Classification of Elements and Periodicity in Properties JEE Advanced previous year questions with solutions

4 solved JEE Advanced questions on Classification of Elements and Periodicity in Properties, free to read — no sign-in needed. The full chapter has 6 questions; sign in to attempt the remaining 2 in the exam simulator.

1. Q1JEE Advanced Adv 2026 (Paper 2)
   The correct statement(s) regarding the periodic properties of elements is(are)
   1. A.Second ionization enthalpy of carbon atom is less than that of boron atom.
   2. B.Increasing order of ionic radii: $\text{Al}^{3+} \lt \text{Mg}^{2+} \lt \text{Na}^+$
   3. C.Under identical conditions, in solid state, the density of potassium metal is more than density of sodium metal.
   4. D.The H$-$H bond is weaker than F$-$F bond.
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   Answer: A,B

   For (A): The electronic configuration of $\text{B}^+$ is $1s^2 2s^2$ and that of $\text{C}^+$ is $1s^2 2s^2 2p^1$. Removing an electron from the fully filled, more penetrating $2s$ orbital of $\text{B}^+$ requires more energy than removing a $2p$ electron from $\text{C}^+$. Thus, the second ionization enthalpy of carbon is less than that of boron. Statement (A) is correct. For (B): $\text{Al}^{3+}$, $\text{Mg}^{2+}$, and $\text{Na}^+$ are isoelectronic species with $10$ electrons each. The nuclear charge increases in the order $\text{Na}^+ (11) \lt \text{Mg}^{2+} (12) \lt \text{Al}^{3+} (13)$. Higher effective nuclear charge leads to a stronger pull on the electrons, resulting in a smaller ionic radius. Thus, the order of ionic radii is $\text{Al}^{3+} \lt \text{Mg}^{2+} \lt \text{Na}^+$. Statement (B) is correct. For (C): The density of potassium ($0.86 \text{ g/cm}^3$) is less than that of sodium ($0.97 \text{ g/cm}^3$) due to an abnormal increase in atomic volume caused by the presence of empty $3d$ orbitals in potassium. Statement (C) is incorrect. For (D): The bond dissociation enthalpy of $\text{H}-\text{H}$ ($436 \text{ kJ/mol}$) is much higher than that of $\text{F}-\text{F}$ ($159 \text{ kJ/mol}$) due to strong interelectronic repulsions between the lone pairs on the small fluorine atoms. Statement (D) is incorrect. Answer: Second ionization enthalpy of carbon atom is less than that of boron atom.; Increasing order of ionic radii: $\text{Al}^{3+} \lt \text{Mg}^{2+} \lt \text{Na}^+$
2. Q2JEE Advanced Adv 2020 (Paper 2)
   The ${1}^{st}$, ${2}^{nd}$ and the ${3}^{rd}$ ionization enthalpies, ${I}_{1},{I}_{2}$ and ${I}_{3}$, of four atoms with atomic numbers $n,n+1,n+2,$ and $n+3,$ where $n\lt 10,$ are tabulated below. What is the value of $n$? Atomic number Ionization Enthalpy (kJ/mol) ${I}_{1}$ ${I}_{2}$ ${I}_{3}$ $n$ $1681$ $3374$ $6050$ $n+1$ $2081$ $3952$ $6122$ $n+2$ $496$ $4562$ $6910$ $n+3$ $738$ $1451$ $7733$
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   Answer: 9

   From given data $(n+2)$ atom is alkali metal which is sodium. As $n+2=11$ So, $n=9$
3. Q3JEE Advanced Adv 2016 (Paper 1)
   The increasing order of atomic radii of the following group $13$ elements is
   1. A.$Al\lt Ga\lt In\lt Tl$
   2. B.$Ga\lt Al\lt In\lt Tl$
   3. C.$Al\lt In\lt Ga\lt Tl$
   4. D.$Al\lt Ga\lt Tl\lt In$
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   Answer: (B)

   The order of radius of $13th$ group elements is $Ga\lt Al\lt In\lt Tl$. Reason$\Rightarrow$Due to poor shielding effect of $d$-orbital, radius of $Ga$ is smaller than $Al$.
4. Q4JEE Advanced Adv 2012 (Paper 1)
   The periodic table consists of 18 groups. An isotope of copper, on bombardment with protons, undergoes a nuclear reaction yielding element $X$ as shown below. To which group, element $X$ belongs in the periodic table? ${ }_{29}^{63} \mathrm{Cu}+{ }_{1}^{1} \mathrm{H} \rightarrow 6_{0}^{1} n+{ }_{2}^{4} \alpha+2{ }_{1}^{1} \mathrm{H}+\mathrm{X}$
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   Answer: 8

   ${ }_{29}^{63} \mathrm{Cu}+{ }_{1}^{1} \mathrm{H} \rightarrow 6{ }_{0}^{1} n+{ }_{2}^{4} \mathrm{He}+2{ }_{1}^{1} \mathrm{H}+{ }_{Z}^{A} X$ Balancing the atomic mass and atomic number $63+1=(6 \times 1)+4+2+A \Rightarrow A=52$ $29+1=(6 \times 0)+2+2+Z \Rightarrow Z=26$ Thus ${ }_{Z}^{A} X={ }_{26}^{52} X$ or ${ }_{26}^{52} \mathrm{Fe}$ Hence, $X$ belongs to group 8 in the periodic table.