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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 121-150 of 426 questions

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121 Medium
121 of 426
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.
122 Medium
122 of 426
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.
123 Medium
123 of 426
Remembering +4 | -1
Atomic number of element Ununnillium is
View answer and explanation
Correct option: B
110
Option B (110) is correct because Ununnillium corresponds to element 110 based on IUPAC nomenclature.
124 Medium
124 of 426
Understanding +4 | -1
Which among the following elements has lowest value of ionisation energy?
View answer and explanation
Correct option: B
Sn\mathrm{Sn}
Option B (Sn) is correct because, within the same group, ionization energy decreases as you move down the group due to increasing atomic size and shielding effect.
125 Easy
125 of 426
Analyzing +4 | -1
The element having maximum value for its 3rd 3^{\text {rd }} ionisation enthalpy is
View answer and explanation
Correct option: B
Be\mathrm{Be}
Option B (Be\mathrm{Be}) is correct because after losing two electrons, Beryllium attains a stable noble gas configuration, making the removal of the third electron require a very high amount of energy.
126 Easy
126 of 426
Understanding +4 | -1
For the alkali metals, which of the following increase with increasing atomic number
View answer and explanation
Correct option: D
Atomic radius
Option D (Atomic radius) is correct because atomic radius increases down the group due to the addition of new electron shells.
127 Medium
127 of 426
Analyzing +4 | -1
Assertion :- Zn\mathrm{Zn} is not a transition element whereas SC\mathrm{SC} is.

Reason :- Configuration of Zn\mathrm{Zn} is 4s2,4p0,4d0.4 \mathrm{s}^2, 4 \mathrm{p}^0, 4 \mathrm{d}^0.
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 Zn is not considered a transition element due to its completely filled d-orbitals, while Sc is a transition element. The reason, however, does not explain the assertion.
128 Medium
128 of 426
Analyzing +4 | -1
Correct relationship for compound A−δ−B+δ\mathrm{A}^{-\delta}-\mathrm{B}^{+\delta} is :-
View answer and explanation
Correct option: B
I.P. A−_\mathrm{A}- E.A.B_\mathrm{B} >> I.P.B−_\mathrm{B}- E.AA_\mathrm{A}
Option B is correct because it correctly relates the ionization potential (IP) and electron affinity (EA) of A and B based on the given compound formation.
129 Easy
129 of 426
Understanding +4 | -1
The consequence of high screening effect of inner electrons of the nucleus for an element “X” is
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Correct option: A
A decrease in the ionization potential
Option A (A decrease in the ionization potential) is correct because a high screening effect reduces the effective nuclear charge experienced by the outer electrons, making them easier to remove.
130 Medium
130 of 426
Understanding +4 | -1
The correct order of the decreasing ionic radii among the following is electronic species are
View answer and explanation
Correct option: C
S2−>Cl−>K+>Ca2+{S}^{2-} > {Cl}^{-} > {K}^{+} > {Ca}^{2+}
Option C (S2- > Cl- > K+ > Ca2+) is correct because, for isoelectronic species, ionic radii decrease with increasing nuclear charge (more protons).
131 Medium
131 of 426
Remembering +4 | -1
Element having atomic no. of 56 belongs to which of the following block of periodic table?
View answer and explanation
Correct option: D
ss-block
Option D (ss-block) is correct because an element with atomic number 56 (Barium) belongs to Group 2, which is part of the ss-block.
132 Easy
132 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 calculated by subtracting the atomic number (number of protons) from the mass number, and the atomic number is equal to the number of electrons plus the charge of the ion.
133 Easy
133 of 426
Analyzing +4 | -1
On which of the following Dobereiner's Triad law is not applicable?
View answer and explanation
Correct option: C
F,Cl,Br{F}, {Cl}, {Br}
Option C (F,Cl,Br{F}, {Cl}, {Br}) is correct because the atomic mass of chlorine is not the arithmetic mean of the atomic masses of fluorine and bromine.
134 Medium
134 of 426
Application +4 | -1
Predict the correct formula of compounds, by the following pair of elements like aluminium and sulphur with the help of oxidation states,
View answer and explanation
Correct option: B
Al2S3\mathrm{Al}_2 \mathrm{S}_3
Option B (Al2S3\mathrm{Al}_2 \mathrm{S}_3) is correct because it reflects the correct balancing of charges between Aluminium (+3) and Sulphur (-2).
135 Easy
135 of 426
Remembering +4 | -1
The atomic number of element Unp is
View answer and explanation
Correct option: A
105
Option A (105) is correct because Unp represents Unnilpentium, where Un = 1, nil = 0, and pent = 5, thus the atomic number is 105.
136 Medium
136 of 426
Understanding +4 | -1
Elements having six electrons in its outermost orbit generally form:
View answer and explanation
Correct option: B
Negative ion
Option B (Negative ion) is correct because elements with six valence electrons tend to gain two electrons to achieve a stable octet, forming a negative ion.
137 Medium
137 of 426
Understanding +4 | -1
Assertion: Electron gain enthalpy can be exothermic or endothermic.
Reason: Electron gain enthalpy provides a measure of the ease with which an atom adds an electron to form anion.
View answer and explanation
Correct option: B
Assertion is correct, reason is correct; reason is not a correct explanation for assertion
Option B (Assertion is correct, reason is correct; reason is not a correct explanation for assertion) is correct because both the assertion and reason are true statements, but the reason does not explain why electron gain enthalpy can be exothermic or endothermic.
138 Medium
138 of 426
Remembering +4 | -1
Element with which of the following atomic number was named by American Society as Rutherfordium, while by Soviet Society it was named as Kurchatovium?
View answer and explanation
Correct option: B
104
Option B (104) is correct because element with atomic number 104 was named Rutherfordium by American Society and Kurchatovium by Soviet Society.
139 Medium
139 of 426
Remembering +4 | -1
The set that represents the pair of neutral oxides nitrogen is
View answer and explanation
Correct option: A
NONO and N2O{N_2}O
Option A (NONO and N2O{N_2}O) is correct because both nitric oxide (NO) and nitrous oxide (N2O{N_2}O) are neutral oxides of nitrogen, meaning they do not react with acids or bases.
140 Medium
140 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.
141 Easy
141 of 426
Remembering +4 | -1
IUPAC name of element having atomic number 108 is
View answer and explanation
Correct option: A
Unniloctium
Option A (Unniloctium) is correct because the IUPAC name for element 108 is Unniloctium, derived from the numerical roots for 1, 0, and 8.
142 Easy
142 of 426
Remembering +4 | -1
The electronic structure (n−1)d1−10ns0−2(n-1) d^{1-10} n s^{0-2} is characteristic of
View answer and explanation
Correct option: D
transition elements
Option D (transition elements) is correct because the electronic configuration (n−1)d1−10ns0−2(n-1) d^{1-10} n s^{0-2} is the general electronic configuration of transition elements.
143 Easy
143 of 426
Analyzing +4 | -1
The electronic configurations of four elements are :

