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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 361-390 of 426 questions

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361 Easy
361 of 426
Analyzing +4 | -1
The first four ionisation energy values of an element are 191, 578, 872 and 5962 kcal. The number of valence electrons in the element is
View answer and explanation
Correct option: C
3
Option C (3) is correct because the large jump in ionization energy between the third and fourth ionization energies indicates that there are 3 valence electrons.
362 Medium
362 of 426
Understanding +4 | -1
Which of the following represents the correct order of increasing negative electron gain enthalpy of the elements?
View answer and explanation
Correct option: C
O<Te<Se<S\mathrm{O} < \mathrm{Te} < \mathrm{Se} < \mathrm{S}
Option C is correct. Electron gain enthalpy becomes less negative down a group, but oxygen has a less negative value than sulfur due to its small size.
363 Medium
363 of 426
Analyzing +4 | -1
Statement I: Ionisation potential of BeBe is less than B.B.
Statement II: The first electron released from BeBe is of s-orbital but that from BB is of p-orbital.
View answer and explanation
Correct option: D
If statement 1 is false but statement 2 is true
Option D is correct because Statement I is false; Be has a higher ionization potential than B due to its stable, filled s-orbital configuration, while Statement II is true.
364 Medium
364 of 426
Understanding +4 | -1
The process requiring the absorption of energy is
View answer and explanation
Correct option: C
O−→O2−{O^ - } \to {O^{2 - }}
The correct answer is C (O−→O2−{O^ - } \to {O^{2 - }}) because adding an electron to an already negatively charged ion requires energy due to electron-electron repulsion.
365 Easy
365 of 426
Remembering +4 | -1
Number of elements presents in the fifth period of periodic table is
View answer and explanation
Correct option: C
18
Option C (18) is correct because the fifth period of the periodic table contains 18 elements, ranging from Rubidium (Rb) to Xenon (Xe).
366 Easy
366 of 426
Understanding +4 | -1
Although electron gain enthalpy of fluorine is less negative as compared to chlorine, fluorine is a stronger oxidising agent then chlorine. This is attributed to
View answer and explanation
Correct option: D
All the above
Option D (All the above) is correct because the lower electron gain enthalpy of fluorine compared to chlorine is compensated by its low bond dissociation enthalpy and high hydration enthalpy, leading to a higher electrode potential.
367 Easy
367 of 426
Remembering +4 | -1
Atomic radii of fluorine and neon in angstrom unit are respectively given by:
View answer and explanation
Correct option: A
0.72,1.600.72,\mathrm{ }1.60
Option A (0.72,1.600.72,\mathrm{ }1.60) is correct because the atomic radius of fluorine is smaller than the van der Waals radius of neon.
368 Easy
368 of 426
Remembering +4 | -1
At present, how many elements are known
View answer and explanation
Correct option: D
118
Option D (118) is correct because, as of now, there are 118 known elements in the periodic table, including both naturally occurring and synthetic elements.
369 Medium
369 of 426
Understanding +4 | -1
Among the elements Ca, Mg, P and Cl, the order of increasing atomic radii is
View answer and explanation
Correct option: D
Cl < P < Mg < Ca
Option D (Cl < P < Mg < Ca) is correct because atomic radius increases down a group and decreases across a period.
370 Easy
370 of 426
Analyzing +4 | -1
The incorrect order is
View answer and explanation
Correct option: C
Cl<F<Br<\mathrm{Cl}<\mathrm{F}<\mathrm{Br}< I (Electronegativity)
Option C (Cl < F < Br < I (Electronegativity)) is correct because electronegativity decreases down the group, so the correct order is F > Cl > Br > I.
371 Medium
371 of 426
Analyzing +4 | -1
Which of the following should be the order of increasing values of second ionisation potential of C, N, O and F
View answer and explanation
Correct option: D
C < N < F < O
Option C (C < F < N < O) is correct because it reflects the increasing trend of second ionization potential considering electronic configurations and stability.
372 Easy
372 of 426
Understanding +4 | -1
Elements having similar outer shell electronic configuration in their atoms are arranged in
View answer and explanation
Correct option: D
All of these
Option D (All of these) is correct because elements with similar outer shell electronic configurations are arranged in groups, vertical columns, and families in the periodic table.
373 Medium
373 of 426
Analyzing +4 | -1
Which of the following has the largest size
View answer and explanation
Correct option: A
N−3{N^{ - 3}}
Option A (N−3{N^{ - 3}}) is correct because, among the given isoelectronic species, the one with the largest negative charge will have the largest size due to increased electron-electron repulsion and decreased effective nuclear charge.
374 Medium
374 of 426
Analyzing +4 | -1
M+3{M^{ + 3}} ion has the electronic configuration as [Xe]  5d104f146s2,\left[ {Xe} \right]\,\,5{d^{10}}4{f^{14}}6{s^2}, hence the element MM belongs to
View answer and explanation
Correct option: B
p-block
Option B (p-block) is correct because after removing three electrons, the electronic configuration ends with 6s² 5d¹⁰ 4f¹⁴ 6p³, indicating that the element M belongs to the p-block.
375 Medium
375 of 426
Analyzing +4 | -1
Which of the following statement(s) is/are true?
View answer and explanation
Correct option: D
All of these
Option D (All of these) is correct because all the given statements regarding ionization energy, ionization energies of Be and Mg, and atomic radii of Nb and Ta are factually true.
376 Easy
376 of 426
Remembering +4 | -1
Electronegativity refers to the tendency of atom to
View answer and explanation
Correct option: D
Attract bonding electrons
Option D (Attract bonding electrons) is correct because electronegativity is defined as the tendency of an atom to attract bonding electrons in a chemical bond.
377 Medium
377 of 426
Understanding +4 | -1
Assertion (A): Na+\mathrm{Na}^{+} and Al3+\mathrm{Al}^{3+} are isoelectronic but the ionic radii of Al3+\mathrm{Al}^{3+} is smaller than Na+\mathrm{Na}^{+}
Reason (R): The nuclear attraction on the outermost shell electrons of Al3+\mathrm{Al}^{3+} is greater than that of Na+\mathrm{Na}^{+}.
View answer and explanation
Correct option: A
Both Assertion and Reason are correct; Reason is the correct explanation of Assertion
Option A (Both Assertion and Reason are correct; Reason is the correct explanation of Assertion) is correct because both ions are isoelectronic, and the higher nuclear charge of Al³⁺ results in a smaller ionic radius.
378 Easy
378 of 426
Analyzing +4 | -1
Read the following three statements and choose the correct option. Here T\mathrm{T} stands for true and F\mathrm{F} stands for false statement.

