractions experienced by outer shell electrons.
From top to bottom of periodic table ions will increase in radii. However, now left to right, the radius is more of a function of the number of proton or nuclear charge. Similarly, $\text{O}^{2-}$ will be larger than $\text{F}^{-}$ as both have 10 electrons but Z = 8 for oxygen and Z = 9 for fluorine.
21 AX is a covalent diatomic molecule, where A and X are second row elements of periodic table. Based on molecular orbital theory, the bond order of AX is 2.5. The total number of electrons in AX is ………. . (Round off to the nearest integer).
[2021, 18 March Shift-I]
Ans. (15)
AX is a covalent diatomic molecule. The molecule is NO (Nitric oxide). Its bond order is 2.5 and it has a total of 15 electrons (7 + 8 = 15).
Here ‘A’ denotes nitrogen (N) and ‘X’ denotes oxygen (O). NO is colourless and toxic gas.
Note : Total number of electrons equal to 13 will also have the 2.5 bond order, but in this case neutral diatomic molecule will not be possible.
22 The set of elements that differ in mutual relationship from those of the other sets is
(a) Li-Mg
(b) B-Si
(c) Be-Al
(d) Li-Na
[2021, 17 March Shift-II]
Ans. (d)
Li-Na pair is different from remaining three options as it does not show diagonal relationship. Both Li-Na belong to same group and are not placed diagonally.
Li-Mg, B-Si and Be-Al show diagonal relationship.
23 Identify the elements X and Y using the ionisation energy values given below.
| Ionisation energy (Ist) | (IInd) | |
|---|---|---|
| X | 495 | 4563 |
| Y | 731 | 1450 |
(a) X = Na, Y = Mg
(b) X = Mg, Y = F
(c) X = Mg, Y = Na
(d) X = F, Y = Mg
[2021, 16 March Shift-II]
Ans. (d)
X is Na ($3s^{1}$) as it has very high second ionisation energy due to removal of electron from highly stable noble gas configuration.
Na $\to$ Na$^{+}$ + e$^{-}$ ; IE$_{1}$ = 495 kJ/mol
$1s^{2}2s^{2}2p^{6}3s^{1}$ $\to$ $1s^{2}2s^{2}2p^{6}$ or [Ne]
Na$^{+}$ $\to$ Na$^{2+}$ + e$^{-}$ ; IE$_{2}$ = 4563 kJ/mol
$1s^{2}2s^{2}2p^{6}$ (Stable noble gas configuration) $\to$ $1s^{2}2s^{2}2p^{5}$ or [F]
In case of Mg, noble gas configuration is achieved after removal of second electron. So, IE$_{2}$ is comparatively less.
Mg $\to$ Mg$^{+}$ + e$^{-}$ ; [Ne]$3s^{2}$ $\to$ [Ne]$3s^{1}$ ; IE$_{1}$ = 731 kJ/mol
Mg$^{+}$ $\to$ Mg$^{2+}$ + e$^{-}$ ; [Ne]$3s^{1}$ $\to$ [Ne] ; IE$_{2}$ = 1450 kJ/mol
24 The first ionisation energy of magnesium is smaller as compared to that of elements X and Y, but higher than that of Z. The elements X, Y and Z, respectively, are
(a) chlorine, lithium and sodium
(b) argon, lithium and sodium
(c) argon, chlorine and sodium
(d) neon, sodium and chlorine
[2021, 18 March Shift-II]
Ans. (c)
The order of first ionisation energy of 3rd period is as follows:
Na < Al < Mg < Si < S < P < Cl < Ar
On moving along a period from left to right, nuclear charge increases that outweighs the shielding effect. As a result, outermost electrons are held more tightly. Hence, ionisation enthalpy increases.
Ionisation enthalpy of group 2 is greater than group 13 and that of group 15 is greater than group 16. This is due to fully filled and half-filled orbital.
∴ X, Y and Z are argon, chlorine and sodium respectively.
25 Which pair of oxides is acidic in nature?
(a) $ \text{B}{_2}\text{O}{_3} $, CaO
(b) $\text{B}{_2}\text{O}{_3}$, $\text{SiO}{_2}$
(c) $\text{N}{_2}\text{O}$, BaO
(d) CaO, $\text{SiO}_{2}$
[2021, 26 Feb Shift-II]
Ans. (b)
(a) $\text{B}_{2}\text{O}_{3}$ → oxide of boron (non-metal): acidic
CaO → oxide of calcium (metal): basic
(b) $\text{B}_{2}\text{O}_{3}$ → oxide of boron (non-metal): acidic
$\text{SiO}_{2}$ → oxide of silicon (non-metal): acidic
(c) $\text{N}_{2}\text{O}$ → oxide of nitrogen (non-metal): neutral
BaO → oxide of barium (metal): basic
(d) CaO → oxide of calcium (metal): basic
$\text{SiO}_{2}$ → oxide of silicon (non-metal): acidic
So, option (b) is the correct answer. $\text{B}_{2}\text{O}_{3}$ and $\text{SiO}_{2}$ are acidic oxides; they react with base to give salt and water.
$\text{B}_{2}\text{O}_{3} + 2\text{NaOH} \to 2\text{NaBO}_{2} + \text{H}_{2}\text{O}$ (Sodium metaborate)
$\text{SiO}_{2} + 2\text{NaOH} \to \text{Na}_{2}\text{SiO}_{3} + \text{H}_{2}\text{O}$ (Sodium silicate)
26 Match List-I with List-II.
| List-I (Electronic configuration of elements) | List-II ($\Delta_{i}H$ in kJ mol$^{-1}$) |
|---|---|
| A. $1s^{2}2s^{2}$ | (i) 801 |
| B. $1s^{2}2s^{2}2p^{4}$ | (ii) 899 |
| C. $1s^{2}2s^{2}2p^{3}$ | (iii) 1314 |
| D. $1s^{2}2s^{2}2p^{1}$ | (iv) 1402 |
Choose the most appropriate answer from the options given below.
