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JEE • CHEMISTRY • CLASSIFICATION OF ELEMENTS AND PERIODICITY IN PROPERTIES

Classification of elements and periodicity in properties practice · Medium

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31 Medium
31 of 227
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.
32 Medium
32 of 227
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.
33 Medium
33 of 227
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)-.
34 Medium
34 of 227
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.
35 Medium
35 of 227
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.
36 Medium
36 of 227
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.
37 Medium
37 of 227
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.
38 Medium
38 of 227
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.
39 Medium
39 of 227
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).
40 Medium
40 of 227
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.
41 Medium
41 of 227
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.
42 Medium
42 of 227
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.
43 Medium
43 of 227
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.
44 Medium
44 of 227
Understanding +4 | -1
Which of the following should be correct for Z1{Z_1} and Z2{Z_2} in the following two processes
M++Z1→ M2++e−{M^ + } + {Z_1}\xrightarrow{\,}{M^{2 + }} + {e^ - }
M2++Z2→ M3++e−{M^{2 + }} + {Z_2}\xrightarrow{\,}{M^{3 + }} + {e^ - }
View answer and explanation
Correct option: D
Z1<Z2{Z_1} < {Z_2}
Option D is correct because the ionization energy increases with each successive ionization, meaning Z2{Z_2} (IE2) will always be greater than Z1{Z_1} (IE1).
45 Medium
45 of 227
Analyzing +4 | -1
Assertion : Size of Ca2+\mathrm{Ca}^{2+} is larger than K+\mathrm{K}^{+}.

Reason : Number of electrons in Ca+2\mathrm{Ca}^{+2} is more than K+\mathrm{K}^{+}.
View answer and explanation
Correct option: D
If both Assertion & Reason are false
The correct answer is If both Assertion & Reason are false because Ca2+ is smaller than K+ due to higher effective nuclear charge, and Ca2+ has fewer electrons than K+.
46 Medium
46 of 227
Understanding +4 | -1
The first ionization enthalpy of Na,Mg\mathrm{Na}, \mathrm{Mg} and Si\mathrm{Si}, respectively, are : 496,737 and 786 kJ mol−1786 \mathrm{~kJ} \mathrm{~mol}^{-1}. The first ionization enthalpy (kJmol−1)\left(\mathrm{kJ} \mathrm{mol}^{-1}\right) of Al\mathrm{Al} is:
View answer and explanation
Correct option: C
577
The correct answer is C (577) because the ionization enthalpy of Al is expected to be between that of Na and Mg, and also less than that of Si due to electronic configuration.
47 Medium
47 of 227
Application +4 | -1
An element having electronic configuration 1s22s22p63s23p64s11 s^2 2 s^2 2 p^6 3 s^2 3 p^6 4 s^1 forms
View answer and explanation
Correct option: B
Basic oxide
The correct answer is B (Basic oxide) because the electronic configuration indicates an alkali metal, which typically forms basic oxides.
48 Medium
48 of 227
Analyzing +4 | -1
The energy evolved is highest for which of the following reactions.
View answer and explanation
Correct option: B
Cl+e−→Cl−Cl + {e^ - } \to C{l^ - }
Option B (Cl+e−→Cl−Cl + {e^ - } \to C{l^ - }) is correct because chlorine has the highest electron affinity among the given options.
49 Medium
49 of 227
Analyzing +4 | -1
S-I: Cu+2\mathrm{Cu}^{+2} paramagnetic.