(i) Boron has a smaller first ionization enthalpy than beryllium.

(ii) Nitrogen has smaller first ionization enthalpy than oxygen.

(iii) The first ionization enthalpy increases across a period.
View answer and explanation
Correct option: B
TFT
Option A (FTT) is incorrect. The correct answer is TFT because Boron has a smaller first ionization enthalpy than beryllium, Nitrogen has larger first ionization enthalpy than oxygen, and the first ionization enthalpy increases across a period.
379 Easy
379 of 426
Understanding +4 | -1
The electron configuration of an element ‘M’ is [Ar]  3d104s24p3.\left[ {Ar} \right]\,\,3{d^{10}}4{s^2}4{p^3}. Then ‘M’ belongs to
View answer and explanation
Correct option: C
VA−group{{\text{V}}_{\text{A}}} - {\text{group}}
Option C (VA−group{{\text{V}}_{\text{A}}} - {\text{group}}) is correct because the element 'M' has the electronic configuration ending with 4s24p34s^2 4p^3, which is the general configuration for group VA elements.
380 Medium
380 of 426
Understanding +4 | -1
Statement I: Ionic radius of O2−\mathrm{O}^{2-} is greater than F−.\mathrm{F}^{-}.

Statement II: O2−\mathrm{O}^{2-} and F−\mathrm{F}^{-}are isoelectronic species.

