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JEE • CHEMISTRY • CLASSIFICATION OF ELEMENTS AND PERIODICITY IN PROPERTIES

Classification of elements and periodicity in properties practice · Mixed difficulty

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These sets use single-correct-answer MCQs. They do not cover every JEE Main or JEE Advanced question format and are not full-length mock papers.

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Showing 1-30 of 426 questions
1 Easy
1 of 426
Understanding +4 | -1
The compounds of the s-block elements, with the exception of lithium and ... XX... are predominantly ionic. Here, XX refers to
View answer and explanation
Correct option: D
Beryllium
The correct answer is D (Beryllium) because, with the exception of lithium and beryllium, compounds of s-block elements are predominantly ionic.
2 Medium
2 of 426
Remembering +4 | -1
Among the following elements, the most electronegative is:
View answer and explanation
Correct option: D
Fluorine
Option D (Fluorine) is correct because electronegativity increases across a period and decreases down a group, making Fluorine the most electronegative element among the given options.
3 Hard
3 of 426
Remembering +4 | -1
The pair of amphoteric hydroxides is
View answer and explanation
Correct option: D
Be(OH)2,Zn(OH)2\mathrm{Be}{\left( {\mathrm{OH}} \right)_2},\mathrm{Zn}{\left( {\mathrm{OH}} \right)_2}
Option D (Be(OH)2,  Zn(OH)2\mathrm{Be}{\left( {\mathrm{OH}} \right)_2},\mathrm{\;Zn}{\left( {\mathrm{OH}} \right)_2}) is correct because both beryllium hydroxide and zinc hydroxide exhibit amphoteric behavior.
4 Easy
4 of 426
Understanding +4 | -1
The element with atomic number 117 if discovered would be placed in
View answer and explanation
Correct option: D
Halogen family
Option A (Basic solution) is correct because ionic hydrides react with water to form a metal hydroxide and hydrogen gas, resulting in a basic solution.
5 Medium
5 of 426
Understanding +4 | -1
The correct order of size among O,O2−,F\mathrm{O}, \mathrm{O}^{2-}, \mathrm{F} and F−\mathrm{F}^{-} is:
View answer and explanation
Correct option: D
O2−>F−>O>F\mathrm{O}^{2-}>\mathrm{F}^{-}>\mathrm{O}>\mathrm{F}
Option D (O2−>F−>O>F\mathrm{O}^{2-}>\mathrm{F}^{-}>\mathrm{O}>\mathrm{F}) is correct because anions are larger than their neutral atoms, and for isoelectronic species, size decreases with increasing nuclear charge.
6 Medium
6 of 426
Analyzing +4 | -1
The electronic configuration of four elements are given below. Which elements does not belong to the same family as others?
View answer and explanation
Correct option: C
[Ne]3s23p5[\mathrm{Ne}] 3 s^2 3 p^5
Option C ([Ne] 3s² 3p⁵) is correct because it is the only element with 7 valence electrons, belonging to Group 17, while the others have 2 valence electrons.
7 Easy
7 of 426
Analyzing +4 | -1
First, second and third IEIE values are 100 eV, 150 eV and 1500 eV. Element can be
