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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 151-180 of 426 questions

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151 Medium
151 of 426
Understanding +4 | -1
The electron affinity of K(g)+K_{(g)}^ + ion is numerically equal to
View answer and explanation
Correct option: D
Ionisation energy of K atom
Option D (Ionisation energy of K atom) is correct because the electron affinity of a cation (K+) is numerically equal to the ionization energy of its neutral atom (K).
152 Medium
152 of 426
Understanding +4 | -1
Among Zn(OH)2,Sn(OH)2,Cu(OH)2,Fe(OH)3, Pb(OH)2,Al(OH)3\mathrm{Zn}(\mathrm{OH})_2, \mathrm{Sn}(\mathrm{OH})_2, \mathrm{Cu}(\mathrm{OH})_2, \mathrm{Fe}(\mathrm{OH})_3, \mathrm{~Pb}(\mathrm{OH})_2, \mathrm{Al}(\mathrm{OH})_3
The number of hydroxides which are soluble in excess of NaOH\mathrm{NaOH} solution is
View answer and explanation
Correct option: A
4
Option A (4) is correct because Zn(OH)₂, Sn(OH)₂, Pb(OH)₂, and Al(OH)₃ are amphoteric hydroxides and thus soluble in excess NaOH.
153 Medium
153 of 426
Understanding +4 | -1
The correct order of ionisation energy is
View answer and explanation
Correct option: C
C<N>OC < N > O
Option C (C<N>OC < N > O) is correct because nitrogen has a stable half-filled electronic configuration, making its ionization energy higher than both carbon and oxygen.
154 Medium
154 of 426
Analyzing +4 | -1
Which of the following statement is incorrect?
View answer and explanation
Correct option: B
The electron affinity of fluorine is greater than that of chlorine
Option B is correct because the electron affinity of fluorine is actually less than that of chlorine due to fluorine's smaller size and higher electron density.
155 Medium
155 of 426
Analyzing +4 | -1
Which one of the following elements has the highest ionisation energy?
View answer and explanation
Correct option: B
[Ne]3s23p3\left[ {\mathrm{Ne}} \right]3{s^2}3{p^3}
Option B ([Ne]3s²3p³) is correct because the half-filled p-orbital configuration ([Ne]3s²3p³) provides extra stability, making it more difficult to remove an electron, thus resulting in the highest ionization energy.
156 Medium
156 of 426
Remembering +4 | -1
The atomic orbitals used for filling in sixth period of periodic table are
View answer and explanation
Correct option: C
6s,4f,5d,6p6 s, 4 f, 5 d, 6 p
Option C (6s,4f,5d,6p6 s, 4 f, 5 d, 6 p) is correct because in the sixth period, the filling order of atomic orbitals is 6s, then 4f, then 5d, and finally 6p.
157 Medium
157 of 426
Understanding +4 | -1
Identify the wrong statement in the following.
View answer and explanation
Correct option: C
Amongst isoelectronic species, smaller the positive charge on the cation, smaller is the ionic radius
Option C (Amongst isoelectronic species, smaller the positive charge on the cation, smaller is the ionic radius) is the wrong statement because, for isoelectronic species, a smaller positive charge on the cation results in a larger ionic radius.
158 Easy
158 of 426
Remembering +4 | -1
......... is measured as the radius of an atom's core which is in contact with the core of an adjacent atom in a bonded situation.
View answer and explanation
Correct option: C
Covalent radius
Option C (Covalent radius) is correct because the covalent radius is defined as half the distance between the nuclei of two covalently bonded atoms.
159 Hard
159 of 426
Understanding +4 | -1
Considering the elements B,C,N,F\mathrm{B}, \mathrm{C}, \mathrm{N}, \mathrm{F} and Si\mathrm{Si}, the correct order of their non-metallic character is
View answer and explanation
Correct option: C
F >> N >> C >> B >> Si
Option C (F >> N >> C >> B >> Si) is correct because non-metallic character increases across a period and decreases down a group.
160 Easy
160 of 426
Remembering +4 | -1
Which of the following is a transition element?
View answer and explanation
Correct option: C
NiNi
Option C (NiNi) is correct because Nickel is a transition element, characterized by having partially filled d-orbitals in its electronic configuration.
161 Medium
161 of 426
Analyzing +4 | -1
Which of the following order is correct?