In light of the above statements, choose the correct answer from the options given below
View answer and explanation
Correct option: A
Both Statement I and Statement II are correct
Option A (Both Statement I and Statement II are correct) is correct because O2−\mathrm{O}^{2-} and F−\mathrm{F}^{-} are isoelectronic, and O2−\mathrm{O}^{2-} has a lower effective nuclear charge, resulting in a larger ionic radius.
381 Easy
381 of 426
Remembering +4 | -1
Atomic number 116 is written in IUPAC as
View answer and explanation
Correct option: A
Ununhexium
Option A (Ununhexium) is correct because according to IUPAC nomenclature, 116 is named Ununhexium.
382 Easy
382 of 426
Understanding +4 | -1
The element with atomic number 36 belongs to …block in the Periodic Table.
View answer and explanation
Correct option: A
p
Option A (p) is correct because the element with atomic number 36 (Krypton) has its last electron in the p-block.
383 Medium
383 of 426
Analyzing +4 | -1
The IP1,IP2,IP3,IP4,and IP5\mathrm{I}{\mathrm{P}_1},\mathrm{I}{\mathrm{P}_2},\mathrm{I}{\mathrm{P}_3},\mathrm{I}{\mathrm{P}_4},\mathrm{ and } \, {\mathrm{IP}_5} of an element are 7.1,14.3,34.57.1,\mathrm{ }14.3,\mathrm{ }34.5, 46.846.8,162.2  eV162.2 \, \, \text{eV} respectively. The element is likely to be:
View answer and explanation
Correct option: B
Si\mathrm{Si}
Option B (Si) is correct because the large jump in IP5 indicates that the element has 4 valence electrons, which is characteristic of Silicon.
384 Easy
384 of 426
Analyzing +4 | -1
In which of the following species 2ndI.E<1stI.E{2^{nd}}I.E\mathrm{ } < \mathrm{ }{1^{st}}I.E
View answer and explanation
Correct option: D
None
Option D (None) is correct because the second ionization energy is always greater than the first ionization energy for any element or ion.
385 Easy
385 of 426
Analyzing +4 | -1
Statement I: FF is more electronegative than Cl.Cl.

Statement II: FF has high electron affinity than Cl.Cl.
View answer and explanation
Correct option: C
If statement 1 is true but statement 2 is false
Option C (If statement 1 is true but statement 2 is false) is correct because Fluorine is more electronegative than Chlorine, but Chlorine has a higher electron affinity than Fluorine.
386 Hard
386 of 426
Understanding +4 | -1
Identify the correct order of solubility of Na2S,CuS\mathrm{N}{\mathrm{a}_2}\mathrm{S},\mathrm{CuS} and ZnS\mathrm{ZnS} in aqueous medium:
View answer and explanation
Correct option: D
Na2S>ZnS>CuS\mathrm{N}{\mathrm{a}_2}\mathrm{S} > ZnS > CuS
Option D (Na2S>ZnS>CuS\mathrm{N}{\mathrm{a}_2}\mathrm{S} > ZnS > CuS) is correct because sodium salts are highly soluble, and the KspK_{sp} value of CuS\mathrm{CuS} is less than ZnS\mathrm{ZnS}, making ZnS\mathrm{ZnS} more soluble than CuS\mathrm{CuS}.
387 Medium
387 of 426
Application +4 | -1
The ionisation energy and electron affinity of an element are 13.0 eV and 3.8 eV resp.
It’s electronegativity is
View answer and explanation
Correct option: B
3
Option A (2.8) is incorrect. Option B (3) is correct because electronegativity is approximately the average of ionization energy and electron affinity, scaled appropriately.
388 Medium
388 of 426
Analyzing +4 | -1
X−O−H→H2OX - O - H\xrightarrow[{{H_2}O}]{}

Y−O−H→H2OY - O - H\xrightarrow[{{H_2}O}]{}

E.N. - Electronegativity value

Incorrect option is:
View answer and explanation
Correct option: C
∣\midE.N. of YY - E.N. of O∣>∣{O}|>|E.N. of OO -E.N. of H∣{H} \mid Acidic
The correct answer is C (∣\midE.N. of YY - E.N. of O∣>∣{O}|>|E.N. of OO -E.N. of H∣{H} \mid Acidic) because for a compound to be acidic, the electronegativity difference between Y and O should be less than that between O and H.
389 Medium
389 of 426
Remembering +4 | -1
The period containing max. number of element is:
View answer and explanation
Correct option: D
6th 6^{\text {th }}
Option D (6th 6^{\text {th }}) is correct because the 6th period contains 32 elements, which is the maximum number of elements in any period.
390 Medium
390 of 426
Understanding +4 | -1
The first ionisation enthalpy of NaNa is 5.1 eV, the value of electron gain enthalpy of Na+N{a^ + } will be
View answer and explanation
Correct option: C
- 5.1 eV
Option C (- 5.1 eV) is correct because the electron gain enthalpy of Na+Na^+ is the negative of the ionization enthalpy of Na.
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