,Portage Learning CHEM 103 General
Chemistry I Module 5 Exam
Question 1
An element with atomic number \(Z = 11\) forms a stable ion. What is the charge and
electronic configuration of this ion?
A) \(-1\), \(1s^{2}2s^{2}2p^{6}3s^{2}\)
B) \(+1\), \(1s^{2}2s^{2}2p^{6}3s^{1}\)
C) \(+1\), \(1s^{2}2s^{2}2p^{6}\)
D) \(+2\), \(1s^{2}2s^{2}2p^{6}\)
Answer: C
Rationale: Atomic number 11 is Sodium (Na). Its neutral ground-state configuration is
\(1s^{2}2s^{2}2p^{6}3s^{1}\). As an alkali metal, it readily loses its single valence
electron from the 3s orbital to achieve a stable, filled octet matching the noble gas
Neon, forming a \(Na^{+}\) cation with a configuration of \(1s^{2}2s^{2}2p^{6}\).
Question 2
Which of the following electron configurations belongs to a stable nitride ion?
A) \(1s^{2}2s^{2}2p^{3}\)
B) \(1s^{2}2s^{2}2p^{6}\)
C) \(1s^{2}2s^{2}2p^{4}\)
D) \(1s^{2}2s^{2}\)
Answer: B
Rationale: Nitrogen (\(Z = 7\)) has a neutral configuration of \(1s^{2}2s^{2}2p^{3}\). To
fulfill the octet rule, it gains 3 valence electrons to completely fill its 2p subshell, forming
a \(N^{3-}\) anion. The resulting stable configuration is \(1s^{2}2s^{2}2p^{6}\), which is
isoelectronic with Neon.
Question 3
How many total valence electrons are available to draw the Lewis structure for the
carbonate ion (\(CO_{3}^{2-}\))?
, A) 22
B) 24
C) 26
D) 18
Answer: B
Rationale: Carbon contributes 4 valence electrons (\(1 \times 4 = 4\)). Each Oxygen
contributes 6 valence electrons (\(3 \times 6 = 18\)). The \(-2\) overall charge indicates
that 2 extra electrons have been gained (\(+2\)). Summing these gives: \(4 + 18 + 2 =
24\) total valence electrons.
Question 4
Using electronegativity trends, which of the following bonds is classified as the most
polar covalent?
A) \(C - H\)
B) \(O - H\)
C) \(F - F\)
D) \(Na - Cl\)
Answer: B
Rationale: Polar covalent bonds occur when the electronegativity difference (\(\Delta
\)EN) is between 0.5 and 1.7. Oxygen and Hydrogen have a large difference (\(\Delta
\)EN \(\approx 1.4\)), making it highly polar covalent. \(C-H\) (\(\Delta \)EN \(\approx
0.4\)) is nonpolar covalent, \(F-F\) (\(\Delta \)EN \(=0\)) is pure nonpolar, and \(Na-Cl\)
(\(\Delta \)EN \(>1.7\)) is predominantly ionic.
Question 5
According to VSEPR theory, what is the molecular geometry of a molecule with 3
bonding pairs and 1 lone pair on the central atom?
A) Trigonal Planar
B) Tetrahedral
C) Trigonal Pyramidal
D) Bent
Answer: C
Rationale: An atom with 4 total electron domains (3 bonding, 1 non-bonding) has a
tetrahedral electron-pair geometry. However, because the lone pair occupies space
without an attached atom, the actual arrangement of the atoms (molecular geometry) is
trigonal pyramidal, like in ammonia (\(NH_{3}\)).
Chemistry I Module 5 Exam
Question 1
An element with atomic number \(Z = 11\) forms a stable ion. What is the charge and
electronic configuration of this ion?
A) \(-1\), \(1s^{2}2s^{2}2p^{6}3s^{2}\)
B) \(+1\), \(1s^{2}2s^{2}2p^{6}3s^{1}\)
C) \(+1\), \(1s^{2}2s^{2}2p^{6}\)
D) \(+2\), \(1s^{2}2s^{2}2p^{6}\)
Answer: C
Rationale: Atomic number 11 is Sodium (Na). Its neutral ground-state configuration is
\(1s^{2}2s^{2}2p^{6}3s^{1}\). As an alkali metal, it readily loses its single valence
electron from the 3s orbital to achieve a stable, filled octet matching the noble gas
Neon, forming a \(Na^{+}\) cation with a configuration of \(1s^{2}2s^{2}2p^{6}\).
Question 2
Which of the following electron configurations belongs to a stable nitride ion?
A) \(1s^{2}2s^{2}2p^{3}\)
B) \(1s^{2}2s^{2}2p^{6}\)
C) \(1s^{2}2s^{2}2p^{4}\)
D) \(1s^{2}2s^{2}\)
Answer: B
Rationale: Nitrogen (\(Z = 7\)) has a neutral configuration of \(1s^{2}2s^{2}2p^{3}\). To
fulfill the octet rule, it gains 3 valence electrons to completely fill its 2p subshell, forming
a \(N^{3-}\) anion. The resulting stable configuration is \(1s^{2}2s^{2}2p^{6}\), which is
isoelectronic with Neon.
Question 3
How many total valence electrons are available to draw the Lewis structure for the
carbonate ion (\(CO_{3}^{2-}\))?
, A) 22
B) 24
C) 26
D) 18
Answer: B
Rationale: Carbon contributes 4 valence electrons (\(1 \times 4 = 4\)). Each Oxygen
contributes 6 valence electrons (\(3 \times 6 = 18\)). The \(-2\) overall charge indicates
that 2 extra electrons have been gained (\(+2\)). Summing these gives: \(4 + 18 + 2 =
24\) total valence electrons.
Question 4
Using electronegativity trends, which of the following bonds is classified as the most
polar covalent?
A) \(C - H\)
B) \(O - H\)
C) \(F - F\)
D) \(Na - Cl\)
Answer: B
Rationale: Polar covalent bonds occur when the electronegativity difference (\(\Delta
\)EN) is between 0.5 and 1.7. Oxygen and Hydrogen have a large difference (\(\Delta
\)EN \(\approx 1.4\)), making it highly polar covalent. \(C-H\) (\(\Delta \)EN \(\approx
0.4\)) is nonpolar covalent, \(F-F\) (\(\Delta \)EN \(=0\)) is pure nonpolar, and \(Na-Cl\)
(\(\Delta \)EN \(>1.7\)) is predominantly ionic.
Question 5
According to VSEPR theory, what is the molecular geometry of a molecule with 3
bonding pairs and 1 lone pair on the central atom?
A) Trigonal Planar
B) Tetrahedral
C) Trigonal Pyramidal
D) Bent
Answer: C
Rationale: An atom with 4 total electron domains (3 bonding, 1 non-bonding) has a
tetrahedral electron-pair geometry. However, because the lone pair occupies space
without an attached atom, the actual arrangement of the atoms (molecular geometry) is
trigonal pyramidal, like in ammonia (\(NH_{3}\)).
