Classification of Elements and Periodicity questions
Questions are transcribed from NTA's official question paper. Diagrams are described in words; open the official PDF to see them. Every answer was solved twice independently and is shown only when both agree. Where they did not, we say so.
2 April 2026 Shift 1, Q59 · Q59 · MCQ · Classification of Elements and Periodicity in Properties · Medium
Given below are two statements:
Statement (I): The first ionisation enthalpy of the elements Na, Mg, Cl and Ar follows the order $Na > Mg > Cl > Ar$.
Statement (II): Among Ca, Al, Fe and B, the third ionisation enthalpy is very high for Ca.
In the light of the above statements, choose the correct answer from the options given below:
(1) Both Statement I and Statement II are true
(2) Both Statement I and Statement II are false
Show answer and solution
Answer: Option (2)
Key idea: Ionization Enthalpy Trends and Electronic Configuration
**Statement I Analysis**: Check first ionization enthalpy trend for Na, Mg, Cl, Ar
Correct order: Ar > Cl > Mg > Na (noble gas highest, then increases across period). Statement I claims Na > Mg > Cl > Ar, which is **FALSE**
**Statement II Analysis**: Check third IE for Ca, Al, Fe, B
Ca: [Ar]4s² → After 2 electrons removed, Ca²⁺=[Ar]. Third IE breaks stable noble gas core, so IE₃(Ca) is **very high**. Statement II is **TRUE**
Since Statement I is false and Statement II is true, but only two options given, answer is **(2)** as Statement I is definitely false
2 April 2026 Shift 2, Q59 · Q59 · MCQ · Classification of Elements and Periodicity in Properties · Hard
Given below are two statements :
**Statement I :** The second ionization enthalpy of B, Al and Ga is in the order of $B > Al > Ga$.
**Statement II :** The correct order in terms of first ionization enthalpy is $Si < Ge < Pb < Sn$.
In the light of the above statements, choose the **correct** answer from the options given below :
(1) Both **Statement I** and **Statement II** are true
(2) Both **Statement I** and **Statement II** are false
(3) **Statement I** is true but **Statement II** is false
(4) **Statement I** is false but **Statement II** is true
Show answer and solution
Answer: Option (2)
Key idea: Periodic trends in ionization enthalpy, d-block and f-block contraction effects
**Statement I Analysis:** For second ionization enthalpy of B, Al, Ga - remove electron from B⁺, Al⁺, Ga⁺
B is smallest so has highest 2nd IE. Ga has d-block contraction making it smaller than Al, so Ga⁺ has higher 2nd IE than Al⁺
Correct order: $B > Ga > Al$ (NOT $B > Al > Ga$). **Statement I is FALSE**
**Statement II Analysis:** For first ionization enthalpy of Group 14 elements (Si, Ge, Sn, Pb)
Actual IE values: Si (786.5) > Ge (762) > Pb (715.6) > Sn (708.6) kJ/mol
Given order Si < Ge < Pb < Sn is completely reversed. **Statement II is FALSE**
Since both statements are false, answer is option (2)
4 April 2026 Shift 1, Q58 · Q58 · MCQ · Classification of Elements and Periodicity in Properties · Medium
A monoatomic anion $(A^-)$ has 45 neutrons and 36 electrons. Atomic mass, group in the periodic table and physical state at room temperature of the element $(A)$ respectively are
(1) 80, 17, liquid
(2) 81, 16, solid
(3) 80, 16, gas
(4) 81, 15, gas
Show answer and solution
Answer: Option (1)
Key idea: Atomic structure, periodic table classification, and properties of halogens
Given: Anion $A^-$ has 45 neutrons and 36 electrons
Since $A^-$ has gained 1 electron, neutral atom $A$ has $36 - 1 = 35$ electrons, so atomic number $Z = 35$ (Bromine)
Electronic configuration of Br: $[Ar] 3d^{10} 4s^2 4p^5$ with 7 valence electrons, so Group 17 (Halogens)
Bromine is one of only two elements that are liquid at room temperature
Answer: Atomic mass = 80, Group = 17, Physical state = liquid
4 April 2026 Shift 2, Q57 · Q57 · MCQ · Classification of Elements and Periodicity in Properties · Easy
The $1^{st}$ ionization enthalpy for Mg is $+737$ kJ/mol. The most probable estimated value of the $2^{nd}$ ionization enthalpy of Mg is _________.
