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Academic session 2026-27, Chapter 2 in the current curriculum.
Welcome to the NCERT Solutions for Class 11 Chemistry - Chapter Structure of Atom. This page offers a step-by-step solution to the specific question from Exercise 1, Question 23:
:(i) Write the electronic configurations of the following ions: (a) H– (b) Na+ (c) O2–(d) F–
(ii) What are the atomic numbers of elements whose outermost electrons are represented by (a) 3s1 (b) 2p3 and (c) 3p5?
(iii) Which atoms are indicated by the following configurations?
(a) [He] 2s1 (b) [Ne] 3s2 3p3 (c) [Ar] 4s2 3d1.
. 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.:(i) Write the electronic configurations of the following ions: (a) H– (b) Na+ (c) O2–(d) F–
(ii) What are the atomic numbers of elements whose outermost electrons are represented by (a) 3s1 (b) 2p3 and (c) 3p5?
(iii) Which atoms are indicated by the following configurations?
(a) [He] 2s1 (b) [Ne] 3s2 3p3 (c) [Ar] 4s2 3d1.
Electronic configuration of an atom is defined as the representation of the position of electrons in the various energy shell & subshells.
Now A negative charge on the species indicates the gain of an electron by it & A positive charge denotes the loss of an electron
(i) (a) H– ion
The electronic configuration of H atom is 1s1.(atomic number = 1)
∴ Electronic configuration of H– = 1s2
(b) Na+ ion
The electronic configuration of Na atom is 1s2 2s2 2p6 3s1.(atomic number = 11)
∴ Electronic configuration of Na+ = 1s2 2s2 2p6 3s0 Or 1s2 2s2 2p6
(c) O2– ion
The electronic configuration of 0 atom is 1s2 2s2 2p4.(atomic number = 8)
∴ Electronic configuration of O2– ion = 1s2 2s2 p6
(d) F– ion
The electronic configuration of F atom is 1s2 2s2 2p5.(atomic number = 9)
∴ Electron configuration of F– ion = 1s2 2s2 2p6
(ii) (a) 3s1
Completing the electron configuration of the element as 1s2 2s2 2p6 3s1.
∴ Number of electrons present in the atom of the element
= 2 + 2 + 6 + 1 = 11
∴ Atomic number of the element = 11(sodium)
(b) 2p3
Completing the electron configuration of the element as 1s2 2s2 2p3.
∴ Number of electrons present in the atom of the element = 2 + 2 + 3 = 7
∴ Atomic number of the element = 7(nitrogen)
(c) 3p5
Completing the electron configuration of the element as 1s2 2s2 2p5.
∴ Number of electrons present in the atom of the element = 2 + 2 + 5 = 9
∴ Atomic number of the element = 9(fluorine)
(iii) (a) [He] 2s1
The electronic configuration of the element is [He] 2s1 = 1s2 2s1.
∴ Atomic number of the element = 3 (lithium , a p-block element)
(b) [Ne] 3s2 3p3
The electronic configuration of the element is [Ne] 3s2 3p3= 1s2 2s2 2p6 3s2 3p3.
∴ Atomic number of the element = 15(phosphorous, a p block element)
(c) [Ar] 4s2 3d1
The electronic configuration of the element is [Ar] 4s2 3d1= 1s2 2s2 2p6 3s2 3p6 4s2 3d1.
∴ Atomic number of the element = 21(scandium , a d block element)
Following results are observed when sodium metal is irradiated with different wavelengths.
Calculate (a) threshold wavelength and, (b) Planck’s constant.
| λ (nm) | 500 | 450 |
400 |
| v × 10–5 (cm s–1) | 2.55 | 4.35 | 5.35 |
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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 23: :(i) Write the electronic configurations of the following ions: (a) H– (b) Na+ (c) O2–(d....
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11th class chemistry in Q-23 2nd part (c) u give the configuration of 2p5 and in quest. it is 3p5