View answer and explanation
Correct option: A
BeBe
The correct answer is A (Be) because the large jump in IE between IE2 and IE3 indicates that the element readily loses two electrons to achieve a stable configuration.
8 Easy
8 of 426
Remembering +4 | -1
Variable valency is exhibited by
View answer and explanation
Correct option: C
Transitional elements
Option C (Transitional elements) is correct because transition elements exhibit variable valency due to the involvement of d-electrons in bonding.
9 Easy
9 of 426
Understanding +4 | -1
Element of highest first negative electron gain enthalpy is
View answer and explanation
Correct option: D
ClCl
Option D (ClCl) is correct because Chlorine has the highest first negative electron gain enthalpy due to its smaller size and effective nuclear charge compared to Fluorine.
10 Medium
10 of 426
Remembering +4 | -1
Which of the following set of oxides are amphoteric
View answer and explanation
Correct option: D
Al2O3,ZnO,SnO2,Sb4O6A{l_2}{O_3},ZnO,Sn{O_2},S{b_4}{O_6}
Option D (Al2O3,ZnO,SnO2,Sb4O6A{l_2}{O_3},ZnO,Sn{O_2},S{b_4}{O_6}) is correct because these oxides exhibit both acidic and basic properties.
11 Easy
11 of 426
Understanding +4 | -1
The oxide of an element whose electronic configuration is 1s2,  2s2,2p6,3s1  1{s^2},\;2{s^2},2{p^6},3{s^1}\; is
View answer and explanation
Correct option: C
Basic
Option C (Basic) is correct because the element with the given electronic configuration is sodium, which forms a basic oxide (Na2O).
12 Easy
12 of 426
Application +4 | -1
E. N of the element A is E1{E_1} and I.E is E2{E_2}. Hence, E. A will be
View answer and explanation
Correct option: A
2E1−E22{E_1} - {E_2}
Option A (2E1−E22{E_1} - {E_2}) is correct because it is derived from the formula relating electronegativity (E.N), ionization energy (I.E), and electron affinity (E.A).
13 Easy
13 of 426
Understanding +4 | -1
Which of the following species has the negative electron gain enthalpy?
View answer and explanation
Correct option: B
O\mathrm{O}
The correct answer is B (O) because oxygen has a high electron affinity and releases energy when it gains an electron.
14 Easy
14 of 426
Analyzing +4 | -1
Which of the following pairs of atomic numbers represents elements belonging to the same group?
View answer and explanation
Correct option: C
13 and 31
Option C (13 and 31) is correct because elements with atomic numbers 13 (Al) and 31 (Ga) both belong to group 13 of the periodic table.
15 Medium
15 of 426
Understanding +4 | -1
Arrange the elements Se,Cl\mathrm{Se}, \mathrm{Cl} and S\mathrm{S} in the increasing order of ionisation energy
View answer and explanation
Correct option: B
Se<S<Cl\mathrm{Se} < \mathrm{S} < \mathrm{Cl}
Option B (Se<S<Cl\mathrm{Se} < \mathrm{S} < \mathrm{Cl}) is correct because ionization energy generally increases across a period and decreases down a group.