(1) $-906$ kJ/mol
(2) $-856$ kJ/mol
(3) $+1450$ kJ/mol
(4) $+590$ kJ/mol
Show answer and solution
Answer: Option (3)
Key idea: Successive Ionization Enthalpies and Periodic Trends
Write the electronic configuration of Mg: $1s^2 2s^2 2p^6 3s^2$
First ionization: $\text{Mg} \rightarrow \text{Mg}^+ + e^-$ removes one $3s$ electron, requiring $+737$ kJ/mol
Second ionization: $\text{Mg}^+ \rightarrow \text{Mg}^{2+} + e^-$ removes the second $3s$ electron
The second ionization enthalpy is always greater than the first because $\text{Mg}^+$ has higher effective nuclear charge and smaller size
Eliminate negative values (options 1, 2) and values less than first IE (option 4)
The second ionization enthalpy $\approx 2 \times$ first IE when both electrons are from the same shell
Answer: $+1450$ kJ/mol (option 3)
4 April 2026 Shift 2, Q58 · Q58 · MCQ · Classification of Elements and Periodicity in Properties · Hard
The electronegativity of a group 13 element 'E' is same as that of Ge (on Pauling scale and upto one decimal point). The CORRECT statements about $E^{3+}$ are
A. It can act as a reducing agent.
B. It can act as an oxidizing agent.
C. $E^{3+}$ is more stable than $E^{-}$.
D. The standard electrode potential value for $E^{3+}/E$ is positive.
Choose the correct answer from the options given below:
(1) A and C Only
(2) B and C Only
(3) B and D Only
(4) A and D Only
Show answer and solution
Answer under review. Our two independent solutions did not agree, so we are not showing an answer yet. Check it against NTA's final answer key.
Key idea: Inert pair effect, electronegativity trends, redox properties of group 13 elements
5 April 2026 Shift 1, Q58 · Q58 · MCQ · Classification of Elements and Periodicity in Properties · Easy
Given below are two statements:
**Statement I:** The correct order of electronegativity of fluorine, oxygen and nitrogen is $F > O > N$.
**Statement II:** The oxidation state of oxygen in $OF_2$ is $+2$ and in $Na_2O$ is $-2$.
In the light of the above statements, choose the *correct* answer from the options given below
(1) Both Statement I and Statement II are true
(2) Both Statement I and Statement II are false
(3) Statement I is true but Statement II is false
(4) Statement I is false but Statement II is true
Show answer and solution
Answer: Option (1)
Key idea: Electronegativity trends and oxidation states
**Statement I Analysis:** Check electronegativity values from Pauling scale: F = 3.98, O = 3.44, N = 3.04
The order $F > O > N$ is correct since fluorine is most electronegative element. Statement I is **TRUE**
**Statement II Analysis:** For $OF_2$, F has oxidation state $-1$ (most electronegative). So O: $x + 2(-1) = 0 \Rightarrow x = +2$ ✓
For $Na_2O$, Na has oxidation state $+1$. So O: $2(+1) + y = 0 \Rightarrow y = -2$ ✓
Statement II is **TRUE** (oxygen can have positive oxidation state when bonded to fluorine)
Since both statements are true, the answer is option **(1)**
In a period, the first ionisation enthalpy of the element at extreme left and the negative electron gain enthalpy of the extreme right element, except noble gases, are respectively.
(1) lowest and lowest
(2) highest and lowest
(3)
(4)
Show answer and solution
Answer: Option (1)
Key idea: Periodic trends: Ionization enthalpy and electron gain enthalpy across a period
Identify extreme left elements in a period: **Alkali metals** (Group 1)
Alkali metals have largest atomic size and lowest effective nuclear charge, so they have **lowest first ionization enthalpy** in the period
Identify extreme right elements (except noble gases): **Halogens** (Group 17)
Halogens have smallest atomic size (except noble gases) and highest effective nuclear charge, with strong tendency to gain electrons
Halogens have **most negative (lowest) electron gain enthalpy** as they release maximum energy when gaining an electron
Therefore: First ionization enthalpy = **lowest**, Electron gain enthalpy = **lowest** (most negative)
FAQs
Frequently asked questions
How many questions come from Classification of Elements and Periodicity in JEE Main?
In the 9 shifts of JEE Main 2026 Session 2 we found 7 questions on it, about 0.8 per shift.
Where can I practise Classification of Elements and Periodicity PYQs for free?
On this page, with solutions. For more, use NTA Abhyas and our free mock tests.
