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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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Showing 211-240 of 426 questions

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211 Hard
211 of 426
Analyzing +4 | -1
The third ionisation enthalpy is minimum for:
View answer and explanation
Correct option: D
FeFe
Option D (FeFe) is correct because Fe2+Fe^{2+} has a 3d63d^6 configuration, and removing the third electron results in the stable 3d53d^5 configuration, requiring less energy.
212 Medium
212 of 426
Analyzing +4 | -1
The electronic configuration having maximum difference in first and second ionization energies is
View answer and explanation
Correct option: D
1s22s22p63s11 s^2 2 s^2 2 p^6 3 s^1
The correct answer is D (1s22s22p63s11 s^2 2 s^2 2 p^6 3 s^1) because removing the second electron from this configuration requires significantly more energy as it involves breaking into a stable, completely filled s-orbital.
213 Medium
213 of 426
Analyzing +4 | -1
Assertion : Hydrated ionic radius of Li+\mathrm{Li}^{+} is maximum in IA group elements.

Reason : Lithium is a metal.
View answer and explanation
Correct option: B
If both Assertion & Reason True but Reason not a correct explanation of the assertion
Option B is correct because both the assertion and reason are true, but the reason does not explain why the hydrated ionic radius of Li+ is maximum.
214 Easy
214 of 426
Understanding +4 | -1
The first ionisation potential (eV) of Be and B respectively are
View answer and explanation
Correct option: D
9.32 eV, 8.29 eV
Option C (Dioxygen is paramagnetic) is correct because dioxygen (O2\mathrm{O_2}) has unpaired electrons in its molecular orbital configuration, resulting in its paramagnetic property.
215 Easy
215 of 426
Understanding +4 | -1
Highest electron affinity is associated with the configuaration
View answer and explanation
Correct option: B
2S22P22{S^2}2{P^2}
Option D (2S22P12{S^2}2{P^1}) is correct because adding an electron to the 2p2p orbital will lead to a more stable configuration.
216 Medium
216 of 426
Understanding +4 | -1
The radii of F2 F−,O\mathrm{F}_2 \mathrm{~F}^{-}, \mathrm{O} and O2−\mathrm{O}^{2-} are in the order
View answer and explanation
Correct option: D
O2−>F−>O>F\mathrm{O}^{2-}>\mathrm{F}^{-}>\mathrm{O}>\mathrm{F}
The correct answer is D (O2−>F−>O>F\mathrm{O}^{2-}>\mathrm{F}^{-}>\mathrm{O}>\mathrm{F}) because ionic radius increases with negative charge and decreases across a period.
217 Medium
217 of 426
Application +4 | -1
An element, which belongs to third period and group 16 in the periodic table has electronic configuration.
View answer and explanation
Correct option: B
1 s2,2 s22p6,3 s23p41 {~s}^2, 2 {~s}^2 2 {p}^6, 3 {~s}^2 3 {p}^4
Option B (1 s2,2 s22p6,3 s23p41 {~s}^2, 2 {~s}^2 2 {p}^6, 3 {~s}^2 3 {p}^4) is correct because it represents an element in the third period (3 shells) and group 16 (6 valence electrons).
218 Easy
218 of 426
Analyzing +4 | -1
Statement-I: In pp and dd block both metals and non-metals are present.
Statement-II: Electronegativity and ionisation enthalpy of metals is greater than non-metals.
View answer and explanation
Correct option: B
Both statement-I and II are incorrect
Option B (Both statement-I and II are incorrect) is correct because d-block contains only metals, and non-metals have higher electronegativity and ionization enthalpy than metals.
219 Medium
219 of 426
Understanding +4 | -1
The correct electron affinity order of N,  O,  S,  ClN,\,\,O,\,\,S,\,\,Cl is
View answer and explanation
Correct option: C
N<O<S<ClN < O < S < Cl
Option C (N<O<S<ClN < O < S < Cl) is correct because electron affinity generally increases across a period and decreases down a group, with exceptions due to electronic configuration.
220 Medium
220 of 426
Analyzing +4 | -1
Electronic configurations of four elements A, B, C\mathrm{C} and D\mathrm{D} are given below :

