MENU

Chapter 2 Structure of The Atom Solutions

Question - 61 : - The quantum numbers of six electrons are givenbelow. Arrange them in order of increasing energies. List if any of thesecombination(s) has/have the same energy
(i) n = 4, l = 2, ml =-2, ms = -1/2
(ii) n = 3, l = 2, ml =1, ms = +1/2
(iii) n = 4, l = 1, ml =0, ms = +1/2
(iv) n = 3, l = 2, ml =-2, ms = -111
(v) n = 3, l = l, ml =-1, ms = +1/2
(vi) n = 4, l = 1, ml =0, ms = +1/2

Answer - 61 : -

The electrons may beassigned to the following orbitals :
(i) 4d
(ii) 3d
(iii) 4p
(iv) 3d
(v) 3p
(vi) 4p.
The increasing order of energy is :
(v) < (ii) = (iv) < (vi), = (iii) < (i)

Question - 62 : - The bromine atom possesses 35 electrons. Itcontains 6 electrons in 2p orbital, 6 electrons in 3p orbital and 5 electronsin 4p orbital. Which of these electrons experiences lowest effective nuclearcharge ?

Answer - 62 : -

4p electronexperiences lowest effective nuclear charge because of the maximum magnitude ofscreening or shielding effect. It is farthest from the nucleus.

Question - 63 : - Among the following pairs of orbitals, whichorbital will experience more effective nuclear charge (i) 2s and 3s (ii) 4d and4f (iii) 3d and 3p ?

Answer - 63 : -

Please note thatgreater the penetration of the electron present in a particular orbital towardsthe nucleus, more will be the magnitude of the effective nuclear charge. Basedupon this,
(i) 2s electron willexperience more effective nuclear charge.
(ii) 4d electron willexperience more effective nuclear charge.
(iii) 3p electron willexperience more effective nuclear charge.

Question - 64 : - The unpaired electrons in A1 and Si arepresent in the 3p orbital. Which electrons will experience more effectivenuclear charge from the nucleus?

Answer - 64 : -

Configuration of thetwo elements are :
A1 (Z = 13) : [Ne]103s23p1 ; Si (Z = 14): [Ne] 103s23p2
The unpaired electrons in silicon (Si) will experience more effective nuclearcharge because the atomic number of the element Si is more than that of A1.

Question - 65 : - Indicate the number of unpaired electrons in :
(a) P (b) Si (c) Cr (d) Fe and (e) Kr.

Answer - 65 : -

(a) P (z=15) : [Ne]103s23p3 ;No. of unpaired electrons = 3
(b) Si (z=14) : [Ne]103s23p2 ;No. of unpaired electrons = 2
(c) Cr (z=24): [Ar]184s13d5 ;No. of unpaired electrons = 6
(d) Fe (z=26): [Ar]184s23d6 ;No. of unpaired electrons = 4
(e) Kr (z=36) : [Ar]184s23d104p6 ;No. of unpaired electrons = Nil.

Question - 66 : - (a) How many sub-shells are associated with n= 4 ?
(b) How many electrons will be presentin the sub-shells having ms value of -1/2 for n = 4 ?

Answer - 66 : -

(a) For n = 4 ; No.of sub-shells = (l = 0, l = 1, l = 2, l = 3) = 4.
(b) Total number oforbitals which can be present = n2 = 42 = 16.
Each orbital can have an electron with ms = – 1/2 -‘. Total no.of electrons with m, = – 1/2 is 16.

Free - Previous Years Question Papers
Any questions? Ask us!
NodeJS Interview Questions Answers
ASP.Net Interview Questions Answers
Interview Questions Answers
×