A B C D
(a) (ii) (iii) (iv) (i)
(b) (i) (iv) (iii) (ii)
(c) (i) (iii) (iv) (ii)
(d) (iv) (i) (ii) (iii)
[2021, 26 Feb Shift-II]
Ans. (d)
Here, (B), (C) and (D) are p-block elements of the 2nd period.
(D) → p¹ configuration → B of group 13
(C) → p³ configuration → N of group 15
(B) → p⁴ configuration → O of group 16
Stability order : p³ > p⁴ >> p¹ (Half-filled > Partially filled)
We know, ionisation enthalpy ($\Delta_{i}H$) ∝ stability of the subshell concerned.
Therefore, half-filled subshell is more stable than partially filled.
∴ $\Delta_{i}H$ order is C > B >> D
(A) is a s-block element (group 2) of 2nd period with s²-configuration → Be of group 2 [fully-filled; stable]
So, the correct order of IE₁ or $\Delta_{i}H_{1}$ (in kJ mol$^{-1}$) of the 2nd period elements will be:
Be (899) < B (801) < O (1314) < N (1402)
27 In general, the property (magnitudes only) that shows an opposite trend in comparison to other properties across a period is
(a) electronegativity
(b) electron gain enthalpy
(c) ionisation enthalpy
(d) atomic radius
[2020, 2 Sep Shift-I]
Ans. (d)
With increase in atomic number (Z) along a period, effective nuclear charge (Z*) of the elements also increases. We know,
Z ∝ Electronegativity (χ)
Z ∝ Electron gain enthalpy or |Δ$_{eg}$H|
Z ∝ Ionisation enthalpy or Δ$_{i}$H
Atomic radius (r) ∝ 1/Z
So, with increase in atomic number across a period, Electronegativity, electron gain enthalpy and ionisation enthalpy increase whereas atomic radius decreases.
28 Three elements X, Y and Z are in the 3rd period of the periodic table. The oxides of X, Y and Z, respectively, are basic, amphoteric and acidic. The correct order of the atomic number of X, Y and Z is
(a) Z < Y < X
(b) X < Y < Z
(c) X < Z < Y
(d) Y < X < Z
[2020, 2 Sep Shift-II]
Ans. (b)
While moving across a period, metallic property of elements changes first into metalloids then into non-metallic nature.
⇒ Metal gives basic oxides.
Metalloid produces amphoteric oxides.
Non-metal gives acidic oxides.
So, X → metal, Y → metalloid and Z → non-metal
Order of atomic number : X < Y < Z
As X, Y and Z belong to 3rd period, X can be Na or Mg, Y can be Al or Si, Z can be P or S or Cl.
29 The ionic radii of $\text{O}^{2-}$, $\text{F}^{-}$, $\text{Na}^{+}$ and $\text{Mg}^{2+}$ are in the order
(a) $\text{O}^{2-} > \text{F}^{-} < \text{Mg}^{2+} > \text{Na}^{+}$
(b) $\text{Mg}^{2+} > \text{Na}^{+} > \text{F}^{-} > \text{O}^{2-}$
(c) $\text{O}^{2-} > \text{F}^{-} > \text{Na}^{+} > \text{Mg}^{2+}$
(d) $\text{F}^{-} > \text{O}^{2-} > \text{Na}^{+} > \text{Mg}^{2+}$
[2020, 4 Sep Shift-I]
Ans. (c)
$\text{O}^{2-}$, $\text{F}^{-}$, $\text{Na}^{+}$ and $\text{Mg}^{2+}$ are isoelectronic species.
| Ionic | $\text{O}^{2-}$ | $\text{F}^{-}$ | $\text{Na}^{+}$ | $\text{Mg}^{2+}$ |
|---|---|---|---|---|
| Number of e⁻ | 10 | 10 | 10 | 10 |
For isoelectronic species,
Size of species ∝ 1/Nuclear charge
Iso-electronic species are those atoms or ions which have the same number of electrons.
Size of species decreases with increasing protons. More is effective nuclear charge (Z$_{eff}$), lesser will be ionic size.
So, correct order of ionic radius is $\text{O}^{2-} > \text{F}^{-} > \text{Na}^{+} > \text{Mg}^{2+}$.
30 The correct order of the ionic radii of $\text{O}^{2-}$, $\text{N}^{3-}$, $\text{F}^{-}$, $\text{Mg}^{2+}$, $\text{Na}^{+}$ and $\text{Al}^{3+}$ is
(a) $\text{N}^{3-} < \text{O}^{2-} < \text{F}^{-} < \text{Na}^{+} < \text{Mg}^{2+} < \text{Al}^{3+}$
(b) $\text{Al}^{3+} < \text{Na}^{+} < \text{Mg}^{2+} < \text{O}^{2-} < \text{F}^{-} < \text{N}^{3-}$
(c) $\text{Al}^{3+} < \text{Mg}^{2+} < \text{Na}^{+} < \text{F}^{-} < \text{O}^{2-} < \text{N}^{3-}$
(d) $\text{N}^{3-} < \text{F}^{-} < \text{O}^{2-} < \text{Mg}^{2+} < \text{Na}^{+} < \text{Al}^{3+}$
[2020, 5 Sep Shift-II]
Ans. (c)
Size of species ∝ 1/Nuclear charge
Iso-electronic species are those atoms or ions which have the same number of electrons.
Size of species decreases with increasing protons. More is effective nuclear charge (Z$_{eff}$), lesser will be ionic size.
Correct order of ionic radii
$\text{Al}^{3+} < \text{Mg}^{2+} < \text{Na}^{+} < \text{F}^{-} < \text{O}^{2-} < \text{N}^{3-}$
31 The electron gain enthalpy (in kJ/mol) of fluorine, chlorine, bromine and iodine, respectively, are
(a) –333, –325, –349 and –296
(b) –296, –325, –333 and –349
(c) –333, –349, –325 and –296
(d) –349, –333, –325 and –296
[2020, 7 Jan Shift-I]
Ans. (c)
Electron gain enthalpy ($\Delta_{eg}H$) is the enthalpy change for converting 1 mol of isolated atoms to anions by adding electrons. All halogens have negative $\Delta_{eg}H$ (exothermic) values. Generally, $\Delta_{eg}H$ becomes less negative when comparing elements of the same group from top to bottom.
