ELECTRONS IN ATOMS

1. GENERAL
CHEMISTRY
ELECTRONS IN ATOMS
PERIODIC TABLE - QUESTIONS
Fuat TOPUZ
Istanbul Technical
University
Department of Chemistry
Week - 2

2. Exercise question:
Which n values are invalid? Why?
a) n = 3
b) n = 0
c) n = -2
d) n = 5005
e) n = 3.75
f) n = ½
g) n = 1
Ans: b, c, e, f

3. Exercise question:
Which l values are invalid for an electron with n = 2? Why?
a) l = 1
b) l = 0
c) l = 3
d) l = 0.8
e) l = -1
f) l = -2
g) l = +∞
Ans: c, d, e, f, g
l = 0 (s orbitals):
l = 1 (p orbitals):

4. Exercise question:
Which orbitals are impossible (invalid) for an electron? Why?
a) 3s
b) 1p
c) 2s
d) 1s
e) 3f
f) 2p
g) 4f
h) 5d
i) 2b
If n = 1, l = 0 (orbital s only)
Ans: b, e, i
If n = 3, l = 0,1,2 (orbitals s,p,d only)
No such thing as b orbital

5. Exercise question:
Which ml values are invalid for an electron with l = 2 ? Why?
a) ml = 1
b) ml = 0
c) ml = 3
d) ml = 0.8
e) ml = -1
f) ml = -2
g) ml = +∞
Ans: c, d, g
l = 2 (d orbitals):
ml = -2, -1, 0, 1, 2 ⇒ 5 orientations

6. Exercise question:
We have a maximum of how many different orientations in f-orbitals?
Ans: 7

7. List the values of n, ℓ, and mℓ for orbitals in the 4d subshell.
Exercise question:

8. Strategy What are the relationships among n, ℓ, and mℓ?
What do “4” and “d” represent in 4d?
Solution As we saw earlier, the number given in the designation of the subshell is the
principal quantum number, so in this case n = 4. The letter designates the type of orbital.
Because we are dealing with d orbitals, ℓ = 2. The values of mℓ can vary from −ℓ to ℓ.
Therefore, mℓ can be −2, −1, 0, 1, or 2.
Check The values of n and ℓ are fixed for 4d, but mℓ can have any one of the five values,
which correspond to the five d orbitals.

9. Exercise question:
a) List the values of n, ℓ, and mℓ for orbitals in the 4f subshell.
b) What is the number of orbitals in 4f subshell?
c) What is the number of orientations of orbitals in 4f subshell?
Ans: a) n=4, l=3, ml = -3,-2,-1,0, 1, 2, 3 b) 7 c)7

10. What is the total number of orbitals associated with the principal quantum number n
= 3?
Exercise question:

11. Strategy To calculate the total number of orbitals for a given n value, we need to first write
the possible values of ℓ. We then determine how many mℓ values are associated with each
value of ℓ. The total number of orbitals is equal to the sum of all the mℓ values.
Solution For n = 3, the possible values of ℓ are 0, 1, and 2. Thus, there is one 3s orbital (n
= 3, ℓ = 0, and mℓ = 0); there are three 3p orbitals (n = 3, ℓ = 1, and mℓ = −1, 0, 1); there are
five 3d orbitals (n = 3, ℓ = 2, and mℓ = −2, −1, 0, 1, 2). The total number of orbitals is 1 + 3 +
5 = 9.
Check The total number of orbitals for a given value of n is n2. So here we have 32 = 9.
Can you prove the validity of this relationship?

12. Exercise question:
a) List the possible subshells for n=4.
(show the letters corresponding to possible l values together with the n value as a prefix)
b) What is the number of orbitals in each subshell?
c) What is the total number of orbitals?
A shortcut method to find the number of orbitals is to use the equation:
Number of orbitals = n2
However, answer this question by listing the orbitals for each subshell and sum them
up. Compare it with the result of the above-given equation.
Ans: a) l=0, 1, 2, 3 (5s, 5p, fd, 5f) b) 1, 3, 5, 7 c) 16

13. A period on the periodic table best corresponds to the _____ in electron
configurations
a) shell
b) subshell
c) orbital
d) spin
Exercise question:
(Ans: Shell)

