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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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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 271-300 of 426 questions

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271 Medium
271 of 426
Remembering +4 | -1
Pauling’s equation for determining the electronegativity of an element, is XA,  XB={X_A},\;{X_B} = electronegativity values of elements and Δ=\Delta = represents polarity of A-B bond
View answer and explanation
Correct option: A
XA−XB=0.208Δ{X_A} - {X_B} = 0.208\sqrt \Delta
Option A (XA−XB=0.208Δ{X_A} - {X_B} = 0.208\sqrt \Delta) is correct because it represents Pauling's equation for electronegativity difference based on bond polarity.
272 Easy
272 of 426
Understanding +4 | -1
All elements in the third period have
View answer and explanation
Correct option: B
Three complete subshells
Option B is correct because elements in the third period have electrons filling the 3s, 3p, and 3d subshells, making them have three complete subshells.
273 Medium
273 of 426
Understanding +4 | -1
Which of the following arrangements represents the correct order of electron gain enthalpy (with –ve sign) of the given atomic species?
View answer and explanation
Correct option: D
O<S<F<ClO < S < F < Cl
Option D (O<S<F<ClO < S < F < Cl) is correct because electron gain enthalpy becomes more negative as we move from left to right across a period and less negative as we move down a group, with some exceptions due to electron-electron repulsion.
274 Easy
274 of 426
Remembering +4 | -1
Which atom has the highest electronegativity?
View answer and explanation
Correct option: C
ClCl
Option C (ClCl) is correct because Chlorine has the highest electronegativity among the given elements.
275 Medium
275 of 426
Analyzing +4 | -1
S-I: The first ionisation energy of BB is less than that of Be{Be}

S-II: The penetration ability of s-electrons is higher than that of p-electrons.

The correct answer is
View answer and explanation
Correct option: A
Both I and II are correct
Option A (Both I and II are correct) is correct because Boron has a lower ionization energy than Beryllium, and s-electrons penetrate better than p-electrons.
276 Medium
276 of 426
Application +4 | -1
The ionisation energy and electron affinity of an element are 13.0 eV and 3.8 eV resp.
It’s electronegativity is
View answer and explanation
Correct option: B
3
Option A (2.8) is incorrect. Option B (3) is correct because electronegativity is approximately the average of ionization energy and electron affinity, scaled appropriately.
277 Easy
277 of 426
Understanding +4 | -1
An atom of an element has electronic configuration 2,8,12,8,1. Which of the following statement is correct?
View answer and explanation
Correct option: D
The element forms basic oxide
Option D (The element forms basic oxide) is correct because the electronic configuration 2,8,1 indicates that the element is a metal, and metals generally form basic oxides.
278 Easy
278 of 426
Understanding +4 | -1
IP2\mathrm{I}{\mathrm{P}_2} for an element is invariably higher than IP1\mathrm{I}{\mathrm{P}_1} because:
View answer and explanation
Correct option: D
All of the above
Option D (All of the above) is correct because the size of the cation is smaller, it's more difficult to remove an electron, and the effective nuclear charge is greater.
279 Medium
279 of 426
Application +4 | -1
The formation of O2−(g){O}^{2-}(\mathrm{g}) starting from O(g){O}(\mathrm{g}) is endothermic by 603 kJ mol−1603 \mathrm{~kJ} \mathrm{~mol}^{-1}. If electron affinity of O(g){O}(\mathrm{g}) is −141 kJ mol−1-141 \mathrm{~kJ} \mathrm{~mol}^{-1}, the second electron affinity of oxygen would be:
View answer and explanation
Correct option: A
+744 kJ mol−1+744 \mathrm{~kJ} \mathrm{~mol}^{-1}
Option A (+744 kJ mol⁻¹) is correct because it represents the energy required to add a second electron to O⁻(g), calculated from the overall endothermic formation of O²⁻(g).
280 Medium
280 of 426
Understanding +4 | -1
Choose the correct coding for following statements. Here T\mathrm{T} stands for True and F\mathrm{F} stands for False statement.

(i) Mendeleev left several gaps in his periodic table for the undiscovered elements.

(ii) The gap under aluminium and a gap under silicon was left and these elements were called Eka aluminium and Eka silicon.

(iii) Germanium was placed in place of Eka-aluminium and gallium was placed in place of Eka silicon.
View answer and explanation
Correct option: C
TTF
Option C (TTF) is correct because Mendeleev did leave gaps, he predicted Eka-aluminum and Eka-silicon, but Germanium replaced Eka-silicon while Gallium replaced Eka-aluminum.
281 Medium
281 of 426
Analyzing +4 | -1
Incorrect statement is/are :

a. Maximum element containing group is third

b. Te is 1st 1^{\text {st }} man made radioactive element

c. Law of octave is valid only upto atomic no. 20.

