Periodic Table is one of the 19 topics from JEE Main Chemistry Syllabus. The topics hold a weightage of 12-16 marks in JEE Main Question Paper with 3-4 questions. Having an accurate knowledge about the concepts of periodic table can help students crack the paper well. 

The periodic table is an arrangement of all the elements with their increasing atomic number and recurring chemical properties. They are arranged in a tabular format, where a row is a period and column is a group. 

Periodic Classification and Periodicity in Properties is one of the most important topics in JEE Main Chemistry Syllabus. Important subtopics include Modern periodic law and present form of the periodic table, s, p, d and f block elements, Periodic trends in properties of elements atomic and ionic radii, ionization enthalpy, Electron gain enthalpy, valence, oxidation states and chemical reactivity.

Students are advised to have a holistic approach towards every aspect in the chapter that has been included. This will give them a clearance of concept as well as help them in making through the examination. 

JEE Main Periodic Table Previous Year Questions and Solutions

1. The first ionization potential of Na is 5.1 eV. The value of electron gain enthalpy of Na+ will be? (2013) 
   1. - 2.55 eV
   2. - 5.1 eV
   3. - 10. 2 eV
   4. + 2.55 eV

Solution : Option 'd' (+ 2.55 eV) is the correct answer. 

2. Electron gain enthalpy with negative sign of fluorine is less than that of chlorine due to ; (2013) 
   1. Smaller size of Chlorine Atom
   2. Bigger size of 2p orbital of fluorine
   3. High ionization enthalpy of fluorine
   4. Small size of fluorine atom

Solution : Option 'd' (Small size of fluorine atom) is the correct answer

3. The radius of La3+ (Atomic number of La=57) is 1.06A. Which one of the following given values will be closest to the radius of Lu3+ (Atomic number of Lu = 71)? (2003)
   1. 40 A
   2. 1.06 A
   3. 0.85 A
   4. 1.60 A

Solution : Option 'c' ( 0.85A) is the correct answer. 

4. According to the Periodic Law of elements, the variation in properties of elements is related to their. (2003)
   1. nuclear masses
   2. numbers
   3. nuclear neutron-proton number ratios
   4. atomic masses

Solution : Option'b' (Atomic numbers) is the correct answer. The properties of elements change with a change in atomic number. 

5. The atomic numbers of vanadium (V),Chromium (Cr), manganese (Mn) and iron (Fe) are respectively 23, 24, 25 and 26. Which one of these may be expected to have the highest second ionization enthalpy ? (2003)
   1. Cr
   2. Mn
   3. Fe
   4. V

Solution : Option 'd' (Cr) is the correct answer. 

6. Among A1203, Si02, P203 and S02 the correct order of acid strength is; (2004)
   1. S02 < P203 < Si02 < A1203
   2. A1203 < Si02 < P203 < S02
   3. A1203 < Si02 < S02 < P203
   4. Si02 < S02 < A1203 < P203

Solution : 

Option 'b' (A1203 < Si02 < P203 < S02) is the correct answer. Acidity of oxides increases with increase in non-metallic character.

7. Beryllium and aluminium exhibit many properties which are similar. But the two elements differ in; (2004)
   1. exhibiting maximum covalency in compound
   2. exhibiting amphoteric nature in their oxides
   3. forming covalent halides
   4. forming polymeric hydrides

Solution : Option 'a' (exhibiting maximum covalency in compound)is the correct answer. Be and Al have a diagonal relationship. They possess similar properties but Be cannot form the polymeric hydrides.

8. The lanthanide contraction is responsible for the fact that ; (2005)
   1. and Y have about the same radius
   2. Zr and Nb have similar oxidation state
   3. Zr and Hf have about the same radius
   4. Zr and Zn have the same oxidation

Solution : Option 'c' (Zr and Hf have about the same radius) is the correct answer. Due to Lanthanide contraction. 

9. In which of the following arrangements the order is NOT according to the property indicated against it? (2005)
   1. Al3+ < Mg2+ < Na+ < F”
      Increasing ionic size
   2. B
      Increasing first ionization enthalpy
   3. I < Br < F < Cl
      Increasing electron gain enthalpy
   4. Li < Na < K < Rb
      Increasing metallic radius

Solution : Option 'b' (B < C < O < N) is the correct answer. Ionisation energy of N is more (p3 config) B < C < O < N. 