But among fluorine and chlorine there is an anomaly because inter-electron repulsion is stronger in fluorine due to its extra small size.
∴ $\Delta_{eg}H$ is less exothermic than expected for F atom.
Thus, the correct values of electron gain enthalpies:
F (–333) < Cl (–349) > Br (–325) > I (–296) kJ mol⁻¹
32 Within each pair of elements F and Cl, S and Se, and Li and Na, respectively, the elements that release more energy upon an electron gain are
(a) F, Se and Na
(b) F, S and Li
(c) Cl, S and Li
(d) Cl, Se and Na
[2020, 7 Jan Shift-II]
Ans. (c)
The first electron gain enthalpy is exothermic (or negative).
Generally, electron gain enthalpy becomes less exothermic (or less negative) when comparing elements of a group from top to bottom.
Therefore, electron gain enthalpy of S > Se and Li > Na.
But there are some exceptions to this. One of them is the case of group 17 elements where electron gain is most negative for Cl instead of F, due to extra small size of fluorine.
∴ Upon an electron gain, energy releases in the order:
Cl > F, S > Se and Li > Na
33 The first ionisation energy (in kJ/mol) of Na, Mg, Al and Si respectively, are :
(a) 496, 577, 737, 786
(b) 786, 737, 577, 496
(c) 496, 577, 786, 737
(d) 496, 737, 577, 786
[2020, 8 Jan Shift-I]
Ans. (d)
All four elements belong to the same period (3rd period) and the general trend in a period from left to right is increase in first ionisation energy, because of increase in nuclear charge and decrease in radii.
Therefore, the expected order of first ionisation enthalpies of given elements should be Na < Mg < Al < Si.
But first ionisation energies of Mg and Al show anomalous order because Mg has stable [Ne] $3s^{2}$ configuration and, therefore higher ionisation energy than expected, whereas Al has [Ne] $3s^{2} \, 3p^{1}$ configuration, which on ionisation becomes stable [Ne] $3s^{2}$.
∴ Its ionisation energy is less than expected.
34 The third ionisation enthalpy is minimum for :
(a) Mn
(b) Ni
(c) Co
(d) Fe
[2020, 8 Jan Shift-I]
Ans. (d)
Fe has minimum value of third ionisation enthalpy. Ground state electronic configuration of Fe is [Ar] $3d^{6} \, 4s^{2}$.
$\text{Fe}^{2+}$ has [Ar] $3d^{6}$ configuration, whereas $\text{Fe}^{3+}$ has [Ar] $3d^{5}$. The latter is a stable configuration and easier to ionise $\text{Fe}^{2+}$ to $\text{Fe}^{3+}$ than expected. Hence possess least value.
Ni has highest value of third ionisation enthalpy due to its greater nuclear charge and smaller size.
Due to the same reason, Co > Fe > Mn should be the order for the remaining elements, but the anomalous order is due to greater stability of $\text{Mn}^{2+}$ having [Ar] $3d^{5}$ configuration than $\text{Mn}^{3+}$ with [Ar] $3d^{4}$.
Whereas for $\text{Co}^{+}$, the electronic configuration is [Ar] $3d^{7}$. Thus, ionisation enthalpy needed to remove third electron is less as compared to that of $\text{Mn}^{2+}$.
35 The increasing order of the atomic radii of the following elements is
(A) C (B) O (C) F (D) Cl (E) Br
(a) (A) < (B) < (C) < (D) < (E)
(b) (C) < (B) < (A) < (D) < (E)
(c) (D) < (C) < (B) < (A) < (E)
(d) (B) < (C) < (D) < (A) < (E)
[2020, 8 Jan Shift-II]
Ans. (b)
Atomic radius generally decreases as we compare elements in a period from left to right,
∴ C > O > F
but elements present in next period are larger in size,
∴ Br > Cl > C > O > F;
So, the correct increasing order of the atomic radii:
C < O < F < Cl < Br
36 B has a smaller first ionisation enthalpy than Be. Consider the following statements :
(I) It is easier to remove 2p electron than 2s electron
(II) 2p electron of B is more shielded from the nucleus by the inner core of electrons than the 2s electrons of Be
(III) 2s electron has more penetration power than 2p electron
(IV) atomic radius of B is more than Be (atomic number B = 5, Be = 4)
The correct statements are
(a) (I), (II) and (III)
(b) (II), (III) and (IV)
(c) (I), (III) and (IV)
(d) (I), (II) and (IV)
[2020, 9 Jan Shift-II]
Ans. (a)
Statements I, II and III are correct, whereas statement IV is incorrect. Its correct form is that atomic radius of B is less than Be.
37 The first and second ionisation enthalpies of a metal are 496 and 4560 kJ mol⁻¹, respectively. How many moles of HCl and H₂SO₄, respectively, will be needed to react completely with 1 mole of the metal hydroxide?
(a) 1 and 2
(b) 1 and 1
(c) 2 and 0.5
(d) 1 and 0.5
[2020, 9 Jan Shift-II]
Ans. (d)
A small first ionisation enthalpy (496 kJ mol⁻¹) and a great difference between first and second ionisation enthalpies (4064 kJ mol⁻¹) characterise an alkali metal (group-1) element. Its hydroxide is a monoacidic base (MOH).
∴ For 1 mol of that base, 1 mol of HCl (a monoacidic acid) and 0.5 mol of H₂SO₄ (a dibasic acid) will be needed for complete reaction.
38 The process that is not endothermic in nature is
(a) $\text{O}^{-}(g) + e^{-} \to \text{O}^{2-}(g)$
(b) $\text{Na}(g) \to \text{Na}^{+}(g) + e^{-}$
(c) $\text{H}(g) + e^{-} \to \text{H}^{-}(g)$
(d) $\text{Ar}(g) + e^{-} \to \text{Ar}^{-}(g)$
[2020, 4 Sep Shift-II]
Ans. (c)
(a) $\text{O}^{-}(g) + e^{-} \to \text{O}^{2-}(g)$ (endothermic)
(b) $\text{Na}(g) \to \text{Na}^{+}(g) + e^{-}$ (endothermic)
(c) $\text{H}(g) + e^{-} \to \text{H}^{-}(g)$ (exothermic)
(d) $\text{Ar}(g) + e^{-} \to \text{Ar}^{-}(g)$ (endothermic)
Reaction (c) is not endothermic. It is exothermic. Electron gain enthalpy of H(g) is negative, due to $1s^{1}$ configuration while that of Ar(g) is positive due to $ns^{2} np^{6}$ configuration.