14. Which of the following can contain 10 electrons?
a) 4s orbital
b) 5p subshell
c) shell 2
d) 3d subshell
Exercise question:
(Ans: 3d subshell)

15. Number of Electrons in Valence Shells for Atoms
Referring to Table 3.2, indicate the number of electrons in the
valence shell of elements in groups IA(1), IIA(2), IIIA(13), and
IVA(14);
The elements in group IA(1) are hydrogen, lithium, sodium, and
potassium;
 Each element has one electron in the valence shell
 Hydrogen belongs in group IA(1) on the basis of its electronic
structure, but its properties differ significantly from other
members
 Group IIA(2) elements are beryllium, magnesium, and calcium
•Each has two electrons in the valence shell
 Group IIIA(13) elements are boron and aluminum
•Both have three electrons in the valence shell

16. How many valence electrons does arsenic (#33) have?
a) 2
b) 3
c) 5
d) 8
Exercise question:
(Ans: 5)

17. Exercise question:
A maximum of how many electrons can be placed in the following subshells?
(What is the electron capacity of the following subshells)
a) 3s
b) 5d
c) 4f
d) n=2, l=0
Ans: 2, 10, 14, 2

18. An oxygen atom has a total of eight electrons. Write the four
quantum numbers for each of the eight electrons in the ground
state.

19. Strategy
We start with n = 1 and proceed to fill orbitals in the order shown in
Figure 7.21.
For each value of n we determine the possible values of ℓ.
For each value of ℓ, we assign the possible values of mℓ.
We can place electrons in the orbitals according to the Pauli exclusion
principle and Hund’s rule.

20. Solution
We start with n = 1, so ℓ = 0, a subshell corresponding to the 1s orbital. This orbital
can accommodate a total of two electrons. Next, n = 2, and / may be either 0 or 1. The
ℓ = 0 subshell contains one 2s orbital, which can accommodate two electrons. The
remaining four electrons are placed in the ℓ = 1 subshell, which contains three 2p
orbitals. The orbital diagram is

21. The results are summarized in the following table:
Of course, the placement of the eighth electron in the orbital labeled mℓ = 1 is completely arbitrary. It
would be equally correct to assign it to mℓ = 0 or mℓ = −1.

22. Exercise question:
Which electron arrangements in an orbital are forbidden (invalid)?
a) (↑ )
b) (↓ )
c) (↑↓)
d) (↓↑)
e) (↑↑)
f) (↓↓)
g) (1/2, -1/2)
h) (1/2, 1/2)
i) (-1/2, -1/2)
Ans: e, f, h, i

23. Exercise question:
Which electron arrangements are not possible for 4 electrons in a 3d subshell?
a) (↑ ) (↑ ) (↑ ) (↑ ) ( )
b) (↓ ) (↓ ) (↓ ) (↓ ) ( )
c) (↑↓) (↑↓) ( ) ( ) ( )
d) (↑↑) (↑↑) ( ) ( ) ( )
e) (↑ ) (↑ ) (↓ ) (↓ ) ( ) Ans: c, d, e

24. An atom of a certain element has 15 electrons. Without consulting a
periodic table, answer the following questions:
(a)What is the ground-state electron configuration of the element?
(b) How should the element be classified?
(c) Is the element diamagnetic or paramagnetic?

25. Strategy
(a) We refer to the building-up principle discussed in Section 7.9
and start writing the electron configuration with principal quantum
number n = 1 and continuing upward until all the electrons are
accounted for.
(b) What are the electron configuration characteristics of
representative elements? transition elements? noble gases?
(c) Examine the pairing scheme of the electrons in the outermost shell.
What determines whether an element is diamagnetic or paramagnetic?

26. Solution
(a)We know that for n = 1 we have a 1s orbital (2 electrons); for n = 2 we
have a 2s orbital (2 electrons) and three 2p orbitals (6 electrons); for n = 3
we have a 3s orbital (2 electrons). The number of electrons left is 15 − 12
= 3 and these three electrons are placed in the 3p orbitals. The electron
configuration is 1s22s22p63s23p3.
(b) Because the 3p subshell is not completely filled, this is a
representative element. Based on the information given, we cannot say
whether it is a metal, a nonmetal, or a metalloid.
(c) According to Hund’s rule, the three electrons in the 3p orbitals have
parallel spins (three unpaired electrons). Therefore, the element is
paramagnetic.