(i) [Xe]6s1[{Xe}] 6 s^1

(ii) [Xe]4f145d16s2[{Xe}] 4 f^{14} 5 d^1 6 s^2

(iii) [Ar]4s24p5[{Ar}] 4 s^2 4 p^5

(iv) [Ar]3d74s2[{Ar}] 3 d^7 4 s^2

Which one of the following statements about these elements is not correct?
View answer and explanation
Correct option: B
(ii) is a d-block element
Option B ((ii) is a d-block element) is not correct because element (ii) is a lanthanide (f-block) element, specifically the last member of the lanthanide series.
144 Medium
144 of 426
Analyzing +4 | -1
The correct order of ionisation energies of F−,Cl−,F{F}^{-}, {Cl}^{-}, {F} and Cl{Cl} is:
View answer and explanation
Correct option: C
F−<Cl−<Cl<F{F}^{-} < {Cl}^{-} < {Cl} < {F}
Option A ([RhCl(CO)(PPh3)(NH3)\mathrm{RhCl(CO)(PPh_3)(NH_3)}]) is correct because it is a tetrahedral complex with different ligands attached to the central metal atom, making it chiral and thus optically inactive.
145 Hard
145 of 426
Application +4 | -1
The energy required to convert all atoms present in 1.2 g1.2 \mathrm{~g} magnesium to Mg2+{Mg}^{2+} ions if IE\mathrm{IE}, and IE2\mathrm{IE}_2 of magnesium are 120 kJ mol−1120 \mathrm{~kJ} \mathrm{~mol}^{-1} and 240 kJ mol−1240 \mathrm{~kJ} \mathrm{~mol}^{-1} respectively
View answer and explanation
Correct option: A
18 kJ18 \mathrm{~kJ}
Option A (18 kJ18 \mathrm{~kJ}) is correct because the total ionization energy for converting 0.05 moles of Mg to Mg²⁺ is calculated by summing IE₁ and IE₂ and then multiplying by the number of moles.
146 Medium
146 of 426
Analyzing +4 | -1
Amongst the elements with following electronic configurations, which one of them may have the highest ionization energy?
View answer and explanation
Correct option: B
Ne[3s23p3]\mathrm{Ne}\left[ {3{s^2}3{p^3}} \right]
Option B (Ne[3s23p3]\mathrm{Ne}\left[ {3{s^2}3{p^3}} \right]) is correct because a half-filled p-orbital configuration results in higher stability and consequently higher ionization energy.
147 Easy
147 of 426
Understanding +4 | -1
The correct order of electronegativity of N, O, F and P is
View answer and explanation
Correct option: A
F>O>N>P\mathrm{F > O > N > P}
Option A (F>O>N>P\mathrm{F > O > N > P}) is correct because electronegativity increases across a period and decreases down a group in the periodic table.
148 Medium
148 of 426
Analyzing +4 | -1
BB has a smaller first ionisation enthalpy than BeBe. Consider the following statements :

I) It is easier to remove 2p2p electron than 2s2s electron

II) 2p2p electron of BB is more shielded from the nucleus by the inner core of electrons than the 2s2s electrons of BeBe

III) 2s2s electron has more penetration power than 2p2p electron
IV) atomic radius of BB is more than BeBe (atomic number B=5B = 5, Be=4Be = 4)
The correct statements are
View answer and explanation
Correct option: A
(I), (II) and (III)
Option A is correct because statements I, II, and III accurately describe the reasons for Boron having a smaller first ionization enthalpy than Beryllium.
149 Medium
149 of 426
Understanding +4 | -1
Correct order of increasing ionic radii is
View answer and explanation
Correct option: D
Al3+<Mg2+<O2−<N3−\mathrm{Al}^{3+} < \mathrm{Mg}^{2+} < \mathrm{O}^{2-} < \mathrm{N}^{3-}
Option D (Al3+<Mg2+<O2−<N3−\mathrm{Al}^{3+} < \mathrm{Mg}^{2+} < \mathrm{O}^{2-} < \mathrm{N}^{3-}) is correct because for isoelectronic species, ionic radii decrease with increasing nuclear charge.
150 Easy
150 of 426
Analyzing +4 | -1
The element having greatest difference between its first and second ionisation energy, is
View answer and explanation
Correct option: D
KK
Option D (K) is correct because Potassium (K) has a large difference between its first and second ionization energies due to its electronic configuration.
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