Second electron gain enthalpy is always positive for an atom.
Ionisation potential of an atom is positive.
39 The element having greatest difference between its first and second ionisation energy, is
(a) Ca
(b) Sc
(c) Ba
(d) K
[2019, 9 April Shift-I]
Ans. (d)
The electronic configurations of given elements are as follows:
K(19) = $1s^{2} \, 2s^{2} \, 2p^{6} \, 3s^{2} \, 3p^{6} \, 4s^{1}$
Mg(12) = $1s^{2} \, 2s^{2} \, 2p^{6} \, 3s^{2}$
Sr(38) = $1s^{2} \, 2s^{2} \, 2p^{6} \, 3s^{2} \, 3p^{6} \, 4s^{2} \, 3d^{10} \, 4p^{6} \, 5s^{2}$
Sc(21) = $1s^{2} \, 2s^{2} \, 2p^{6} \, 3s^{2} \, 3p^{6} \, 4s^{2} \, 3d^{1}$
First ionisation enthalpy (I.E.) of K is lowest among the given options. Here, the energy required to remove an electron from $4s^{1}$ is least as only one electron is present in the outermost shell. I.E.(I) is comparatively high for Mg and Sr and two electrons (fully-filled) are placed in s-orbital.
Second ionisation enthalpy of K is highest among the given options. Now, removal of an electron occurs from p⁶ (fully-filled). So, high energy is required to remove the electron. From the above discussion, it can be concluded that (I.E.₂ – I.E.₁) value is maximum for K (potassium).
40 In comparison to boron, beryllium has
(a) lesser nuclear charge and lesser first ionisation enthalpy
(b) greater nuclear charge and lesser first ionisation enthalpy
(c) greater nuclear charge and greater first ionisation enthalpy
(d) lesser nuclear charge and greater first ionisation enthalpy
[2019, 12 April Shift-II]
Ans. (d)
In comparison to boron, beryllium has lesser nuclear charge and greater first ionisation enthalpy.
Electronic configuration of Be(4) = $1s^{2} \, 2s^{2}$.
It possesses completely filled s-orbitals. Hence, high amount of energy is required to pull the electron from the gaseous atom. Beryllium (4) lies left to the boron (5) and on moving from left to right an electron is added due to which nuclear charge increases from Be to B.
41 In general, the properties that decrease and increase down a group in the periodic table, respectively are
(a) electronegativity and atomic radius
(b) electronegativity and electron gain enthalpy
(c) electron gain enthalpy and electronegativity
(d) atomic radius and electronegativity
[2019, 9 Jan Shift-I]
Ans. (a)
The summary of variation of periodic properties is given in table below:
| S. No. | Periodic property | Along a period | Down a group |
|---|---|---|---|
| 1. | Atomic radius | Decreases | Increases |
| 2. | Electron gain enthalpy | Increases | Decreases |
| 3. | Electronegativity | Increases | Decreases |
Thus, electronegativity decreases and atomic radius increases down a group in the periodic table.
42 The correct option with respect to the Pauling electronegativity values of the elements is
(a) P > S
(b) Si < Al
(c) Te > Se
(d) Ga < Ge
[2019, 11 Jan Shift-II]
Ans. (d)
The electronegativity values of given elements on the Pauling scale can be shown as follows:
| Period No. | Group 13 | Group 14 | Group 15 | Group 16 |
|---|---|---|---|---|
| 3 | Al (1.5) | Si (1.8) | P (2.1) | S (2.5) |
| 4 | Ga (1.6) | Ge (1.8) | Se (2.4) | |
| 5 | Te (2.01) |
On moving from left to right across a period, i.e. from Ga to Se, the effective nuclear charge increases and size decreases.
As a result, the value of electronegativity increases due to increase in the attraction between the outer electrons and the nucleus. Whereas on moving down the group, (i.e. from Se to Te), the atomic size increases.
As a result, the force of attraction between the outer electron and the nucleus decreases. Hence, the electronegativity decreases.
43 Which of the following atoms has the highest first ionisation energy?
(a) Na
(b) K
(c) Sc
(d) Rb
[JEE Main 2016]
Ans. (c)
Order of first ionisation energy is Sc > Na > K > Rb.
Due to poor shielding effect, removal of one electron from 4s orbital is difficult as compared to 3s-orbital.
44 The correct order of the atomic radii of C, Cs, Al and S is
(a) C < S < Al < Cs
(b) C < S < Cs < Al
(c) S < C < Cs < Al
(d) S < C < Al < Cs
[2019, 11 Jan Shift-I]
Ans. (d)
| Element | Period No. | Group No. |
|---|---|---|
| C | 2nd | 14 |
| Al | 3rd | 13 |
| S | 3rd | 16 |
| Cs | 6th | 1 |
Along the period atomic radius decreases, so, radii : Al > S.
With the addition of a new shell, period number as well as atomic radius increases. It is because of the successive addition of one extra shell of electrons. So, the order of the atomic radii of the given elements will be:
C < S < Al < Cs
45 The ionic radii (in Å) of $\text{N}^{3-}$, $\text{O}^{2-}$ and $\text{F}^{-}$ respectively are
(a) 1.36, 1.40 and 1.71
(b) 1.36, 1.71 and 1.40
(c) 1.71, 1.40 and 1.36
(d) 1.71, 1.36 and 1.40
[JEE Main 2015]
Ans. (c)
Number of electrons in $\text{N}^{3-}$ = 7 + 3 = 10
Number of electrons in $\text{O}^{2-}$ = 8 + 2 = 10
Number of electrons in $\text{F}^{-}$ = 9 + 1 = 10
Since, all the three species have each 10 electrons hence they are isoelectronic species.