(i) 1s22s22p61 s^2 2 s^2 2 p^6

(ii) 1s22s22p41 s^2 2 s^2 2 p^4

(iii) 1s22s22p63s11 s^2 2 s^2 2 p^6 3 s^1

(iv) 1s22s22p51 s^2 2 s^2 2 p^5

Which of the following is the correct order of increasing tendency to gain electron?
View answer and explanation
Correct option: A
(i) << (iii) << (ii) << (iv)
Option A ((i) < (iii) < (ii) < (iv)) is correct because the electron gain enthalpy increases as the effective nuclear charge increases and noble gases have the least tendency to gain electrons.
221 Easy
221 of 426
Understanding +4 | -1
Which of the following ion with the smallest ionic radius is
View answer and explanation
Correct option: D
Ti+4T{i^{ + 4}}
Option D (Ti4+Ti^{4+}) has the smallest ionic radius because, for isoelectronic species, the ionic radius decreases with increasing nuclear charge (number of protons).
222 Medium
222 of 426
Remembering +4 | -1
The atomic number of the element unnilennium is
View answer and explanation
Correct option: A
109
Option A (109) is correct because the element unnilennium corresponds to atomic number 109 based on IUPAC nomenclature rules.
223 Easy
223 of 426
Remembering +4 | -1
In solid argon, the atoms are held together by
View answer and explanation
Correct option: C
Vander Waals forces
Option C (Vander Waals forces) is correct because in solid argon, the atoms are held together by weak van der Waals forces due to temporary fluctuations in electron distribution.
224 Easy
224 of 426
Remembering +4 | -1
The elements in which 4 f- orbitals are progressively filled up are called as
View answer and explanation
Correct option: C
Lanthanoids
Option C (Lanthanoids) is correct because lanthanoids are the elements in which the 4f orbitals are progressively filled.
225 Medium
225 of 426
Understanding +4 | -1
Arrange N, O and S in order of decreasing electron affinity
View answer and explanation
Correct option: A
S > O > N
Option A (S > O > N) is correct because electron affinity generally increases across a period and decreases down a group, but oxygen has a lower electron affinity than sulfur due to its smaller size and higher electron density.

