Chapter 2 Structure of The Atom Solutions
Question - 21 : - The mass of an electron is 9.1 × 10-31 kg.If its kinetic energy is 3.0 × 10-25 J, calculate itswavelength.
Answer - 21 : - 
Question - 22 : - Which of the following are iso-electronicspecies ?
Na+, K+, Mg2+,Ca2+, S2-, Ar.
Answer - 22 : -
Na+ andMg2+ are iso-electronic species (have 10 electrons) K+,Ca2+ , S2- are iso-electronic species (have 18electrons)
Question - 23 : - (i) Write the electronic configuration of thefollowing ions : (a) H (b) Na+ (c) 02~ (d) F–.
(ii) What are the atomic numbers of theelements whose outermost electronic configurations are represented by :
(a) 3s1 (b) Ip3 and(c) 3d6 ?
(iii) Which atoms are indicated by thefollowing configurations ?
(a) [He]2s1 (b) [Ne] 3s2 3p3 (c)[Ar] 4s2 3d1.
Answer - 23 : -
(i)(a) 1s2
(b) 1s2 2s2 2p6
(c) 1s22s22p6
(d) 1s22s22p6.
(ii) (a) Na (Z = 11) hasoutermost electronic configuration = 3s1
(b) N (Z = 7) has outermostelectronic configuration = 2p3
(c) Fe (Z = 26) has outermostelectronic configuration = 3d6
(iii) (a) Li
(b) P
(c) Sc
Question - 24 : - What is the lowest value of n which allows ‘g’orbital to exist ?
Answer - 24 : -
The lowest value of lw’here ‘g’ orbital can be present = 4
The lowest value of n where ‘g’ orbital can be present = 4+1=5.
Question - 25 : - An electron is in one of the 3d orbitals. Givethe possible values of n, l and nil for the electron.
Answer - 25 : -
For electron in 3dorbital, n = 3, l = 2, mi = -2, -1, 0, +1, +2.
Question - 26 : - An atom of an element contains 29 electronsand 35 neutrons. Deduce (i) the number of protons and (ii) the electronicconfiguration of the element.
Answer - 26 : -
No. of protons in aneutral atom = No. of electrons = 29
Electronic configuration = 1s2 2s2 2p6 3s2 3p6 3d10 4s1.
Question - 27 : - Give the number of electrons in the species :H2+, H2 and 02+.
Answer - 27 : -
H2+ =one ; H2 = two ; 02+ = 15
Question - 28 : - (i) An atomic orbital has n = 3. What are thepossible values of l and ml ?
(ii) List the quantum numbers ml andl of electron in 3rd orbital.
(iii) Which of the following orbitalsare possible ?
1p, 2s, 2p and 3f.
Answer - 28 : -
(i) For n = 3; l =0, 1 and 2.
For l = 0 ; ml = 0
For l = 1; ml = +1, 0, -1
For l = 2 ; ml = +2, +1,0, +1, + 2
(ii) For an electron in 3rdorbital ; n = 3; l = 2 ; ml can have any of the values -2, -1,0,
+ 1, +2.
(iii) 1p and 3f orbitalsare not possible.
Question - 29 : - Using s, p and d notations, describe theorbitals with follow ing quantum numbers :
(a) n = 1, l = 0
(b) n = 4, l = 3
(c) n = 3, l = 1
(d) n = 4, l = 2
Answer - 29 : -
(a) 1s orbital
(b) 4f orbital
(c) 3p orbital
(d) 4d orbital
Question - 30 : - From the following sets of quantum numbers,state which are possible. Explain why the others are not possible.
(i) n = 0, l = 0, ml =0, ms = +1/2
(ii) n = 1, l = 0, ml =0, ms – – 1/2
(iii) n = 1, l = 1, ml =0, ms= +1/2
(iv) n = 1, l = 0, ml =+1, ms= +1/2
(v) n = 3, l = 3, ml =-3, ms = +1/2
(vi) n = 3, l = 1, ml =0, ms= +1/2
Answer - 30 : -
(i) The set ofquantum numbers is not possible because the minimum value of n can be 1 and notzero.
(ii) The set of quantumnumbers is possible.
(iii) The set of quantumnumbers is not possible because, for n = 1, l can not be equal to 1. It canhave 0 value.
(iv) The set of quantumnumbers is not possible because for l = 0. mt cannot be + 1. It must be zero.
(v) The set of quantumnumbers is not possible because, for n = 3, l ≠ 3.
(vi) The set of quantumnumbers is possible.