It is considered that, in case of isoelectronic species as the negative charge increases, ionic radii increases and therefore the value of ionic radii are
$\text{N}^{3-}$ = 1.71 (highest among the three)
$\text{O}^{2-}$ = 1.40
$\text{F}^{-}$ = 1.36 (lowest among the three)
Time Saving Technique There is no need to mug up the radius values for different ions. This particular question can be solved through following time saving trick:
The charges on the ions indicate the size as $\text{N}^{3-} > \text{O}^{2-} > \text{F}^{-}$. Thus, you have to look for the option in which the above trend is followed. Option (c) is the only one in which this trend is followed. Hence, it is the correct answer.
46 Which of the following represents the correct order of increasing first ionisation enthalpy for Ca, Ba, S, Se and Ar ?
(a) Ca < S < Ba < Se < Ar
(b) S < Se < Ca < Ba < Ar
(c) Ba < Ca < Se < S < Ar
(d) Ca < Ba < S < Se < Ar
[JEE Main 2013]
Ans. (c)
Ionisation energy increases along a period from left to right and decreases down a group. The position of given elements in the periodic table is as
2 16 18
Ca S Ar
Ba Se
Thus, the order of increasing $\Delta H_{1e}$ is Ba < Ca < Se < S < Ar
47 The first ionisation potential of Na is 5.1 eV. The value of electron gain enthalpy of Na⁺ will be
(a) –2.55 eV
(b) –5.1 eV
(c) –10.2 eV
(d) +2.55 eV
[JEE Main 2013]
Ans. (b)
Na → Na⁺ + e⁻ (First IE)
Na⁺ + e⁻ → Na (Electron gain enthalpy of Na⁺)
Because reaction is reverse so
$\Delta H_{eg} = -5.1$ eV
48 The increasing order of the ionic radii of the given isoelectronic species is
(a) $\text{Cl}^{-}$, $\text{Ca}^{2+}$, $\text{K}^{+}$, $\text{S}^{2-}$
(b) $\text{S}^{2-}$, $\text{Cl}^{-}$, $\text{Ca}^{2+}$, $\text{K}^{+}$
(c) $\text{Ca}^{2+}$, $\text{K}^{+}$, $\text{Cl}^{-}$, $\text{S}^{2-}$
(d) $\text{K}^{+}$, $\text{S}^{2-}$, $\text{Ca}^{2+}$, $\text{Cl}^{-}$
[AIEEE 2012]
Ans. (c)
For isoelectronic species, $r_{i} \propto 1/Z$ (Here, Z is atomic number)
| Species | Z | Electrons |
|---|---|---|
| $\text{Cl}^{-}$ | 17 | 18 |
| $\text{Ca}^{2+}$ | 20 | 18 |
| $\text{K}^{+}$ | 19 | 18 |
| $\text{S}^{2-}$ | 16 | 18 |
Thus, ionic size is in order
$\text{Ca}^{2+} < \text{K}^{+} < \text{Cl}^{-} < \text{S}^{2-}$
49 Which one of the following orders presents the correct sequence of the increasing basic nature of the given oxides?
(a) $\text{Al}_{2}\text{O}_{3} < \text{MgO} < \text{Na}_{2}\text{O} < \text{K}_{2}\text{O}$
(b) $\text{MgO} < \text{K}_{2}\text{O} < \text{Al}_{2}\text{O}_{3} < \text{Na}_{2}\text{O}$
(c) $\text{Na}_{2}\text{O} < \text{K}_{2}\text{O} < \text{MgO} < \text{Al}_{2}\text{O}{3}$
(d) $\text{K}_{2}\text{O} < \text{Na}_{2}\text{O} < \text{Al}_{2}\text{O}_{3} < \text{MgO}$
[AIEEE 2011]
Ans. (a)
The basicity of an oxide depends on the metallic character of the element it is bonded to. As metallic character increases, the ability of the oxide to donate electrons or react with water to form a strong base increases.
1. Periodic Trend (Left to Right):
Across a period, metallic character decreases and electronegativity increases. Consequently, the oxides shift from basic to amphoteric to acidic.
Sodium ($\text{Na}$) and Magnesium ($\text{Mg}$) are in Group 1 and 2 respectively. $\text{Na}$ is more metallic than $\text{Mg}$, making $\text{Na}_{2}\text{O}$ more basic than $\text{MgO}$.
Aluminum ($\text{Al}$) is further to the right in Group 13. Its oxide, $\text{Al}_{2}\text{O}_{3}$, is amphoteric (acting as both an acid and a base), making it the least basic in this set.
2. Group Trend (Top to Bottom):
Down a group, metallic character increases as the atomic size increases and ionization enthalpy decreases.
Potassium ($\text{K}$) is below Sodium ($\text{Na}$) in Group 1. Therefore, $\text{K}$ is more electropositive than $\text{Na}$, making $\text{K}_{2}\text{O}$ more basic than $\text{Na}_{2}\text{O}$.
$\text{Al}_{2}\text{O}_{3}$: Amphoteric (Lowest basicity)
$\text{MgO}$: Basic
$\text{Na}_{2}\text{O}$: Strongly Basic
$\text{K}_{2}\text{O}$: Most Strongly Basic (Highest basicity)
Hence, basic nature of oxides are as follows
$\text{Al}_{2}\text{O}_{3} < \text{MgO} < \text{Na}_{2}\text{O} < \text{K}_{2}\text{O}$
50 The correct order of electron gain enthalpy with negative sign of F, Cl, Br and I, having atomic number 9, 17, 35 and 53 respectively, is
(a) I > Br > Cl > F
(b) F > Cl > Br > I
(c) Cl > F > Br > I
(d) Br > Cl > I > F
[AIEEE 2011]
Ans. (c)
As we go down the group in periodic table, atomic size increases, force of attraction for the added electron decreases. Thus, the added electron becomes loosely bound. Hence, electron gain enthalpy decreases down a group.
$X(g) + e^{-} \to X^{-}(g)$
Actual order is Cl > F > Br > I
The fact that fluorine has a less electron gain enthalpy than chlorine seems to be due to the relatively greater effectiveness of 2p-electron in the small F-atom to repel the additional electron entering the atom than 3p-electrons in the larger Cl-atom.
