CHEM 122 - GENERAL CHEMISTRY II -
InOrganic Chemistry Question Bank
Question 1
Step-by-step solution: 1. Coordination number in inorganic chemistry refers
to the number of atoms or ions that surround a central metal atom or ion in
a complex. 2. The coordination number indicates the total number of bonds
formed by the ligands to the central metal atom or ion. 3. The coordination
number often determines the geometry of the complex, with common coordina-
tion numbers including 4 (tetrahedral), 6 (octahedral), and 2 (linear). 4. In gen-
eral, coordination number is an important concept in understanding the struc-
ture and properties of coordination complexes in inorganic chemistry.Question
1: Define coordination number in the context of inorganic chemistry.
Step-by-step solution: 1. Coordination number in inorganic chem-
istry refers to the number of atoms or ions that surround a central
metal atom or ion in a complex. 2. The coordination number indi-
cates the total number of bonds formed by the ligands to the central
metal atom or ion. 3. The coordination number often determines
the geometry of the complex, with common coordination numbers
including 4 (tetrahedral), 6 (octahedral), and 2 (linear). 4. In gen-
eral, coordination number is an important concept in understanding
the structure and properties of coordination complexes in inorganic
chemistry.
Question 2
Solution:
1. Molecular geometry and bond angles of BF3:
Step 1: Draw the Lewis structure of BF3.
B
F
F F
1
Step 2: Determine the steric number of the central atom (Boron).
Steric number =Number of lone pairs on the central atom
+Number of atoms bonded to the central atom
= 0 + 3
= 3
Step 3: Determine the molecular geometry based on the steric
number.
For a steric number of 3, the molecular geometry is trigonal
planar.
Step 4: Determine the bond angles in BF3.
The bond angles in BF3are approximately 120.
2. Molecular geometry and bond angles of CH4:
(You can follow similar steps as above for CH4and NH3.)
Question 2: Determine the molecular geometry and bond angles
for the following molecules:
1. BF3
2. CH4
3. NH3
Solution:
1. Molecular geometry and bond angles of BF3:
Step 1: Draw the Lewis structure of BF3.
B
F
F F
Step 2: Determine the steric number of the central atom (Boron).
Steric number =Number of lone pairs on the central atom
+Number of atoms bonded to the central atom
= 0 + 3
= 3
Step 3: Determine the molecular geometry based on the steric
number.
2
For a steric number of 3, the molecular geometry is trigonal
planar.
Step 4: Determine the bond angles in BF3.
The bond angles in BF3are approximately 120.
2. Molecular geometry and bond angles of CH4:
(You can follow similar steps as above for CH4and NH3.)
Question 3
Solution: To balance the chemical equation, we need to make sure
that the number of atoms of each element is the same on both sides
of the equation.
Let’s count the number of atoms for each element:
Iron (Fe): 1 atom on the left side, 2 atoms on the right side
Oxygen (O): 2 atoms on the left side (1 in Fe and 1 in O2), 3
atoms on the right side
The equation is currently unbalanced because the number of atoms
of Fe and O are not the same on both sides.
To balance the equation, we can start by adding a coefficient in
front of Fe:
4Fe +O2→Fe2O3
Now let’s count the number of atoms again:
Iron (Fe): 4 atoms on the left side, 2 atoms on the right side
Oxygen (O): 2 atoms on the left side (1 in Fe and 1 in O2), 3
atoms on the right side
The number of Fe atoms is now balanced, but the number of O
atoms is still not the same on both sides.
To balance the number of O atoms, we can add a coefficient in
front of Fe2O3:
4Fe + 3O2→2Fe2O3
Now the chemical equation is balanced with 4 Fe atoms and 6 O
atoms on both sides.Question 3: Balance the following chemical equa-
tion:
Fe +O2→Fe2O3
Solution: To balance the chemical equation, we need to make sure
that the number of atoms of each element is the same on both sides
of the equation.
Let’s count the number of atoms for each element:
3
Iron (Fe): 1 atom on the left side, 2 atoms on the right side
Oxygen (O): 2 atoms on the left side (1 in Fe and 1 in O2), 3
atoms on the right side
The equation is currently unbalanced because the number of atoms
of Fe and O are not the same on both sides.
To balance the equation, we can start by adding a coefficient in
front of Fe:
4Fe +O2→Fe2O3
Now let’s count the number of atoms again:
Iron (Fe): 4 atoms on the left side, 2 atoms on the right side
Oxygen (O): 2 atoms on the left side (1 in Fe and 1 in O2), 3
atoms on the right side
The number of Fe atoms is now balanced, but the number of O
atoms is still not the same on both sides.
To balance the number of O atoms, we can add a coefficient in
front of Fe2O3:
4Fe + 3O2→2Fe2O3
Now the chemical equation is balanced with 4 Fe atoms and 6 O
atoms on both sides.
