Skip to main content
Back to CLASSIFICATION OF ELEMENTS AND PERIODICITY IN PROPERTIES

Your Progress

0 / 426 Answered
0
Correct
0
Incorrect
0%
Accuracy
JEE • CHEMISTRY • CLASSIFICATION OF ELEMENTS AND PERIODICITY IN PROPERTIES

Classification of elements and periodicity in properties practice · Mixed difficulty

Try a question, then open “View answer” to check it. Sign in to submit answers and save progress.

Practice format and chapter quiz

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.

Take a chapter quiz →
Showing 61-90 of 426 questions
61 Medium
61 of 426
Analyzing +4 | -1
Assertion(A) Electron gain enthalpy of oxygen is less than that of fluorine but greater than that of nitrogen.

Reason(R) Ionisation enthalpy is as follows: N>O>F{N}>{O}>{F}
View answer and explanation
Correct option: C
If (A) is correct and (R) is incorrect.
The correct answer is C (If (A) is correct and (R) is incorrect) because electron gain enthalpy of oxygen is less negative than fluorine but more negative than nitrogen, and the ionization enthalpy order is incorrect.
62 Easy
62 of 426
Remembering +4 | -1
Which is not a transition metal?
View answer and explanation
Correct option: B
PbPb
Option B (PbPb) is correct because Lead (PbPb) is a p-block element and not a transition metal.
63 Medium
63 of 426
Understanding +4 | -1
The element with the electronic configuration as [Ar]3d104s24p3\left[ {\mathrm{Ar}} \right]3{d^{10}}4{s^2}4{p^3} represents a
View answer and explanation
Correct option: C
Metalloid
Option C is correct because the element with the given electronic configuration ([Ar]3d10 4s2 4p3) is Arsenic (As), which is a metalloid.
64 Easy
64 of 426
Remembering +4 | -1
The longest and shortest periods are
View answer and explanation
Correct option: C
6 & 1
Option C (6 & 1) is correct because the 6th period has 32 elements and is the longest, while the 1st period has only 2 elements and is the shortest.
65 Medium
65 of 426
Remembering +4 | -1
Which of the following relationship proposed graphically by Lothar Meyer?
View answer and explanation
Correct option: B
Physical properties vs atomic weight
The correct answer is B (Physical properties vs atomic weight) because Lothar Meyer plotted physical properties against atomic weight to observe periodic patterns.
66 Easy
66 of 426
Remembering +4 | -1
Pauling’s electronegativity is based on
View answer and explanation
Correct option: D
Bond energies
Option D (Bond energies) is correct because Pauling's electronegativity scale is based on bond energies.
67 Medium
67 of 426
Remembering +4 | -1
Alkaline earth metals form ions of the formula:
View answer and explanation
Correct option: A
M2+{M^{2 + }}
Option A (M2+{M^{2 + }}) is correct because alkaline earth metals have two valence electrons which they lose to form +2 ions.
68 Medium
68 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.
69 Easy
69 of 426
Understanding +4 | -1
The element with the lowest ionisation potential is
View answer and explanation
Correct option: D
Cs
Option D (Cs) is correct because ionization potential decreases down the group due to increasing atomic size and weaker nuclear attraction.
70 Medium
70 of 426
Remembering +4 | -1
Which of the following is not a member of actinoid series?
View answer and explanation
Correct option: D
Europium
Option D (Europium) is correct because Europium (Eu) is a lanthanoid, not an actinoid.
71 Easy
71 of 426
Understanding +4 | -1
Which element has maximum electron affinity?
View answer and explanation
Correct option: D
S
The correct answer is S because sulfur has the highest electron affinity among the given elements due to its position in the periodic table.
72 Medium
72 of 426
Analyzing +4 | -1
Assertion : Na+{Na}^{+} and Cl−{Cl}^{-} have similar ionic radius.