10. Which of the following factors may be regarded as the main cause of lanthanide contraction ?(2005)
    1. Poor shielding of one of 4f electron by another in the subshell
    2. Effective shielding of one of 4f electrons by another in the subshell
    3. Poorer shielding of 5d electrons by 4f electrons
    4. Greater shielding of 5d electrons by 4f electrons

Solution : Option 'a' ( Poor shielding of one of 4f electron by another in the subshell) is the correct answer. 

11. Following statements regarding the periodic trends of chemical reactivity of the alkali metals and the halogens are given. Which of these statements gives the correct picture?(2006)
    1. The reactivity decreases in the alkali metals but increases in the halogens with increase in atomic number down the group
    2. In both the alkali metals and the halogens the chemical reactivity decreases with increase in atomic number down the group
    3. Chemical reactivity increases with increase in atomic number down the group in both the alkali metals and halogens
    4. In alkali metals the reactivity increases but in the halogens it decreases with increase in atomic number down the group. 

Solution : Option 'd' is the correct answer. Most reactive metal is Cs and most reactive non metal is F.

12. The charge/size ratio of a cation determines its polarizing power. Which one of the follow­ing sequences represents the increasing order of the polarizing order of the polarizing power of the cationic species, K+, Ca2+, Mg2+, Be2+ ? (2007)
    1. Mg2+,Be2+,K+,Ca2+
    2. Be2+, K+, Ca2+, Mg2+
    3. K+,Ca2+,Mg2+,Be2+
    4. Ca2+,Mg2+,Be2+,K+

Solution : Option 'c' (K+,Ca2+,Mg2+,Be2+) is the correct answer. Higher the charge / size ratio, more is the polarizing power.

13. Which one of the following orders presents the correct sequence of the increasing basic nature of the given oxides ? (2011)
    1. IS^O < I^O < MgO < A/203
    2. I^O < Na20 < A/203 < MgO
    3. A/203 < MgO < NajO < K^O
    4. MgO < K^O < AZ203 < Na20

Solution : Option 'c' (A/203 < MgO < NajO < K^O) is the correct answer.As electropositive character increases basic nature metal oxides increases. 

14. Iron exhibits +2 and +3 oxidation states.Which of the following statements about iron is incorrect ? (2012)
    1. Ferrous oxide is more basic in nature than the ferric oxide. 
    2. Ferrous compounds are relatively more ionic than the corresponding ferric compounds. 
    3. Ferrous compounds are less volatile than the corresponding ferric compounds. 
    4. Ferrous compounds are more easily hydrolysed than the corresponding ferric compounds. 

Solution : Option 'd' is the correct answer.With increasing the oxidation state, acidic strength increases. Basic strength : FeO > Fe203. With increasing the positive charge of cation, ionic nature decreases. Thus, ferrous compounds are more ionic than ferric compounds.Ferrous compounds are less easily hydrolysed than ferric compounds.

15. Which of the following represents the correct order of increasing first ionization enthalpy for Ca, Ba, S, Se and Ar? (2013)
    1. Ca< S < Ba < Se < Ar
    2. S < Se < Ca < Ba < Ar
    3. Ba < Ca < Se < S < Ar
    4. Ca < Ba < S < Se < Ar

Solution : Option 'c' (Ba < Ca < Se < S < Ar) is the correct answer. First ionisation enthalpy is highest for Ar and least for Ba. 

16. The ionic radii (in )of N respectively is ; 
    1. 36, 1.40 and 1.71
    2. 36, 1.71 and 1.40
    3. 71, 1.40 and 1.36
    4. 71 and 1.36 and 1.40

Solution : Option 'c' is the correct answer. N-3 > CL2 > F­In isoelectronic series size decreases as increases.

JEE Main Periodic Table- Quick Formulas

Given below are some important points and formulas for JEE Main Test Series on Periodic Table:

• General formula for p-block elements is ns2np1-6 , Where n=2 to 6.
• General formula for d-block elements is ns2p6d1-10, where n=4 to 7.
• General formula for f-block elements is (n-2)s2p6d10f(1-14)(n-1)s2p6d0-1s2 where n =6 , 7.
• Effective nuclear charge (Zeff) = Z-σ where Z= atomic number,σ is the screening constant.
  • For 1s electron,σ = 0.30,
  • 0.35 for ns and np electron,
  • 0.85 for (n-1) penultimate orbit s,p,d electrons,
  • 1 for (n-2) and inner all the electrons present in s ,p, d, f.
• rA=(dA-A / 2 )
• Mullikan’s scale of electronegativity
  Electronegativity= (electron affinity +ionization potential)/2

Modern Form of Periodic Table