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Academic session 2026-27, Chapter 3 in the current curriculum.
Welcome to the NCERT Solutions for Class 11 Chemistry - Chapter Classification of Elements and Periodicity in Properties. This page offers a step-by-step solution to the specific question from Exercise 1, Question 30:
Assign the position of the element having outer electronic configuration
(i) ns2 np4 for n = 3 (ii) (n - 1)d2 ns2 for n = 4, and (iii) (n - 2) f7 (n - 1)d1 ns2 for n = 6, in the periodic table.
. With detailed answers and explanations for each chapter, students can strengthen their understanding and prepare confidently for exams. Ideal for CBSE and other board students, this resource will simplify your study experience.Assign the position of the element having outer electronic configuration
(i) ns2 np4 for n = 3 (ii) (n - 1)d2 ns2 for n = 4, and (iii) (n - 2) f7 (n - 1)d1 ns2 for n = 6, in the periodic table.
(i) Since n = 3, the element belongs to the 3rd period. It is a p–block element since the last electron occupies the p–orbital.
There are four electrons in the p–orbital. Thus, the corresponding group of the element
= Number of s–block groups (3s2) + number of d–block groups ([Ne]10 + number of p–electrons(3p4)
= 2 + 10 + 4
= 16
Therefore, the element belongs to the 3rd period and 16th group of the periodic table. Hence, the element is Sulphur ([Ne]103s23p4)
(ii) Since n = 4, the element belongs to the 4th period. It is a d–block element as d–orbitals are incompletely filled.
There are 2 electrons in the d–orbital.
Thus, the corresponding group of the element
= Number of s–block groups (4s2) + number of d–block groups (3d2)
= 2 + 2
= 4
Therefore, it is a 4th period and 4th group element. Hence, the element is Titanium ([Ar]183d24s2.
(iii) Since n = 6, the element is present in the 6th period. It is an f –block element as the last electron occupies the f–orbital. It belongs to group 3 of the periodic table since all f-block elements belong to group 3. Its electronic configuration is [Xe]54 4f7 5d1 6s2. Thus, its atomic number is 54 + 7 + 2 + 1 = 64. Hence, the element is Gadolinium.
The first (ΔiH1) and the second (ΔiH) ionization enthalpies (in kJ mol–1) and the (ΔegH) electron gain enthalpy (in kJ mol–1) of a few elements are given below:
| Elements | ΔiH1 | ΔiH | ΔegH |
| I | 520 | 7300 | -60 |
| II | 419 | 3051 | -48 |
| III | 1681 | 3374 | -328 |
| IV | 1008 | 1846 | -295 |
| V | 2372 | 5251 | +48 |
| VI | 738 | 1451 | -40 |
Which of the above elements is likely to be :
(a) the least reactive element.
(b) the most reactive metal.
(c) the most reactive non-metal.
(d) the least reactive non-metal.
(e) the metal which can form a stable binary halide of the formula MX2, (X=halogen).
(f) the metal which can form a predominantly stable covalent halide of the formula MX (X=halogen)?
NCERT questions are designed to test your understanding of the concepts and theories discussed in the chapter. Here are some tips to help you answer NCERT questions effectively:
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Welcome to the NCERT Solutions for Class 11 Chemistry - Chapter . This page offers a step-by-step solution to the specific question from Excercise 1 , Question 30: Assign the position of the element having outer electronic configuration (i) ns2 np4 for n = 3 (i....
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Comments
It is a good source of niformation to the childern of 11 (Che)we expect you to keep more and more like this
For (n-1)d1ns2
What is oxidation number of carbon's in CH3CH2O4
Really helpful. Thanks