Indicate the number of unpaired electron | Class 11 Chemistry Chapter Structure of Atom, Structure of Atom NCERT Solutions

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 66: . 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.

Question 66:

Indicate the number of unpaired electrons in: (a) P, (b) Si, (c) Cr, (d) Fe and (e) Kr.

Answer:

(a) Phosphorus (P): 1s2 2s2 2p6 3s2 3p3

No of unpaired electron = 3

 

(b) Silicon (Si): 1s2 2s2 2p6 3s2 3p2

No of unpaired electron = 2 (since p orbital can have maximum 6 electron )

 

(c) Chromium (Cr): 1s2 2s2 2p6 3s2 3p6 4s1 3d5

No of unpaired electrons = 6 (since 1 electron is to be added to 4s & 5 electron to be added to 3d orbital.

 

(d) Iron (Fe): 1s2 2s2 2p6 3s2 3p6 4s2 3d6

No of unpaired electrons = 4

 

(e) Krypton (Kr): 1s2 2s2 2p6 3s2 3p6 4s2 3d10 4p6

Since all orbitals are fully occupied, there are no unpaired electrons in krypton.


Study Tips for Answering NCERT Questions:

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  • Read the question carefully and focus on the core concept being asked.
  • Reference examples and data from the chapter when answering questions about Structure of Atom.
  • Review previous year question papers to get an idea of how such questions may be framed in exams.
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Comments

  • Saurav kumar chaturvedi
  • Jul 18, 2018

It is very good trick. But my question is that. What amount of cyclohexane will contain same number of bond as the number of unpaired electron in 16.42liter o Oxygen gas measured as 1.2 atmospheric pressure and 27°c


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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 66: Indicate the number of unpaired electrons in: (a) P, (b) Si, (c) Cr, (d) Fe and (e) Kr.....