S-II: All the orbitals present in Cu+2\mathrm{Cu}^{+2} ion are doubly occupied.
View answer and explanation
Correct option: C
Only I is true
Option A (I and II are true) is incorrect because Statement I is true, but Statement II is false. Cu+2\mathrm{Cu}^{+2} is paramagnetic, but not all orbitals are doubly occupied.
50 Medium
50 of 227
Analyzing +4 | -1
The electronic configuration, 1s2,  2s22p6,  3s2  3p63d91{s^2},\;2{s^2}2{p^6},\;3{s^2}\;3{p^6}3{d^9} represents a
View answer and explanation
Correct option: D
Metallic cation
Option D is correct because the electronic configuration represents a copper(II) ion (Cu2+), which is a metallic cation.
51 Medium
51 of 227
Analyzing +4 | -1
Four successive members of first row transition elements are listed below with their atomic numbers. Which one of them is expected to have the highest third ionization enthalpy
View answer and explanation
Correct option: C
Manganese (Z = 25)
Option C is correct because Manganese (Z = 25) has an electronic configuration of 3d⁵4s², and removing the third electron requires removing an electron from the stable, half-filled 3d orbital, resulting in a significantly higher ionization enthalpy.
52 Medium
52 of 227
Analyzing +4 | -1
The ionic radius of ‘Cr’ is minimum in which of the following compounds?
View answer and explanation
Correct option: B
K2CrO4{\mathrm{K}_2}\mathrm{Cr}{\mathrm{O}_4}
Option B is correct because Cr has the highest oxidation state (+6) in K2CrO4, leading to the smallest ionic radius due to increased effective nuclear charge.
53 Medium
53 of 227
Application +4 | -1
Aqueous solutions of two compounds M−O−HM-{O}-{H} and M′−O−HM^{\prime}-{O}-{H} have been prepared in two different beakers. If the electronegativity of M=3.5,M′=1.72M=3.5, M^{\prime}=1.72, O=3.0{O}=3.0 and H=2.1{H}=2.1, then the solutions respectively are
View answer and explanation
Correct option: B
acidic, basic
Option B (acidic, basic) is correct because the acidity or basicity of the solutions depends on the electronegativity difference between M/M' and oxygen.
54 Medium
54 of 227
Application +4 | -1
The first, second and third ionisation energies of Al{Al} are 578,1817 and 2745 kJ mol−12745 \mathrm{~kJ} \mathrm{~mol}^{-1} respectively. Calculate the energy required to convert all the atoms of Al{Al} to Al+3{Al}^{+3} present in 270mg270{mg} of Al{Al} vapours
View answer and explanation
Correct option: B
51.40 kJ51.40 \mathrm{~kJ}
The correct answer is 51.40 kJ because the energy required to convert all Al atoms to Al+3 in 270 mg of Al vapor is 51.40 kJ, calculated by summing the ionization energies and multiplying by the number of moles of Al.
55 Medium
55 of 227
Understanding +4 | -1
Arrange the following elements in increasing order of electronegativity:

N,O,F,C,Si\mathrm{N}, \mathrm{O}, \mathrm{F}, \mathrm{C}, \mathrm{Si}

Choose the correct answer from the options given below:
View answer and explanation
Correct option: D
Si<C<N<O<F\text{Si}<\mathrm{C}< \mathrm{N}< \mathrm{O}< \mathrm{F}
The correct answer is D (Si<C<N<O<F\text{Si}<\mathrm{C}< \mathrm{N}< \mathrm{O}< \mathrm{F}) because electronegativity increases across a period and decreases down a group.
56 Medium
56 of 227
Understanding +4 | -1
Higher values of ionisation energies of the 5d-transition elements are consistent with the :
View answer and explanation
Correct option: B
relative smaller size of their atoms
Option B (relative smaller size of their atoms) is correct because lanthanide contraction leads to similar sizes for 4d and 5d series, impacting ionization energy.
57 Medium
57 of 227
Understanding +4 | -1
A large difference between the third and fourth ionization energies indicates the presence of :
View answer and explanation
Correct option: C
3 valence electrons in an atom
Option C is correct because a large jump between the third and fourth ionization energies indicates that the element readily loses three electrons to achieve a stable configuration.
58 Medium
58 of 227
Analyzing +4 | -1
Which of the following does not represent the correct order of the properties indicated?
View answer and explanation
Correct option: D
Li<Na>K>Cs(E)1\mathrm{Li} < \mathrm{Na} > \mathrm{K} > \mathrm{Cs}(\mathrm{E})_1
Option D is correct. The electron affinity of Li is not less than Na. The correct order for E1 is Li > Na > K > Cs.
59 Medium
59 of 227
Understanding +4 | -1
The second ionization enthalpies of elements are always higher than their first ionization enthalpies because:
View answer and explanation
Correct option: D
the cation is smaller than its parent atom
Option D (the cation is smaller than its parent atom) is correct because removing an electron increases the effective nuclear charge, shrinking the ion.
60 Medium
60 of 227
Analyzing +4 | -1
S-I: The second ionisation energy of 'OO' is greater than that of 'NN'.

S-II: The half filled p-orbitals cause greater stability.
View answer and explanation
Correct option: A
Both I and II are correct
Option A (Both I and II are correct) is correct because the second ionization energy of oxygen is indeed greater than that of nitrogen, and half-filled p-orbitals do contribute to greater stability.
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