Question 4
Solution: To write the formula for the ionic compound formed
between magnesium and oxygen, we first determine the charges of
the ions. Magnesium (Mg) is a metal that typically forms a 2+
cation, while oxygen (O) is a non-metal that typically forms a 2-
anion. Therefore, the formula for the ionic compound between mag-
nesium and oxygen is Mg2+ and O2−.
To write the formula, we need to balance the charges. Since the
charges are equal in magnitude, we only need one of each ion to
balance the charges. Therefore, the formula for the compound is
MgO.Question 4: Write the formula for the ionic compound formed
between magnesium and oxygen.
Solution: To write the formula for the ionic compound formed
between magnesium and oxygen, we first determine the charges of
the ions. Magnesium (Mg) is a metal that typically forms a 2+
cation, while oxygen (O) is a non-metal that typically forms a 2-
anion. Therefore, the formula for the ionic compound between mag-
nesium and oxygen is Mg2+ and O2−.
To write the formula, we need to balance the charges. Since the
charges are equal in magnitude, we only need one of each ion to
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balance the charges. Therefore, the formula for the compound is
MgO.
Question 5
Balance the following chemical equation:
Al +O2→Al2O3
Step-by-step solution:
1. Write down the unbalanced chemical equation provided:
Al +O2→Al2O3
2. Count the number of atoms for each element on both the re-
actant and product sides of the equation: Reactant side: Aluminum
(Al): 1 atom Oxygen (O): 2 atoms
Product side: Aluminum (Al): 2 atoms Oxygen (O): 3 atoms
3. In order to balance the equation, start by adjusting the coeffi-
cients in front of the compounds. Let’s balance the number of atoms
for Aluminum first:
Al +O2→2Al2O3
Now, check the number of atoms for Oxygen: Reactant side: 2
atoms Product side: 2×3 = 6 atoms
4. Since the number of oxygen atoms is not equal, adjust the
coefficient in front of O2:
4Al + 3O2→2Al2O3
Now, double-check the number of atoms for each element: Reac-
tant side: Aluminum (Al): 4×1 = 4 atoms Oxygen (O): 3×2 = 6
atoms
Product side: Aluminum (Al): 2×2 = 4 atoms Oxygen (O): 2×3=6
atoms
5. The balanced chemical equation is:
4Al + 3O2→2Al2O3
Question 5:
Balance the following chemical equation:
Al +O2→Al2O3
Step-by-step solution:
1. Write down the unbalanced chemical equation provided:
Al +O2→Al2O3
5
2. Count the number of atoms for each element on both the re-
actant and product sides of the equation: Reactant side: Aluminum
(Al): 1 atom Oxygen (O): 2 atoms
Product side: Aluminum (Al): 2 atoms Oxygen (O): 3 atoms
3. In order to balance the equation, start by adjusting the coeffi-
cients in front of the compounds. Let’s balance the number of atoms
for Aluminum first:
Al +O2→2Al2O3
Now, check the number of atoms for Oxygen: Reactant side: 2
atoms Product side: 2×3 = 6 atoms
4. Since the number of oxygen atoms is not equal, adjust the
coefficient in front of O2:
4Al + 3O2→2Al2O3
Now, double-check the number of atoms for each element: Reac-
tant side: Aluminum (Al): 4×1 = 4 atoms Oxygen (O): 3×2 = 6
atoms
Product side: Aluminum (Al): 2×2 = 4 atoms Oxygen (O): 2×3=6
atoms
5. The balanced chemical equation is:
4Al + 3O2→2Al2O3
Question 6
Balance the following chemical equation:
F eS2+O2→F e2O3+SO2
Step-by-step Solution:
1. Write down the unbalanced equation:
F eS2+O2→F e2O3+SO2
2. Count the number of atoms of each element on both sides of
the equation: - For Fe: 1 iron on the left, 2 iron on the right - For
S: 2 sulfur on the left, 3 sulfur on the right - For O: 2 oxygen on the
left, 5 oxygen on the right
3. Begin by balancing the most complex molecule first. In this
case, balance Fe in F e2O3:
F eS2+O2→2F e2O3+SO2
4. Check if other elements are balanced. In this case, check Sulfur
(S) next: - 2 sulfur on the left, 2 sulfur on the right
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5. Lastly, balance Oxygen (O) by adjusting the coefficient of O2
on the reactant side:
4F eS2+ 11O2→2F e2O3+ 8SO2
The balanced chemical equation is:
4F eS2+ 11O2→2F e2O3+ 8SO2
Question 6:
Balance the following chemical equation:
F eS2+O2→F e2O3+SO2
Step-by-step Solution:
1. Write down the unbalanced equation:
F eS2+O2→F e2O3+SO2
2. Count the number of atoms of each element on both sides of
the equation: - For Fe: 1 iron on the left, 2 iron on the right - For
S: 2 sulfur on the left, 3 sulfur on the right - For O: 2 oxygen on the
left, 5 oxygen on the right
3. Begin by balancing the most complex molecule first. In this
case, balance Fe in F e2O3:
F eS2+O2→2F e2O3+SO2
4. Check if other elements are balanced. In this case, check Sulfur
(S) next: - 2 sulfur on the left, 2 sulfur on the right
5. Lastly, balance Oxygen (O) by adjusting the coefficient of O2
on the reactant side:
4F eS2+ 11O2→2F e2O3+ 8SO2
The balanced chemical equation is:
4F eS2+ 11O2→2F e2O3+ 8SO2
Question 7
Explain the VSEPR theory and its role in predicting the geome-
tries of molecules.