27. Exercise question:
a) Write the electron configuration for fluorine, neon and sodium
b) Write the “abbreviated electron configuration for sodium.
Test yourself:
a) Write the electron configuration for iodine (I), xenon (Xe) and cesium
(Cs)
b) Write the “abbreviated electron configuration for cesium.
Ans: a) F: 1s2 2s2 2p5 Ne: 1s2 2s2 2p6 Na:1s2 2s2 2p6 3s1 b) [Ne] 3s1

28. Exercise question:
Write the ground-state electron configuration and abbreviated electron configuration
for the following elements by using the periodic table.
a) Ge
b) Hg
Ans: Practice by yourself for part a and electron configuration for part b. Here is the abbreviated electron configuration for part b: [Xe] 6s2 4f14 5d10

29. •Which of the following electron configurations matches aluminum?
a) [Ne]3s23p1
b) [Ne]3p3
c) [Ar]3p5
d) [Ar]3s22p1
Exercise question:
(Ans: [Ne] 3s2 3p1)
29

30. •How many unpaired electrons does selenium (#34) have?
a) 1
b) 2
c) 3
d) 4
Exercise question:
(Ans: 2)
30

31. Exercise question:
A fluorine atom has a total of nine electrons.
a) Write the electron configuration.
b) Write the orbital diagram.
c) Is fluorine paramagnetic or diamagnetic?
d) Write the four quantum numbers for the electrons in 2p. Ans:
a) 1s2 2s2 2p5
b) vvvvvvvvvvvvvvv
1s2 2s2 2p5 xxxx
(↿⇂) (↿⇂) (↿⇂)(↿⇂)(↿ )
c) Paramagnetic
d) n, l, ml ms
2, 0, -1, +1/2
2, 0, -1, -1/2
2, 0, 0, +1/2
2, 0, 0, -1/2
2, 0, +1, +1/2

32. Write the four quantum numbers for an electron in a 3p orbital.
Exercise question:

33. Strategy
What do the “3” and “p” designate in 3p?
How many orbitals (values of mℓ) are there in a 3p subshell?
What are the possible values of electron spin quantum number?
Solution To start with, we know that the principal quantum number n is 3 and the angular
momentum quantum number ℓ must be 1 (because we are dealing with a p orbital). For ℓ =
1, there are three values of mℓ given by −1, 0, and 1. Because the electron spin quantum
number ms can be either +½ or −½, we conclude that there are six possible ways to
designate the electron using the (n, ℓ , mℓ, ms) notation.

34. These are:
Check In these six designations we see that the values of n and ℓ are
constant, but the values of mℓ and ms can vary.

35. 35
Exercise question:
Write the possible four quantum numbers for an electron in
a 4p orbital.
Ans: See Example 7.8, solve similarly.

36. What is the maximum number of electrons that can be present in the
principal level for which n = 3?
Exercise question:

37. Strategy We are given the principal quantum number (n) so we can determine all
the possible values of the angular momentum quantum number (ℓ). The preceding
rule shows that the number of orbitals for each value of ℓ is (2 ℓ + 1). Thus, we can
determine the total number of orbitals. How many electrons can each orbital
accommodate?
Solution When n = 3, ℓ = 0, 1, and 2. The number of orbitals for each value of ℓ is
given by

38. Answer
The total number of orbitals is nine. Because each orbital can accommodate two
electrons, the maximum number of electrons that can reside in the orbitals is 2 × 9, or
18.
Check If we use the formula (n2) in Example 7.7, we find that the total number of
orbitals is 32 and the total number of electrons is 2(32) or 18. In general, the number of
electrons in a given principal energy level n is 2n2.

39. Exercise question:
What is the maximum number of electrons that can be present
in the principal level for which n = 4?

40. Groups and Periods of the Periodic Table
1. Identify the group and period to which each of the following belongs
a)P
b)Cr
c)Element number 30
d)Element number 53
2. How many elements are found in period 6 of the periodic table?
3. How many elements are found in group VA (15) of the periodic table?
Question

41. Exercise Question:
•Lithium belongs to group _____ and period _____
a) 1, 2
b) 2, 1
c) 1, 3
d) 3, 2