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JEE • CHEMISTRY • CLASSIFICATION OF ELEMENTS AND PERIODICITY IN PROPERTIES

Classification of elements and periodicity in properties practice · Easy

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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 151-180 of 184 questions

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151 Easy
151 of 184
Remembering +4 | -1
Which element has the highest electronegativity?
View answer and explanation
Correct option: B
O
Option B (O) is correct because oxygen has the highest electronegativity among the given elements due to its small atomic size and high effective nuclear charge.
152 Easy
152 of 184
Understanding +4 | -1
Arrange the following in increasing order of ionic radii? C4−,N3−,F−,O2−{C}^{4-}, {N}^{3-}, {F}^{-}, {O}^{2-}
View answer and explanation
Correct option: C
F−<O2−<N3−<C4−{F}^{-} < {O}^{2-} < {N}^{3-} < {C}^{4-}
Option C (F−<O2−<N3−<C4−{F}^{-} < {O}^{2-} < {N}^{3-} < {C}^{4-}) is correct because for isoelectronic species, ionic radii increase with increasing negative charge.
153 Easy
153 of 184
Application +4 | -1
If Aufbau rule is not followed, K-19 will be placed in
View answer and explanation
Correct option: C
d-block
Option C (d-block) is correct because if the Aufbau principle is not followed, the 19th electron would enter the 3d subshell instead of the 4s subshell.
154 Easy
154 of 184
Remembering +4 | -1
Which one of the elements has the maximum electron affinity?
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Correct option: B
Cl
Option B (Cl) is correct because Chlorine has the highest electron affinity due to its electronic configuration and effective nuclear charge.
155 Easy
155 of 184
Understanding +4 | -1
Correct order of screening effect for electrons present in different sub shell will be :
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Correct option: A
s>p>d>fs>p>d>f
Option A (s>p>d>fs>p>d>f) is correct because the screening effect decreases as the orbital shape becomes more diffused, leading to a weaker shielding of the nuclear charge.
156 Easy
156 of 184
Analyzing +4 | -1
Four elements are given, choose the correct order of their first ionisation potential :

a. 1s2,2s22p63s11 s^2, 2 s^2 2 p^6 3 s^1

b. 1s2,2s22p63s23p61 s^2, 2 s^2 2 p^6 3 s^2 3 p^6

c. 1s2,2s22p63s23p21 s^2, 2 s^2 2 p^6 3 s^2 3 p^2

d. 1s2,2s22p63s23p11 s^2, 2 s^2 2 p^6 3 s^2 3 p^1
View answer and explanation
Correct option: C
b >> c >> d >> a
Option C (b > c > d > a) is correct because ionization potential generally increases across a period and decreases down a group. Noble gases have the highest ionization potential, followed by elements with stable electronic configurations.
157 Easy
157 of 184
Remembering +4 | -1
The discovery of which of the following group of elements gave death blow to the Newland&rsquo;s law of octaves?
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Correct option: A
Inert gases
Option A is correct because the discovery of inert gases disrupted Newland's law of octaves, as they introduced elements that didn't fit the pattern.
158 Easy
158 of 184
Remembering +4 | -1
Pauling’s electronegativity scale is based upon experimental values of
View answer and explanation
Correct option: B
Bond energies
Option B (Bond energies) is correct because Pauling's electronegativity scale is based on experimental bond energies.
159 Easy
159 of 184
Analyzing +4 | -1
Which of the following is a false statement?
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Correct option: C
Lithium is amphoteric
The correct answer is Lithium is amphoteric because Lithium is basic, not amphoteric. Amphoteric substances can act as both acid and base.
160 Easy
160 of 184
Understanding +4 | -1
Which of the following compounds is/are amphoteric in nature?
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Correct option: D
Both (2) and (3)
Option D is correct because both Al₂O₃ and As₂O₃ can act as both acids and bases, thus exhibiting amphoteric behavior.
161 Easy
161 of 184
Understanding +4 | -1
Order of atomic radii of Ti, Zr\mathrm{Ti},\,\mathrm{Zr} and Hf\mathrm{Hf} is:
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Correct option: C
Ti<Zr=Hf\mathrm{Ti} < \mathrm{Zr}=\mathrm{Hf}
Option C (Ti < Zr=Hf) is correct because atomic radii increase down the group, but Zr and Hf have similar radii due to lanthanide contraction.
162 Easy
162 of 184
Understanding +4 | -1
Mendeleef’s Periodic Table is upset by the fact that
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Correct option: A
Many elements has several isotopes
Option A (Many elements has several isotopes) is correct because the existence of isotopes with different atomic masses challenged Mendeleev's periodic law, which was based on atomic weights.
163 Easy
163 of 184
Remembering +4 | -1
The starting element of fifth period is
View answer and explanation
Correct option: B
Rb
Option B (Rb) is correct because Rubidium (Rb) is the first element in the fifth period of the periodic table.
164 Easy
164 of 184
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.
165 Easy
165 of 184
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).
166 Easy
166 of 184
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.
167 Easy
167 of 184
Remembering +4 | -1
Atomic radii of fluorine and neon in angstrom unit are respectively given by:
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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.
168 Easy
168 of 184
Remembering +4 | -1
At present, how many elements are known
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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.
169 Easy
169 of 184
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.
170 Easy
170 of 184
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.
171 Easy
171 of 184
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.
172 Easy
172 of 184
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.
173 Easy
173 of 184
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.
174 Easy
174 of 184
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.
175 Easy
175 of 184
Understanding +4 | -1
The element with atomic number 36 belongs to …block in the Periodic Table.
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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.
176 Easy
176 of 184
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.
177 Easy
177 of 184
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.
178 Easy
178 of 184
Understanding +4 | -1
The factors that influence ionisation energies are
View answer and explanation
Correct option: D
All of these
Option D (All of these) is correct because the size of the atom, the charge on the nucleus, and the screening effect all influence the ionization energy.
179 Easy
179 of 184
Remembering +4 | -1
Of the following pairs, the one containing examples of metalloid elements is :
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Correct option: D
As and Sb
Option D (As and Sb) is correct because Arsenic (As) and Antimony (Sb) are both metalloids, exhibiting properties of both metals and nonmetals.
180 Easy
180 of 184
Remembering +4 | -1
The charges on the colloidal CdSCdS sol and TiO2Ti{O_2} sol are, respectively
View answer and explanation
Correct option: D
negative and positive
Option D (negative and positive) is correct because sulfide sols like CdS are typically negatively charged, while oxide sols like TiO2 are typically positively charged.
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