7.3: Electron Configuration of Atoms
- Page ID
- 169580
Electron Configuration
Exercise \(\PageIndex{1}\)
Which of the following elements is found in the d-block of the periodic table?
- Ir
- Tb
- Li
- Cl
- None of these
- Answer
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a Ir
Exercise \(\PageIndex{2}\)
An element that has the same ground state valence-shell electron configuration as thallium is
a. gallium b. carbon c. krypton d. cesium e. magnesium
- Answer
-
a. gallium
Exercise \(\PageIndex{3}\)
How many valence electrons does an arsenic atom have?
a. 5 b. 8 c. 7 d. 2 e. 33
- Answer
-
a. 5
Exercise \(\PageIndex{4}\)
How many unpaired electrons are found in the ground state electron configuration of Barium (Ba)?
a. 0 b. 1 c. 2 d. 3 e. 5
- Answer
-
a. 0
Exercise \(\PageIndex{5}\)
Which of the following orbital diagrams represents a paramagnetic atom?
1s 2s 2p
a. 1 only b. 2 only c. 3 only d. 1 and 2 e. 2 and 3
- Answer
-
e. 2 and 3
Exercise \(\PageIndex{6}\)
Which of the following atoms is diamagnetic in its ground state?
- Mercury (Hg)
- Tin (Sn)
- Rhenium (Re)
- Berkelium (Bk)
- Phosphorus (P)
- Answer
-
a. Mercury (Hg)
Exercise \(\PageIndex{7}\)
Which of the following orbital diagrams represents a diamagnetic atom?
1s 2s 2p
- Answer
-
a.
Exercise \(\PageIndex{8}\)
Which atom has the ground state electronic configuration 1s22s22p63s23p64s23d3?
a. Ga b. V c. As d. Nb e. none
- Answer
-
b. V
Exercise \(\PageIndex{9}\)
Which of the following elements has the ground state electron configuration [Ar]3d104s1?
a. Cu b. Zn c. Ge d. Ag e. Cd
- Answer
-
a. Cu
Exercise \(\PageIndex{10}\)
What is the ground-state electron configuration of sulfur (S)?
a. [Ne]3s33sp3 b. [Ar]3s23p4 c. [Ar]3p6 d. [Ne]3s23p4 e. [Ar]3p6
- Answer
-
d. [Ne]3s23p4
Exercise \(\PageIndex{11}\)
Which of the following electron configurations corresponds to the ground state of an atom of a transition element?
- 1s22s22p1
- 1s22s22p63s23p63d104s24p3
- 1s22s22p63s23p63d14s2
- 1s22s22p63s23p64s1
- 1s22s22p63s23p4
- Answer
-
c. 1s22s22p63s23p63d14s2
Exercise \(\PageIndex{12}\)
The complete electron configuration of tin is _____.
- 1s22s22p63s23p64s23d104p65s24d105p2
- 1s22s22p63s23p64s23d104p65s24d105p2
- 1s22s22p63s23p64s23d104d104p2
- 1s22s22p63s23p64s23d104p65s24d105d105p2
- None of these
- Answer
-
a. 1s22s22p63s23p64s23d104p65s24d105p2
Exercise \(\PageIndex{13}\)
Hund's rule states that the most stable arrangement of electrons (for a ground state electron configuration)
- has a filled valence shell of electrons.
- has three electrons per orbital, each with identical spins.
-
has values greater than or equal to +1.
-
has the maximum number of unpaired electrons, all with the same spin.
-
has two electrons per orbital, each with opposing spins.
- Answer
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d. has the maximum number of unpaired electrons, all with the same spin.
Exercise \(\PageIndex{14}\)
All of the following ground-state electron configurations are correct except
- V: [Ar]4s24d3
- K: [Ar]4s1
- Sb: [Kr]4d105s25p3
- Cr: [Ar]3d54s1
- Te: [Kr]4d105s25p4
- Answer
-
a. V: [Ar]4s24d3
Exercise \(\PageIndex{15}\)
What noble gas core precedes the valence shell ground state electron configuration for potassium (K)?
a. [Ar] b. [Rn] c. [Kr] d. [Ne] e. [Xe]
- Answer
-
a. [Ar]
Exercise \(\PageIndex{16}\)
Which ground-state electron configuration is incorrect?
- Br: [Ar]3d104s24p5
- K: [Ar]4s1
- Ni: [Ar]3d5
- Mg: 1s22s22p63s2
- Co: [Ar]3d74s2
- Answer
-
c. Ni: [Ar]3d5
Exercise \(\PageIndex{17}\)
Which element has the following ground state electron configuration?
a. Be b. O c. Li d. Si e. N
- Answer
-
e. N
Exercise \(\PageIndex{18}\)
Which element has the following ground state electron configuration?
3d 4s
[Ar]
a. Sc b. Ni c. Co d. Fe e. V
- Answer
-
a. Sc
Exercise \(\PageIndex{19}\)
Which element has the following ground state electron configuration?
a. In b. Y c. Nb d. Tl e. Ga
- Answer
-
a. In
Exercise \(\PageIndex{20}\)
Which is the correct valence shell orbital box notation for the ground state electron configuration of Fe?
3d 4s
- Answer
-
a.
Exercise \(\PageIndex{21}\)
What is a possible set of quantum numbers for an unpaired electron in the orbital box diagram below?
- n = 1, l = 1, = –1, ms = \(+\frac{1}{2}\)
- n = 4, l = 2, = –1, ms = \(-1\frac{1}{2}\)
- n = 5, l = 2, = –2, ms = \(+\frac{1}{2}\)
- n = 5, l = 0, = 0, ms = \(-1\frac{1}{2}\)
- n = 5, l = 1, = –1, ms = \(+\frac{1}{2}\)
- Answer
-
e. n = 5, l = 1, = –1, ms = \(+\frac{1}{2}\)