Reason : Zeff{Z}_{\text {eff}} in Na+{Na}^{+} and Cl−{Cl}^{-} are same.
View answer and explanation
Correct option: D
If both Assertion & Reason are false
Option D is correct. Both the Assertion and Reason are false because Na+{Na}^{+} and Cl−{Cl}^{-} do not have similar ionic radii, and their effective nuclear charges are different.
73 Medium
73 of 426
Analyzing +4 | -1
The value of IE1,IE2,IE3\mathrm{IE}_1, \mathrm{IE}_2, \mathrm{IE}_3 and IE4\mathrm{IE}_4 of an atom are 7.5 eV,25.6 eV,48.6 eV7.5 \mathrm{~eV}, 25.6 \mathrm{~eV}, 48.6 \mathrm{~eV} and 170.6 eV170.6 \mathrm{~eV} respectively.
The electronic configuration of the atom will be:-
View answer and explanation
Correct option: B
1 s22 s22p63 s23p11 \mathrm{~s}^2 2 \mathrm{~s}^2 2 \mathrm{p}^6 3 \mathrm{~s}^2 3 \mathrm{p}^1
Option D (Both (A) and (R) are false) is correct because solid PCl5 exists as an ionic compound, not a molecular solid, and the ions are (PCl4)+ and (PCl6)-.
74 Easy
74 of 426
Remembering +4 | -1
Which of the following element has less electro negativity
View answer and explanation
Correct option: B
S
Option B (S) is correct because electronegativity decreases down the group, and sulfur (S) is below oxygen (O) in the same group.
75 Medium
75 of 426
Analyzing +4 | -1
Consider the following elements In, Tl, Al, Pb, Sn and Ge.

The most stable oxidation states of elements with highest and lowest first ionisation enthalpies, respectively, are
View answer and explanation
Correct option: B
#ERROR!
The correct answer is B (+4 and +3) because Germanium (Ge) has the highest first ionization enthalpy and its most stable oxidation state is +4, while Indium (In) has the lowest first ionization enthalpy and its most stable oxidation state is +3.
76 Medium
76 of 426
Remembering +4 | -1
Which of the following elements are found in pitch blende?
View answer and explanation
Correct option: D
Both (1) and (2)
Option D (Both (1) and (2)) is correct because pitchblende contains Actinium, Protoactinium, Neptunium, and Plutonium.
77 Medium
77 of 426
Analyzing +4 | -1
Consider the following elements.
Group↓ C′ D′A′ B′→Period{\text{Group}} \downarrow \,_{C'\,D'}^{A'\,B' \to {\text{Period}}}
Which of the following is/are true about A′,B′,C′A^{\prime}, B^{\prime}, C^{\prime} and D′D^{\prime} ?

A. Order of atomic radii: B′<A′<D′<C′\mathrm{B}^{\prime} < \mathrm{A}^{\prime} < \mathrm{D}^{\prime} < \mathrm{C}^{\prime}

B. Order of metallic character: B′<A′<D′<C′B^{\prime} < A^{\prime} < D^{\prime} < C^{\prime}

C. Size of the element: D′<C′<B′<A′D^{\prime} < C^{\prime} < B^{\prime} < A^{\prime}

D. Order of ionic radii : B′+<A′+<D′+<C′+\mathrm{B}^{\prime+} < \mathrm{A}^{\prime+} < \mathrm{D}^{\prime+} < \mathrm{C}^{\prime+}

Choose the correct answer from the option given below:
View answer and explanation
Correct option: D
A, B and D only
The correct answer is D (A, B and D only) because the atomic radii, metallic character, and ionic radii orders are correctly described in statements A, B, and D.
78 Medium
78 of 426
Understanding +4 | -1
Which among the following elements has highest negative electron gain enthalpy?
View answer and explanation
Correct option: B
Cl\mathrm{Cl}
Option B is correct because chlorine has the highest negative electron gain enthalpy due to its smaller size and effective nuclear charge compared to bromine and iodine, and less electron-electron repulsion than fluorine.
79 Medium
79 of 426
Understanding +4 | -1
The IP1\mathrm{I}{\mathrm{P}_1} is maximum for:
View answer and explanation
Correct option: D
He\mathrm{He}
Option D (He) is correct because inert gases have the highest ionization potential due to their stable electronic configurations.
80 Easy
80 of 426
Remembering +4 | -1
Ionization potential is lowest for:
View answer and explanation
Correct option: D
Alkali metals
Option D (Alkali metals) is correct because alkali metals have the lowest ionization potential due to their electronic configuration (ns¹), making it easier to remove an electron.
81 Medium
81 of 426
Analyzing +4 | -1
Maximum I.E is for
View answer and explanation
Correct option: B
Li+L{i^ + }
Option B (Li+Li^+) is correct because it has the highest ionization energy due to its stable electronic configuration (1s21s^2).
82 Easy
82 of 426
Analyzing +4 | -1
Which of the following have the same number of electrons in outermost shell?
View answer and explanation
Correct option: A
Elements with atomic number 30, 48, 80
Option A is correct because elements with atomic numbers 30 (Zn), 48 (Cd), and 80 (Hg) all belong to the same group (group 12) and have 2 electrons in their outermost shell.
83 Medium
83 of 426
Understanding +4 | -1
Which of the following option is the correct order for the basic strength of metallic hydroxides?
View answer and explanation
Correct option: A
Al(OH)3<Lu(OH)3<Ce(OH)3<Ca(OH)2{Al}({OH})_3 < {Lu}({OH})_3 < {Ce}({OH})_3 < {Ca}({OH})_2
Option A (Al(OH)3<Lu(OH)3<Ce(OH)3<Ca(OH)2{Al}({OH})_3 < {Lu}({OH})_3 < {Ce}({OH})_3 < {Ca}({OH})_2) is correct because basic strength of metallic hydroxides increases with decreasing charge density and increasing electropositive character.
84 Easy
84 of 426
Analyzing +4 | -1
Which among the following element has highest value of second ionization enthalpy?
View answer and explanation
Correct option: A
LiLi
Option A (Li) is correct because after losing one electron, Lithium attains a stable noble gas configuration, requiring significantly more energy to remove the second electron.
85 Medium
85 of 426
Analyzing +4 | -1
Assertion : IP of FF >> EA of F.F.