51 The correct sequence which shows decreasing order of the ionic radii of the elements is
(a) $\text{Al}^{3+} > \text{Mg}^{2+} > \text{Na}^{+} > \text{F}^{-} > \text{O}^{2-}$
(b) $\text{Na}^{+} > \text{Mg}^{2+} > \text{Al}^{3+} > \text{O}^{2-} > \text{F}^{-}$
(c) $\text{Na}^{+} > \text{F}^{-} > \text{Mg}^{2+} > \text{O}^{2-} > \text{Al}^{3+}$
(d) $\text{O}^{2-} > \text{F}^{-} > \text{Na}^{+} > \text{Mg}^{2+} > \text{Al}^{3+}$
[AIEEE 2010]
Ans. (d)
$\text{O}^{2-}$, $\text{F}^{-}$, $\text{Na}^{+}$, $\text{Mg}^{2+}$ and $\text{Al}^{3+}$ are isoelectronic species.
For isoelectronic species, $r \propto 1/Z$
Thus, larger the atomic number (effective nuclear charge) smaller is the ionic radii.
52 The set representing the correct order of ionic radius is
(a) $\text{Li}^{+} > \text{Be}^{2+} > \text{Na}^{+} > \text{Mg}^{2+}$
(b) $\text{Na}^{+} > \text{Li}^{+} > \text{Mg}^{2+} > \text{Be}^{2+}$
(c) $\text{Li}^{2+} > \text{Na}^{+} > \text{Mg}^{2+} > \text{Be}^{2+}$
(d) $\text{Mg}^{2+} > \text{Be}^{2+} > \text{Li}^{+} > \text{Na}^{+}$
[AIEEE 2009]
Ans. (b)
The ionic radius in general increases moving top to bottom and further decreases moving left to right. So the correct order is
$\text{Na}^{+} > \text{Li}^{+} > \text{Mg}^{2+} > \text{Be}^{2+}$
53 The ionic mobility of alkali metal ions in aqueous solution is maximum for
(a) $\text{K}^{+}$
(b) $\text{Rb}^{+}$
(c) $\text{Li}^{+}$
(d) $\text{Na}^{+}$
[AIEEE 2006]
Ans. (b)
$\text{Li}^{+}$ is having largest hydrated ionic size while $\text{Rb}^{+}$ is having smallest. Smaller the size, greater is the mobility.
54 The increasing order of the first ionisation enthalpies of the elements B, P, S and F (lowest first) is
(a) F < S < P < B
(b) P < S < B < F
(c) B < P < S < F
(d) B < S < P < F
[AIEEE 2006]
Ans. (d)
To determine the order of first ionization enthalpies ($IE_1$), we must analyze the atomic structure and periodic trends of these four elements: Boron (B), Phosphorus (P), Sulfur (S), and Fluorine (F).
1. General Periodic Trends
Across a Period (Left to Right): Ionization enthalpy generally increases because the effective nuclear charge increases and the atomic radius decreases, making it harder to remove an electron.
Down a Group (Top to Bottom): Ionization enthalpy decreases because the atomic size increases and the outermost electrons are further from the nucleus, experiencing more shielding.
2. Comparing the Elements
Step 1: Boron (B) vs. the rest
Boron is in the 2nd period but in Group 13. It has the largest atomic radius and the lowest nuclear charge among these four. Therefore, it has the lowest ionization enthalpy.
Step 2: Fluorine (F) vs. the rest
Fluorine is in the 2nd period and is the most electronegative element in Group 17. Its small size and high nuclear charge give it the highest ionization enthalpy in this set.
Step 3: The Phosphorus (P) vs. Sulfur (S) Exception
This is the most critical part of the problem. Normally, $IE_1$ increases from left to right, which would suggest $P < S$. However, we must look at their electronic configurations:
Phosphorus (P): $[Ne] 3s^2 3p^3$ (Exactly half-filled $p$-subshell)
Sulfur (S): $[Ne] 3s^2 3p^4$ (One electron more than half-filled)
A half-filled subshell is exceptionally stable due to symmetry and exchange energy. Removing an electron from Phosphorus requires more energy than removing the fourth $p$-electron from Sulfur (which, once removed, actually leaves Sulfur with a stable half-filled state).
Rule: $IE_1$ of Group 15 (half-filled) $>$ $IE_1$ of Group 16.
3. Conclusion of the Order
B is the lowest (Group 13, 2nd period).
S comes next (Group 16, 3rd period).
P is higher than S (Group 15, 3rd period) due to its stable half-filled configuration.
F is the highest (Group 17, 2nd period).
Increasing Order: $B < S < P < F$
55 Following statements regarding the periodic trends of chemical reactivity of the alkali metals and the halogens are given. Which of these statements give the correct picture?
(a) The reactivity decreases in the alkali metals but increases in the halogens with increase in atomic number down the group
(b) In both the alkali metals and the halogens the chemical reactivity decreases with increase in atomic number down the group
(c) Chemical reactivity increases with increase in atomic number down the group in both the alkali metals and halogens
(d) In alkali metals, the reactivity increases but in the halogens it decreases with increase in atomic number down the group
[AIEEE 2006]
Ans. (d)
The chemical reactivity of alkali metals is due to their ability to lose extra electron more readily than other elements. Since, the tendency to lose electron by alkali metals increases down the group, therefore, their chemical reactivity increases down the group.
On the other hand, chemical reactivity of halogens is due to their ability to gain extra electron more readily than other elements. Since the tendency to gain electron by halogens decreases down the group, hence their chemical reactivity decreases down the group.
56 In which of the following arrangements the order is not according to the property indicated against it ?
(a) Li < Na < K < Rb Increasing metallic radius
(b) I < Br < F < Cl Increasing electron gain enthalpy (with negative sign)
(c) B < C < N < O Increasing first ionisation enthalpy
(d) $\text{Al}^{3+} < \text{Mg}^{2+} < \text{Na}^{+} < \text{F}^{-}$ Increasing ionic size
[AIEEE 2005]
Ans. (c)
(a) Metallic radii increase in a group from top to bottom. Thus, Li < Na < K < Rb — True
(b) Electron gain enthalpy of Cl > F and decreases along a group. Thus, I < Br < F < Cl is true.