A) NH3<PH3<AsH3N{H_3} < P{H_3} < As{H_3} (Reducing nature)
B) Na>Mg>Al>SiNa > Mg > Al > Si (Radius)
C) Li<Be<B<C(I.E1)Li < Be < B < C\left( {\mathrm{I}\mathrm{.}{\mathrm{E}_1}} \right)
D) O<F<S<Cl   (EA)O < F < S < Cl\,\,\,(EA)
View answer and explanation
Correct option: A
A, B Only
Option A is correct because 0.1 mole of CO2 has the smallest number of molecules compared to the other options.
162 Easy
162 of 426
Understanding +4 | -1
Which of the following set of elements cannot be a triad
View answer and explanation
Correct option: B
B, Al, Ga
Option B is correct because B, Al, and Ga do not form a triad as the atomic weight of Al is not the arithmetic mean of the atomic weights of B and Ga.
163 Easy
163 of 426
Understanding +4 | -1
Correct statements among the following is /are:-
View answer and explanation
Correct option: D
All of these
Option D (All of these) is correct because all the statements regarding electron gain enthalpy, electronegativity, and amphoteric oxides are true.
164 Easy
164 of 426
Remembering +4 | -1
The number of elements which can be accommodated in the present set-up of the long form of the periodic table is:
View answer and explanation
Correct option: B
118
The correct answer is B (118) because the current long form of the periodic table accommodates 118 elements, from Hydrogen (atomic number 1) to Oganesson (atomic number 118).
165 Easy
165 of 426
Understanding +4 | -1
The element with electronic configuration 1s22s22p63s21{s^2}2{s^2}2{p^6}3{s^2} is
View answer and explanation
Correct option: C
an alkaline earth metal
Option C (an alkaline earth metal) is correct because the electronic configuration 1s22s22p63s21s^22s^22p^63s^2 indicates two valence electrons, which is characteristic of alkaline earth metals.
166 Easy
166 of 426
Understanding +4 | -1
Which of the following is the correct order of electron gain enthalpy
View answer and explanation
Correct option: A
P>O>S>FP > O > S > F
Option C (F > S > P > O) is correct because electron gain enthalpy generally increases across a period and decreases down a group, with exceptions due to electron-electron repulsion.
167 Medium
167 of 426
Remembering +4 | -1
Lothar Meyer plotted the physical properties such as atomic volume, melting point and ...XX... against atomic weight. Here, X{X} refers to
View answer and explanation
Correct option: B
Boiling point
Option B (Boiling point) is correct because Lothar Meyer plotted atomic volume, melting point, and boiling point against atomic weight.
168 Easy
168 of 426
Remembering +4 | -1
Which important property did Mendeleev use to classify the elements in his periodic table?
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Correct option: A
Atomic weight
Option A (Atomic weight) is correct because Mendeleev used atomic weight to classify elements.
169 Easy
169 of 426
Understanding +4 | -1
If the ionisation potential is IE, electron affinity is EA and electronegativity is X, which of the following relation is correct
View answer and explanation
Correct option: A
X=IE+EA2X = \frac{{IE + EA}}{2}
Option A is correct because Mulliken's electronegativity (X) is defined as the average of the ionization energy (IE) and electron affinity (EA): X=IE+EA2X = \frac{{IE + EA}}{2}.
170 Easy
170 of 426
Understanding +4 | -1
When an electron is removed from an atom, its energy
View answer and explanation
Correct option: A
Increase
Option A (Increase) is correct because removing an electron requires energy input, thus increasing the energy of the resulting ion.
171 Medium
171 of 426
Understanding +4 | -1
Screening by inner electrons will be more effective in
View answer and explanation
Correct option: D
CsCs
Option D is correct. Cesium (Cs) has the largest atomic size and the most inner electrons among the given options, leading to the most effective screening of the nuclear charge by the inner electrons.
172 Medium
172 of 426
Analyzing +4 | -1
Which of the following is incorrect?
View answer and explanation
Correct option: B
IE1IE_1 of Be>IE1Be > IE_1 of BB
Option A is correct because the first ionization energy of Lithium (Li) is actually less than that of Beryllium (Be) due to Be having a filled s-orbital.
173 Easy
173 of 426
Remembering +4 | -1
Which among the following is a representative element?
View answer and explanation
Correct option: A
Se\mathrm{Se}
Option A (Se\mathrm{Se}) is correct because Selenium (Se) is a p-block element and s-block and p-block elements are classified as representative elements.