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16 Medium
16 of 426
Analyzing +4 | -1
The decreasing order of the ionization potential of the following elements is
View answer and explanation
Correct option: B
Ne>Cl>P>S>Mg>Al\mathrm{Ne}>\mathrm{Cl}>\mathrm{P}>\mathrm{S}>\mathrm{Mg}>\mathrm{Al}
Option A (Ne > Cl > P > S > Al > Mg) is correct because it accurately reflects the general trend of ionization potential, considering exceptions due to electronic configurations.
17 Easy
17 of 426
Remembering +4 | -1
Which of the following electronic configuration represents noble gas?
View answer and explanation
Correct option: A
ns2np6n{s^2}n{p^6}
Option A (ns2np6n{s^2}n{p^6}) is correct because noble gases have a completely filled valence shell, represented by this electronic configuration.
18 Easy
18 of 426
Remembering +4 | -1
The ability of an atom in a chemical compound to attract shared electron is termed as
View answer and explanation
Correct option: D
Electronegativity
Option D (Electronegativity) is correct because electronegativity is the measure of an atom's ability to attract shared electrons in a chemical bond.
19 Medium
19 of 426
Understanding +4 | -1
Select the ion of largest size in gas phase among the given species
View answer and explanation
Correct option: D
N3−{N}^{3-}
Option D (N3−{N}^{3-}) is correct because for isoelectronic species, the size increases with increasing negative charge.
20 Easy
20 of 426
Remembering +4 | -1
Eka-aluminium was later called :
View answer and explanation
Correct option: C
Gallium
Option C (Gallium) is correct because Eka-aluminium, predicted by Mendeleev, was later identified as Gallium.
21 Easy
21 of 426
Understanding +4 | -1
The element with highest electron affinity will belong to
View answer and explanation
Correct option: B
Period 3, group 17
Option B (Period 3, group 17) is correct because chlorine (Cl), in period 3 and group 17, has the highest electron affinity due to its electronic configuration.
22 Medium
22 of 426
Remembering +4 | -1
The pair of amphoteric oxides among the following is
View answer and explanation
Correct option: C
ZnO,Al2O3{ZnO}, {Al}_2 {O}_3
Option C (ZnO,Al2O3{ZnO}, {Al}_2 {O}_3) is correct because both zinc oxide and aluminum oxide can act as both acids and bases.
23 Easy
23 of 426
Remembering +4 | -1
By what name the elements after Uranium are called :
View answer and explanation
Correct option: D
Transuranium elements
Option D (Transuranium elements) is correct because elements with atomic numbers greater than that of uranium are called transuranium elements.
24 Easy
24 of 426
Analyzing +4 | -1
The electronic configuration of the atom having maximum difference in first and second ionisation energies is
View answer and explanation
Correct option: A
1s2,2s2,2p6,3s11{s^2},2{s^2},2{p^6},3{s^1}
Option A (1s2,2s2,2p6,3s11{s^2},2{s^2},2{p^6},3{s^1}) is correct because after removing one electron, it achieves a stable noble gas configuration, resulting in a significantly higher second ionization energy.
25 Easy
25 of 426
Analyzing +4 | -1
In which of the following arrangements the order is not correct according to property indicated against it?
View answer and explanation
Correct option: B
Increasing IE1  :B<C<N<OI{E_1}\mathrm{\;}:\mathrm{B} < C < N < O
The correct answer is B (Increasing IE1  :  B<C<N<OI{E_1}{\;}:{\;}\mathrm{B} < C < N < O) because the first ionization energy of Nitrogen is greater than Oxygen due to the stable half-filled electronic configuration of Nitrogen.
26 Medium
26 of 426
Understanding +4 | -1
The correct order of electron affinities is:
View answer and explanation
Correct option: A
Cl>Si>Na>ArCl > Si > Na > Ar
Option A is correct because electron affinity generally increases across a period and decreases down a group, with noble gases having very low electron affinities.
27 Medium
27 of 426
Analyzing +4 | -1
The five successive ionisation enthalpies of an element are 800,2427,3658,25024  and  32824  kJ mol−1.\mathrm{800, 2427, 3658, 25024 \,\,and\,\, 32824\,\, kJ \,mo}{\mathrm{l}^{ - 1}}. The number of valence electrons in the element is
View answer and explanation
Correct option: A
3
Option A (3) is correct because there is a significant jump in ionization enthalpy after the third ionization, indicating that the element has 3 valence electrons.
28 Medium
28 of 426
Analyzing +4 | -1
Assertion :The 1st1^{\text {st}} IP of Be\mathrm{Be} is greater than that of B\mathrm{B}.

Reason : 2p orbital is lower in energy than 2s.
View answer and explanation
Correct option: C
If Assertion is True but the Reason is False
Option C is correct. The first ionization potential (IP) of Be is greater than that of B because Be has a stable, fully filled 2s orbital, while B has a partially filled 2p orbital. The reason is false because 2s orbital is lower in energy than 2p.
29 Medium
29 of 426
Analyzing +4 | -1
In which of the following have higher difference in the value of IInd and IIIrd I.P.
View answer and explanation
Correct option: C
1s22s22p63s11{s}^2 2 {s}^2 2 {p}^6 3 {s}^1
Option C (1s22s22p63s11{s}^2 2 {s}^2 2 {p}^6 3 {s}^1) is correct because removing the second electron from Na+Na^+ requires breaking the stable noble gas configuration, leading to a large jump in ionization potential.
30 Easy
30 of 426
Analyzing +4 | -1
Which of the following ions is most unlikely to exist?
View answer and explanation
Correct option: A
Li−{Li}^{-}
The correct answer is A (Li-) because lithium is an electropositive metal and does not readily form anions.
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