(c) Ionisation enthalpy increases along a period from left to right but due to presence of half filled orbitals in N, ionisation enthalpy of N > O. Thus, B < C < N < O is incorrect.
(d) Higher the negative charges, larger is their ionic radii. $\text{Al}^{3+} < \text{Mg}^{2+} < \text{Na}^{+} < \text{F}^{-}$ is also true.
57 Based on lattice energy and other considerations which one of the following alkali metal chlorides is expected to have the highest melting point ?
(a) RbCl
(b) KCl
(c) NaCl
(d) LiCl
[AIEEE 2005]
Ans. (c)
As we go down in the group, ionic character increases, hence melting point of halides should increase but NaCl has the highest melting point (800°C) due to its high lattice energy.
58 Which one of the following ions has the highest value of ionic radius ?
(a) $\text{Li}^{+}$
(b) $\text{B}^{3+}$
(c) $\text{O}^{2-}$
(d) $\text{F}^{-}$
[AIEEE 2004]
Ans. (c)
All the ions belong to same period, thus for them cations will be smaller than anions. Now, $\text{O}^{2-}$ and $\text{F}^{-}$ are isoelectronic and $r_{i} \propto 1/Z$
Thus, ionic radius of $\text{O}^{2-}$ (Z=8) > $\text{F}^{-}$ (Z=9)
59 The formation of the oxide ion $\text{O}^{2-}(g)$ requires first an exothermic and then an endothermic step as shown below
$\text{O}(g) + e^{-} = \text{O}^{-}(g) \quad \Delta H^{\circ} = -142 \, \text{kJ mol}^{-1}$
$\text{O}^{-}(g) + e^{-} = \text{O}^{2-}(g) \quad \Delta H^{\circ} = 844 \, \text{kJ mol}^{-1}$
This is because
(a) oxygen is more electronegative
(b) oxygen has high electron affinity
(c) $\text{O}^{-}$ ion will tend to resist the addition of another electron
(d) $\text{O}^{-}$ ion has comparatively larger size than oxygen atom
[AIEEE 2004]
Ans. (c)
$\text{O}^{-}(g) + e^{-} \to \text{O}^{2-}(g) \quad \Delta H^{\circ} = 844 \, \text{kJ mol}^{-1}$
This process is unfavourable in the gas phase because the resulting increase in electron-electron repulsion outweighs the stability gained by achieving the noble gas configuration.
60 Which among the following factors is the most important in making fluorine the strongest oxidising agent ?
(a) Electron affinity
(b) Ionisation enthalpy
(c) Hydration enthalpy
(d) Bond dissociation energy
[AIEEE 2004]
Ans. (c)
The most important factor making fluorine the strongest oxidizing agent is (c) Hydration enthalpy.
While it might seem counterintuitive, the ability of an element to act as an oxidizing agent in aqueous solution is determined by its standard reduction potential, which is the sum of several energetic steps (the Born-Haber cycle for hydration).
To turn $F_2$ gas into $F^-$ ions in water, the following energy changes occur:
Bond Dissociation Energy ($+ \Delta H_{diss}$): The $F-F$ bond must be broken. Fluorine has an unusually low bond dissociation energy due to the strong repulsion between the lone pairs of its small atoms.
Electron Gain Enthalpy ($-\Delta H_{eg}$): An electron is added to the fluorine atom. Interestingly, Chlorine actually has a higher electron affinity than Fluorine. If this were the only factor, Chlorine would be the stronger oxidizer.
Hydration Enthalpy ($-\Delta H_{hyd}$): The resulting $F^-$ ion is surrounded by water molecules. Because the fluoride ion is very small, it has an extremely high charge density. This leads to very strong attractions with water molecules, releasing a massive amount of energy.
The extremely high (negative) Hydration Enthalpy of the small fluoride ion more than compensates for its slightly lower electron affinity compared to chlorine.
Key Point : The total energy released during the process ($F_2 \rightarrow 2F^-_{aq}$) is highest for Fluorine because of its small ionic size, which results in the most exothermic hydration step.
Comparison Table
| Factor | Fluorine (F) | Chlorine (Cl) |
| Electron Affinity | High | Highest |
| Bond Dissociation | Low (Easy to break) | High |
| Hydration Enthalpy | Extremely High | Moderate |
| Result | Strongest Oxidizer | Strong Oxidizer |
Therefore, the exceptionally large heat of hydration is the deciding factor that pushes Fluorine’s reduction potential to the top of the scale.
61 Among $\text{Al}_{2}\text{O}_{3}$, $\text{SiO}_{2}$, $\text{P}_{2}\text{O}_{5}$ and $\text{SO}_{2}$ the correct order of acid strength is
(a) $\text{SO}_{2} < \text{P}_{2}\text{O}_{5} < \text{SiO}_{2} < \text{Al}_{2}\text{O}_{3}$
(b) $\text{SiO}_{2} < \text{SO}_{2} < \text{Al}_{2}\text{O}_{3} < \text{P}_{2}\text{O}_{5}$
(c) $\text{Al}_{2}\text{O}_{3} < \text{SiO}_{2} < \text{SO}_{2} < \text{P}_{2}\text{O}_{5}$
(d) $\text{Al}_{2}\text{O}_{3} < \text{SiO}_{2} < \text{P}_{2}\text{O}_{5} < \text{SO}_{2}$
[AIEEE 2004]
Ans. (d)
1. Periodic Trend: Left to Right
As we move from left to right across a period:
Electronegativity increases: The central atom becomes better at attracting electrons.
Non-metallic character increases: Non-metals form acidic oxides, while metals form basic oxides.
Oxidation state increases: In this specific series, the oxidation states are $\text{Al}(+3)$, $\text{Si}(+4)$, $\text{P}(+5)$, and $\text{S}(+4/+6)$. Generally, a higher oxidation state and higher electronegativity increase the ability of the oxide to react with water to form an acid (or to accept electron pairs).