d. Li-Mg pair shows diagonal relationship.
View answer and explanation
Correct option: D
Only b
Option D (Only b) is correct because Technetium (Tc) is the first man-made radioactive element, not Tellurium (Te).
282 Easy
282 of 426
Analyzing +4 | -1
Identify the least stable ion among the following
View answer and explanation
Correct option: B
Be−B{e^ - }
Option B (Be−B{e^ - }) is correct because Beryllium has a stable electronic configuration, and adding an electron makes it unstable.
283 Medium
283 of 426
Analyzing +4 | -1
In which of the following arrangements, the order is NOT according to the property indicated against it
View answer and explanation
Correct option: C
B<C<N<O\mathrm{B < C < N < O} (Increasing first IP)
Option B is correct because the electron affinity of Cl is more than F due to small size and high electron density of F.
284 Easy
284 of 426
Understanding +4 | -1
The bond length of LiF will be
View answer and explanation
Correct option: D
Less than that of NaF
Option D (Dependent) is correct because the half-life of a reaction is dependent on the rate constant, which is temperature-dependent according to the Arrhenius equation.
285 Easy
285 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, so the atomic number is 105.
286 Easy
286 of 426
Remembering +4 | -1
Which of the following set of elements follows dobereiner law of triads
View answer and explanation
Correct option: B
Li, Na, K
Option B (Li, Na, K) is correct because the atomic weight of Na is approximately the average of the atomic weights of Li and K, satisfying Doberiener's law of triads.
287 Medium
287 of 426
Understanding +4 | -1
The atomic radii of transition elements from CrCr to CuCu are almost equal because
View answer and explanation
Correct option: B
Increased effective nuclear charge is balanced by increased screening effect of (n−1) d(n - 1)\,d-orbitals
Option B (Increased effective nuclear charge is balanced by increased screening effect of (n−1) d(n - 1)\,d-orbitals) is correct because the balance between these two effects results in relatively constant atomic radii across the transition series.
288 Easy
288 of 426
Understanding +4 | -1
Among the following oxides, the one which is most basic is
View answer and explanation
Correct option: B
MgO{MgO}
Option B (MgO{MgO}) is correct because oxides of alkaline earth metals are generally basic, and MgO is more basic than the amphoteric oxides ZnO and Al2O3 and the acidic oxide N2O5.
289 Easy
289 of 426
Analyzing +4 | -1
Find the formula of halide of a metal whose successive ionization enthalpies are x,2x,5x,100x kJ mol−1x, 2 x, 5 x, 100 x \mathrm{~kJ} \mathrm{~mol}^{-1} respectively
View answer and explanation
Correct option: C
MX3{MX}_3
The correct answer is C (MX3{MX}_3) because the large difference between the 3rd and 4th ionization enthalpies indicates a stable +3 oxidation state for the metal.
290 Medium
290 of 426
Understanding +4 | -1
Arrange the following elements in increasing
order of first ionization enthalpy:

Li, Be, B, C, N, O

Choose the correct answer from the options given
below:
View answer and explanation
Correct option: B
Li << B << Be << C << O << N
Option B (Li << B << Be << C << O << N) is correct because it accounts for the exceptions in the general trend of increasing ionization enthalpy across a period, specifically the lower ionization enthalpy of Boron compared to Beryllium and Oxygen compared to Nitrogen.
291 Easy
291 of 426
Remembering +4 | -1
The element which occupies peak of Lother Meyer curve is
View answer and explanation
Correct option: C
KK
Option C (KK) is correct because potassium (K) occupies the peak of the Lother Meyer curve due to its large atomic volume.
292 Easy
292 of 426
Remembering +4 | -1
An element ' XX ' with atomic number 114 has recently been discovered. Its IUPAC name is:
View answer and explanation
Correct option: C
Ununquadium
Option C (Ununquadium) is correct because the IUPAC name for element 114 is Ununquadium.
293 Medium
293 of 426
Understanding +4 | -1
Which of the following properties is inversely related to electronegativity?
View answer and explanation
Correct option: D
Metallic properties
Option D is correct because electronegativity and metallic properties have an inverse relationship; as electronegativity increases, metallic properties decrease.
294 Easy
294 of 426
Remembering +4 | -1
The symbol of element with atomic number Z=109Z=109
View answer and explanation
Correct option: D
Une
Option D (Une) is correct because it represents the systematic IUPAC nomenclature for element 109, where Unnilennium is shortened to Une.
295 Medium
295 of 426
Understanding +4 | -1
The set representing the correct order of first ionisation potential is:
View answer and explanation
Correct option: B
Be>Mg>Ca\mathrm{Be} > Mg > Ca
Option B (Be>Mg>Ca\mathrm{Be} > Mg > Ca) is correct because ionization potential generally decreases down a group, but Be has a higher ionization energy than expected due to its small size and high effective nuclear charge.
296 Medium
296 of 426
Analyzing +4 | -1
Successive first five ionisation energies of a particular element is 7.1, 14.3, 34.5, 46.8, 162.2 (in eV) respectively. The element is likely to be
View answer and explanation
Correct option: B
Si
The correct answer is Si because there is a significant jump between the 4th and 5th ionization energies, indicating that the 4th electron is being removed from a stable, noble gas-like configuration.
297 Easy
297 of 426
Understanding +4 | -1
The ionization enthalpy will be highest when the electron is to be removed from .............. if other factors are equal :
View answer and explanation
Correct option: A
s-orbital
Option A (s-orbital) is correct because s-electrons are closest to the nucleus and experience the highest effective nuclear charge, making them the most difficult to remove.
298 Medium
298 of 426
Analyzing +4 | -1
The five successive ionisation energies of the element are 800,2427,3658,25024800,2427,3658,25024 and 32824 kJ mol−132824 \mathrm{~kJ} \mathrm{~mol}^{-1} respectively. The number of valence electrons is :
View answer and explanation
Correct option: A
3
Option A is correct. The large jump in ionization energy between the 3rd and 4th ionization energies indicates that the element has 3 valence electrons.
299 Easy
299 of 426
Understanding +4 | -1
Highest amount of energy will be required for the removal of the electron from
View answer and explanation
Correct option: A
s−s -subshell
Option A (s−s -subshell) is correct because electrons in s-subshells are closest to the nucleus and experience the highest effective nuclear charge, requiring more energy for removal.
300 Medium
300 of 426
Understanding +4 | -1
Which of the following groups of elements have highly negative electron gain enthalpy?
View answer and explanation
Correct option: D
Both (1) and (2)
The correct answer is D (Both (1) and (2)) because both Group-16 (chalcogens) and Group-17 (halogens) elements have a high tendency to gain electrons due to their electronic configurations.
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