Anand Classes provides detailed NCERT Solutions for Class 11 Chemistry Chapter 3 – Classification of Elements and Periodicity in Properties. In this section, we cover questions 3.36 to 3.40 with step-by-step explanations, focusing on concepts like isoelectronic species, ionization enthalpy, metallic and non-metallic character, and oxidizing properties of elements. These solutions help students strengthen their conceptual clarity and prepare effectively for school exams and competitive entrance tests. Click the print button to download study material and notes.
NCERT Solutions of Problems 3.36, 3.37, 3.38, 3.39, 3.40 of Chapter Classification of Elements & Periodicity in Properties Class 11 Chemistry
NCERT 3.36 : The size of isoelectronic species F⁻, Ne and Na⁺ is affected by
(a) nuclear charge (Z)
(b) valence principal quantum number (n)
(c) electron–electron interaction in the outer orbitals
(d) none of the factors because their size is the same.
Answer :
The size of isoelectronic ions depends upon the nuclear charge (Z).
- As the nuclear charge increases, the size decreases.
For example:
$$F^- \; (Z=9) \; > \; Ne \; (Z=10) \; > \; Na^+ \; (Z=11)$$
Therefore, statement (a) is correct, while all other statements are wrong.
NCERT 3.37 : Which one of the following statements is incorrect in relation to ionization enthalpy?
(a) Ionization enthalpy increases for each successive electron.
(b) The greatest increase in ionization enthalpy is experienced on removal of electron from core noble gas configuration.
(c) End of valence electrons is marked by a big jump in ionization enthalpy.
(d) Removal of electron from orbitals bearing lower n value is easier than from orbital having higher n value.
Answer :
Statement (d) is incorrect.
The correct statement is:
- Removal of electron from orbitals bearing lower n value is more difficult than from orbitals having higher n value.
All other statements are correct.
NCERT 3.38 : Considering the elements B, Al, Mg and K, the correct order of their metallic character is :
(a) B > Al > Mg > K
(b) Al > Mg > B > K
(c) Mg > Al > K > B
(d) K > Mg > Al > B
Answer
In a period, metallic character increases as we move from right to left.
Therefore, metallic character of K, Mg and Al decreases in the order:
$$ K > Mg > Al $$
Within a group, metallic character increases from top to bottom.
Thus, Al is more metallic than B.
Hence, the correct sequence of decreasing metallic character is:
$$ K > Mg > Al > B $$
Therefore, the correct option is (d).
NCERT 3.39 : Considering the elements B, C, N, F and Si, the correct order of their non–metallic character is
(a) B > C > Si > N > F
(b) Si > C > B > N > F
(c) F > N > C > B > Si
(d) F > N > C > Si > B
Answer :
In a period, non–metallic character decreases from right to left.
Thus, among B, C, N and F, the order is:
$$ F > N > C > B $$
Within a group, non–metallic character decreases from top to bottom.
Therefore, C is more non–metallic than Si.
Hence, the correct sequence of decreasing non–metallic character is:
$$ F > N > C > B > Si $$
Thus, the correct option is (c).
NCERT 3.40 : Considering the elements F, Cl, O and N the correct order of their chemical reactivity in terms of oxidizing property is:
(a) F > Cl > O > N
(b) F > O > Cl > N
(c) Cl > F > O > N
(d) O > F > N > Cl
Answer :
Within a period, the oxidizing character increases from left to right.
Therefore, among F, O and N, the oxidizing power decreases in the order:
$$ F > O > N $$
Within a group, the oxidizing power decreases from top to bottom.
Thus, F is a stronger oxidizing agent than Cl.
Since O is more electronegative than Cl, oxygen is also a stronger oxidizing agent than chlorine.
Hence, the overall decreasing order of oxidizing power is:
$$ F > O > Cl > N $$
Thus, the correct option is (b).
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