Reason : F{F} is a highly electronegative element.
View answer and explanation
Correct option: B
If both Assertion & Reason are True but Reason is not a correct explanation of the Assertion
Option B is correct because both the assertion and the reason are true, but the reason (electronegativity) is not the direct explanation for why the ionization potential of F is greater than its electron affinity.
86 Easy
86 of 426
Analyzing +4 | -1
The electronic configurations for some neutral atoms are given below :

(A) 1s2,2s22p6,3s21 s^2, 2 s^2 2 p^6, 3 s^2

(B) 1s2,2s22p6,3s11 s^2, 2 s^2 2 p^6, 3 s^1

(C) 1s2,2s22p6,3s23p21 s^2, 2 s^2 2 p^6, 3 s^2 3 p^2

(D) 1s2,2s22p6,3s23p31 s^2, 2 s^2 2 p^6, 3 s^2 3 p^3

Which of these is expected to have the highest second ionisation enthalpy?
View answer and explanation
Correct option: B
(B)
Option B ((B)) is correct because after losing one electron, element B achieves a stable noble gas configuration, requiring significantly more energy for the second ionization.
87 Medium
87 of 426
Analyzing +4 | -1
Which of the following metal is expected to have the highest third ionisation energy ?
View answer and explanation
Correct option: C
Mn(Z=25)\mathrm{Mn}(\mathrm{Z}=25)
Option C (Mn(Z=25)) is correct because removing the third electron from Mn requires removing an electron from a stable half-filled d-orbital configuration.
88 Easy
88 of 426
Understanding +4 | -1
Which of the following element has higher ionisation energy?
View answer and explanation
Correct option: D
Nitrogen
Option D (Nitrogen) is correct because nitrogen has a higher ionization energy than boron, carbon, and oxygen due to its stable half-filled p-orbital configuration.
89 Easy
89 of 426
Analyzing +4 | -1
Identify the elements XX and YY using the ionisation energy values given below.
        Ionisation  energy (Ist)        KJ/mol  (IInd)X4954563Y7311450\begin{array}{cccccccccccccccccccc} {}&{\,\,\,\,\,\,\,\,\mathrm{Ionisation}{\mkern 1mu} {\mkern 1mu} \mathrm{energy}{\mkern 1mu} (\mathrm{Ist})}&{\,\,\,\,\,\,\,\,KJ/mol{\mkern 1mu} {\mkern 1mu} (\mathrm{IInd})}\\ X&{\mathrm{495}}&{\mathrm{4563}}\\ Y&{\mathrm{731}}&{\mathrm{1450}} \end{array}
View answer and explanation
Correct option: A
X = Na,  Y = MgX\, = \,Na,\,\,Y\, = \,Mg
Option A (X=Na,Y=MgX = Na, Y = Mg) is correct because Na has a significantly higher second ionization energy due to the removal of an electron from a stable noble gas configuration, while Mg's second ionization energy is comparatively lower.
90 Medium
90 of 426
Understanding +4 | -1
Which of the following has the smallest size?
View answer and explanation
Correct option: A
Al3+{Al}^{3+}
Option A (Al³⁺) is correct because, for isoelectronic species, the ion with the highest nuclear charge (most protons) has the smallest size.
Revision notes and key takeaways

Review CLASSIFICATION OF ELEMENTS AND PERIODICITY IN PROPERTIES in your textbook or class notes, then use this set to check your understanding.

Showing 61-90 of 426 questions
0 /426
0✓ 0✗
0% Accuracy
Type at least 2 characters or jump directly:
🔓

Save your learning progress

Sign in with a free account to save your practice answers and quiz results, then return to them when you revise.