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