2. Characterizing the Oxides
$\text{Al}_{2}\text{O}_{3}$ (Amphoteric): It sits on the border between metals and non-metals. It can react with both acids and bases, making it the least acidic in this group.
$\text{SiO}_{2}$ (Weakly Acidic): Silicon is a metalloid/weak non-metal. Its oxide is acidic but requires strong bases to react.
$\text{P}_{2}\text{O}_{5}$ (Strongly Acidic): Phosphorus is a non-metal. It reacts with water to form phosphoric acid ($\text{H}_{3}\text{PO}_{4}$).
$\text{SO}_{2}$ (Very Strongly Acidic): Sulfur is further to the right than Phosphorus. It is highly electronegative and its oxide reacts with water to form sulfurous acid ($\text{H}_{2}\text{SO}_{3}$).
Note: As non-metallic character increases from $\text{Al} \rightarrow \text{Si} \rightarrow \text{P} \rightarrow \text{S}$, the acidic nature of their respective oxides increases accordingly.
The correct answer is (d).
Order: $\text{Al}_{2}\text{O}_{3} < \text{SiO}_{2} < \text{P}_{2}\text{O}_{5} < \text{SO}_{2}$
| Oxide | Central Element | Nature | Acid Formed with H2O |
| $\text{Al}_{2}\text{O}_{3}$ | Aluminum | Amphoteric | None (insoluble) |
| $\text{SiO}_{2}$ | Silicon | Weakly Acidic | Silicic Acid |
| $\text{P}_{2}\text{O}_{5}$ | Phosphorus | Strongly Acidic | Phosphoric Acid |
| $\text{SO}_{2}$ | Sulfur | Very Strongly Acidic | Sulfurous Acid |
62 The atomic numbers of vanadium (V), chromium (Cr), manganese (Mn) and iron (Fe) are, respectively 23, 24, 25 and 26. Which one of these may be expected to have the highest second ionisation enthalpy?
(a) V
(b) Cr
(c) Mn
(d) Fe
[AIEEE 2003]
Ans. (b)
$\text{Fe}^{+} (26) = [\text{Ar}] \, 3d^{6} \, 4s^{1}$
$\text{Mn}^{+} (25) = [\text{Ar}] \, 3d^{5} \, 4s^{1}$
$\text{V}^{+} (23) = [\text{Ar}] \, 3d^{4} \, 4s^{2}$
$\text{Cr}^{+} (24) = [\text{Ar}] \, 3d^{5} \, 4s^{0}$ (By first IP)
The electronic configuration of $\text{Cr}^{+}$ is most stable, hence formation of $\text{Cr}^{2+}$ by second IP requires maximum enthalpy.
63 $\text{Ce}^{3+}$, $\text{La}^{3+}$, $\text{Pm}^{3+}$ and $\text{Yb}^{3+}$ have ionic radii in the increasing order as
(a) $\text{La}^{3+} < \text{Ce}^{3+} < \text{Pm}^{3+} < \text{Yb}^{3+}$
(b) $\text{Yb}^{3+} < \text{Pm}^{3+} < \text{Ce}^{3+} < \text{La}^{3+}$
(c) $\text{La}^{3+} < \text{Ce}^{3+} < \text{Pm}^{3+} < \text{Yb}^{3+}$
(d) $\text{Yb}^{3+} < \text{Pm}^{3+} < \text{La}^{3+} < \text{Ce}^{3+}$
[AIEEE 2002]
Ans. (b)
$r_{i}$ (radius) ∝ 1/Z
Thus, $\text{Yb}^{3+} < \text{Pm}^{3+} < \text{Ce}^{3+} < \text{La}^{3+}$.
Summary
To build a strong foundation in this topic, students should also study Modern Periodic Law and Modern Periodic Table, along with Structure of Atom, which explains the electronic configuration behind periodic trends. For complete preparation, explore Class 11 Chemistry Chapter-wise Notes and practice more JEE Main PYQs from related topics available on this website. Regular revision and practice will help improve speed, accuracy, and confidence in the exam.
FAQs – JEE Main PYQs (Previous Year Questions (MCQs) : Classification of Elements and Periodicity in Properties
Q1. What does this PYQ section cover?
This section covers chapter-wise and topic-wise JEE Main Previous Year Questions (MCQs) from Classification of Elements and Periodicity in Properties, including trends like atomic radius, ionization enthalpy, electronegativity, and periodic behavior.
Q2. Who has written these PYQs and study material?
These questions and solutions are prepared by Er Neeraj Anand, Chemistry Anand Classes, under Anand Classes, and are published by Anand Technical Publishers.
Q3. Why are PYQs important for JEE Main preparation?
PYQs help students understand the exam pattern, frequently asked concepts, and difficulty level. They improve accuracy, speed, and confidence for solving MCQs in the actual exam.
Q4. Are detailed solutions provided with the questions?
Yes, each question is supported with clear and step-by-step solutions to help students understand the concept and approach required to solve it.
Q5. Which topics are most important in this chapter for JEE Main?
Important topics include:
Atomic and ionic radii
Ionization enthalpy
Electron gain enthalpy
Electronegativity
Periodic trends and variations
Q6. Is this material suitable for beginners?
Yes, the content is designed in a simple and structured way, making it suitable for both beginners and advanced learners.
Q7. How should students practice these PYQs effectively?
Students should first revise theory, then attempt PYQs topic-wise, analyze mistakes, and repeat practice to improve speed and accuracy.
Q8. Does this section follow the latest JEE Main syllabus?
Yes, all questions are selected and organized according to the latest JEE Main syllabus and exam pattern.
Q9. Can PYQs help in scoring high marks in JEE Main?
Yes, many questions in JEE Main are directly or indirectly based on previous year patterns, so practicing PYQs can significantly boost scores.
Q10. Is this resource useful for other exams?
Yes, it is also helpful for exams like NEET and other entrance tests where similar concepts of periodicity are asked.
Q11. How is this content different from other study material?
This content focuses on concept clarity, exam-oriented preparation, and structured learning, making it easy to understand and revise.
Q12. Where can students find more related study material?
Students can explore other Class 11 Chemistry chapters, topic-wise notes, and additional PYQs on this website for complete preparation.