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226 Medium
226 of 426
Understanding +4 | -1
Which of the following transition involves maximum amount of energy
View answer and explanation
Correct option: D
M(g)2+→M(g)3+ M_{\left( g \right)}^{2 + } \to M_{\left( g \right)}^{3 + }\,
Option D (M(g)2+→M(g)3+ M_{\left( g \right)}^{2 + } \to M_{\left( g \right)}^{3 + }\,) is correct because ionization enthalpy increases with each successive ionization.
227 Medium
227 of 426
Remembering +4 | -1
The electronic configuration of inner transition elements is
View answer and explanation
Correct option: D
ns2(n−1)d0−1(n−2)f1−14n s^2(n-1) d^{0-1}(n-2) f^{1-14}
Option D (ns2(n−1)d0−1(n−2)f1−14n s^2(n-1) d^{0-1}(n-2) f^{1-14}) is correct because it accurately represents the filling of the (n-2)f orbitals in inner transition elements.
228 Easy
228 of 426
Application +4 | -1
The ratio of radius of 2nd and 3rd Bohr orbit is
View answer and explanation
Correct option: D
0.172916667
Option D (04:09) is correct because the radius of a Bohr orbit is proportional to n2, where n is the principal quantum number.
229 Easy
229 of 426
Analyzing +4 | -1
The type of oxide formed by the element among
Li, Na, Be, Mg, B and Al that has the least atomic
radius is:
View answer and explanation
Correct option: A
A2O3\mathrm{A}_2 \mathrm{O}_3
Option A (A2O3\mathrm{A}_2 \mathrm{O}_3) is correct because Boron (B) has the smallest atomic radius among the given elements and forms an oxide with the formula B2O3\mathrm{B}_2 \mathrm{O}_3.
230 Easy
230 of 426
Analyzing +4 | -1
Which of the following does not represent the correct order of properties indicated?
View answer and explanation
Correct option: D
Li<Be<B<C(IE1)\mathrm{Li} < \mathrm{Be} < \mathrm{B} < \mathrm{C}\left(\mathrm{IE}_1\right)
Option D is correct because the ionization energy (IE1) generally increases across a period, but there are exceptions due to electronic configurations; Be has a higher IE1 than B.
231 Medium
231 of 426
Understanding +4 | -1
The electron density between 1s and 2s is
View answer and explanation
Correct option: C
Zero
Option C (Zero) is correct because there is a node between the 1s and 2s orbitals where the probability of finding an electron is zero.
232 Medium
232 of 426
Understanding +4 | -1
Electronic configuration of four elements A, B, C and D are given below.
A) 1S22S22P61{S^2}2{S^2}2{P^6} B) 1S22S22P41{S^2}2{S^2}2{P^4} C) 1S22S22P63S11{S^2}2{S^2}2{P^6}3{S^1} D) 1S22S22P51{S^2}2{S^2}2{P^5}
Which of the following is the correct order of increasing tendency to gain electron.
View answer and explanation
Correct option: A
A < C < B < D
Option A (A < C < B < D) is correct because it represents the increasing order of electron gain tendency based on electronic configurations and noble gas stability.
233 Hard
233 of 426
Analyzing +4 | -1
The correct order of ionic radius is
View answer and explanation
Correct option: D
P3+>P5+{\mathrm{P}^{3 + }} > {\mathrm{P}^{5 + }}
The correct answer is D (P3+>P5+){\mathrm{P}^{3 + }} > {\mathrm{P}^{5 + }}) because ionic radius is inversely proportional to effective nuclear charge, and P³⁺ has a lower effective nuclear charge than P⁵⁺.
234 Medium
234 of 426
Analyzing +4 | -1
The first second and third ionisation energies (E1,E2  and  E3)\left( {{E_1},{E_2}\,\,and\,\,{E_3}} \right) for an element are 7  eV,12.5  eV7\,\,eV,12.5\,\,eV and 42.5  eV42.5\,\,eV respectively. The most stable oxidation state of the element will be
View answer and explanation
Correct option: D
+2+ 2
Option D (+2) is correct because there is a large jump in ionization energy between E2 and E3, indicating that the element readily loses two electrons to achieve a stable configuration.
235 Easy
235 of 426
Understanding +4 | -1
The I.E of Al is smaller than that of Mg because
View answer and explanation
Correct option: C
I.E in Al pertains to the removal of p – electron which is relatively easy
Option C (I.E in Al pertains to the removal of p – electron which is relatively easy) is correct because Aluminum's outermost electron is in a p-orbital, which is easier to remove than Magnesium's s-orbital electrons.
236 Medium
236 of 426
Understanding +4 | -1
Which among the following elements has lowest value of ionisation energy?
View answer and explanation
Correct option: B
Sn
Option B (Sn) is correct because, down the group, ionization energy decreases, and due to the inert pair effect, Pb has a higher ionization energy than Sn.
237 Easy
237 of 426
Application +4 | -1
Which pair of the atomic numbers represents s-block elements?
View answer and explanation
Correct option: A
3, 12
Option A (3, 12) is correct because elements with atomic numbers 3 and 12 have their last electron entering the s-orbital, making them s-block elements.
238 Easy
238 of 426
Application +4 | -1
A dipositive ion has number of e−{e^ - } equal to 24 and mass number 56, the number of neutrons present is
View answer and explanation
Correct option: C
30
Option C (30) is correct because the number of neutrons is the difference between the mass number and the atomic number (number of protons).
239 Easy
239 of 426
Analyzing +4 | -1
Which of the following sequence correctly represents the decreasing acidic nature of oxides?
View answer and explanation
Correct option: C
N2O3>CO2>B2O3>BeO>Li2O{\mathrm{N}_2}{\mathrm{O}_3} > \mathrm{C}{\mathrm{O}_2} > {\mathrm{B}_2}{\mathrm{O}_3} > \mathrm{BeO} > \mathrm{L}{\mathrm{i}_2}\mathrm{O}
Option C (N2O3>CO2>B2O3>BeO>Li2O{\mathrm{N}_2}{\mathrm{O}_3} > \mathrm{C}{\mathrm{O}_2} > {\mathrm{B}_2}{\mathrm{O}_3} > \mathrm{BeO} > \mathrm{L}{\mathrm{i}_2}\mathrm{O}) is correct because acidic character increases with electronegativity across a period.
240 Medium
240 of 426
Understanding +4 | -1
For elements B, C, N Li, Be, O and F, the correct order of first ionization enthalpy is
View answer and explanation
Correct option: D
Li<B<Be<C<O<N<F\mathrm{Li} < \mathrm{B} < \mathrm{Be} < \mathrm{C} < \mathrm{O} < \mathrm{N} < \mathrm{F}
Option D (Li<B<Be<C<O<N<F\mathrm{Li} < \mathrm{B} < \mathrm{Be} < \mathrm{C} < \mathrm{O} < \mathrm{N} < \mathrm{F}) is correct because it accounts for the exceptions due to stable electronic configurations in Be and N.
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