174 Medium
174 of 426
Analyzing +4 | -1
Assertion :- H2Se{H}_2 {Se} is less acidic than H2S.{H}_2 {S}.

Reason :- SS is less electronegative than Se.Se.
View answer and explanation
Correct option: D
If both Assertion & Reason are false
The correct answer is D (If both Assertion & Reason are false) because H2Se is more acidic than H2S, and S is more electronegative than Se.
175 Medium
175 of 426
Analyzing +4 | -1
Four successive members of the first row transition elements are listed below with their atomic numbers. Which one of them is expected to have the highest third ionisation enthalpy?
View answer and explanation
Correct option: C
Manganese (Z=25)(Z=25)
The correct answer is Manganese (Z=25) because after removing two electrons, it attains a stable half-filled d-orbital configuration, resulting in a higher third ionization enthalpy.
176 Hard
176 of 426
Analyzing +4 | -1
The successive ionisation energy values for an element &lsquo;X&rsquo; are given below XV. Ist ionisation energy =410 kJ mol−1\mathrm{mo}{\mathrm{l}^{ - 1}} XVI. 2nd ionisation energy =820 kJ mol−1\mathrm{mo}{\mathrm{l}^{ - 1}} XVII. 3rd ionisation energy =1100 kJ mol−1\mathrm{mo}{\mathrm{l}^{ - 1}} XVIII. 4th ionisation energy =1500 kJ mol−1\mathrm{mo}{\mathrm{l}^{ - 1}} XIX. 5th ionisation energy =3200 kJ mol−1\mathrm{mo}{\mathrm{l}^{ - 1}} Find out the number of valence electron for the atom &lsquo;X &rsquo;
View answer and explanation
Correct option: A
4
Option A is correct because there is a significant jump in ionization energy between the 4th and 5th ionization energies, indicating 4 valence electrons.
177 Medium
177 of 426
Understanding +4 | -1
Given below are two statements :
Statement (I) : Both metal and non-metal exist in p and d-block elements
Statement (II) : Non-metals have higher ionisation enthalpy and higher electronegativity than the metals.
In the light of the above statements, choose the most appropriate answer from the option given below:
View answer and explanation
Correct option: B
Statement I is false but Statement II is true
Option B (Statement I is false but Statement II is true) is correct because d-block elements are primarily metals, while non-metals generally have higher ionization enthalpy and electronegativity than metals.
178 Medium
178 of 426
Application +4 | -1
First and second ionisation energies of magnesium are 7.646 and 15.035eV15.035 \mathrm{eV} respectively. The amount of energy in kJ\mathrm{kJ} needed to convert all the atoms of magnesium into Mg2+\mathrm{Mg}^{2+} ions present in 12mg12 \mathrm{mg} of magnesium vapours is :

[Given : 1eV=96.5 kJ mol−11 \mathrm{eV}=96.5 \mathrm{~kJ} \mathrm{~mol}^{-1} ]
View answer and explanation
Correct option: A
1.1
Option A (1.1) is correct because the total energy needed to convert all Mg atoms to Mg2+ ions is calculated using the ionization energies and the number of moles of Mg.
179 Medium
179 of 426
Understanding +4 | -1
Which of the following process requires largest amount of energy
View answer and explanation
Correct option: B
Al(g)2+→Al(g)3++e−Al_{(g)}^{2 + } \to Al_{(g)}^{3 + } + {e^ - }
Option B (Al(g)2+→Al(g)3++e−Al_{(g)}^{2 + } \to Al_{(g)}^{3 + } + {e^ - }) is correct because removing an electron from a 2+ ion requires more energy than from a neutral atom or a 1+ ion due to increased effective nuclear charge.
180 Medium
180 of 426
Understanding +4 | -1
The metallic radius of sodium is 1.86  Ao,1.86\,\,\mathop \mathrm{A}\limits^\mathrm{o} , where as its cation radius from its vapour state is
View answer and explanation
Correct option: C
1.54  Ao1.54\,\,\mathop \mathrm{A}\limits^\mathrm{o}
Option C (1.54  Ao1.54 \,\,\mathop \mathrm{A}\limits^\mathrm{o}) is correct because a cation is formed by the loss of electrons, resulting in a smaller ionic radius compared to the metallic radius of the neutral atom.
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