Solution:
The VSEPR theory, which stands for Valence Shell Electron Pair
Repulsion theory, is used in chemistry to predict the shapes of molecules
based on the repulsions between electron pairs in the valence shell
of an atom. The theory assumes that electron pairs (bonding and
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non-bonding) around a central atom repel each other and try to get
as far away from each other as possible to minimize repulsion, thus
determining the molecular geometry.
The basic steps to use VSEPR theory to predict the geometry of
a molecule are as follows:
1. Determine the central atom in the molecule. 2. Count the total
number of electron pairs around the central atom (bonding pairs and
lone pairs). 3. Predict the arrangement of these electron pairs around
the central atom based on minimizing repulsions. 4. Determine the
molecular geometry based on the arrangement of electron pairs.
For example, let’s consider the molecule NH3:
1. The central atom is nitrogen. 2. There are a total of 4 electron
pairs around nitrogen (3 bonding pairs and 1 lone pair). 3. The
electron pairs will arrange themselves in a trigonal pyramidal shape
to minimize repulsion. 4. Therefore, the molecular geometry of NH3
is trigonal pyramidal.
This is how the VSEPR theory helps in predicting the geometries
of molecules based on the arrangement of electron pairs. “‘“‘latex
Question 7:
Explain the VSEPR theory and its role in predicting the geome-
tries of molecules.
Solution:
The VSEPR theory, which stands for Valence Shell Electron Pair
Repulsion theory, is used in chemistry to predict the shapes of molecules
based on the repulsions between electron pairs in the valence shell
of an atom. The theory assumes that electron pairs (bonding and
non-bonding) around a central atom repel each other and try to get
as far away from each other as possible to minimize repulsion, thus
determining the molecular geometry.
The basic steps to use VSEPR theory to predict the geometry of
a molecule are as follows:
1. Determine the central atom in the molecule. 2. Count the total
number of electron pairs around the central atom (bonding pairs and
lone pairs). 3. Predict the arrangement of these electron pairs around
the central atom based on minimizing repulsions. 4. Determine the
molecular geometry based on the arrangement of electron pairs.
For example, let’s consider the molecule NH3:
1. The central atom is nitrogen. 2. There are a total of 4 electron
pairs around nitrogen (3 bonding pairs and 1 lone pair). 3. The
electron pairs will arrange themselves in a trigonal pyramidal shape
to minimize repulsion. 4. Therefore, the molecular geometry of NH3
is trigonal pyramidal.
This is how the VSEPR theory helps in predicting the geometries
of molecules based on the arrangement of electron pairs. “‘
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Question 8
F eCl3+N aOH →F e(OH)3+N aC l
Solution:
To balance the chemical equation, we need to ensure that the
number of atoms of each element is the same on both the reactant
and product sides. Let’s start by counting the atoms of each element
on both sides:
Reactants:
F e : 1
Cl : 3
N a : 1
O: 1
H: 1
Products:
F e : 1
Cl : 1
N a : 1
O: 3
H: 3
Now, let’s balance the equation:
F eCl3+ 3N aOH →F e(OH)3+ 3N aC l
This balanced equation ensures that the number of atoms of each
element is the same on both sides.
Therefore, the balanced chemical equation is:
F eCl3+ 3N aOH →F e(OH)3+ 3N aC l
Question 8: Balance the following chemical equation:
F eCl3+N aOH →F e(OH)3+N aC l
Solution:
To balance the chemical equation, we need to ensure that the
number of atoms of each element is the same on both the reactant
and product sides. Let’s start by counting the atoms of each element
on both sides:
Reactants:
F e : 1
Cl : 3
N a : 1
O: 1
H: 1
Products:
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F e : 1
Cl : 1
N a : 1
O: 3
H: 3
Now, let’s balance the equation:
F eCl3+ 3N aOH →F e(OH)3+ 3N aC l
This balanced equation ensures that the number of atoms of each
element is the same on both sides.
Therefore, the balanced chemical equation is:
F eCl3+ 3N aOH →F e(OH)3+ 3N aC l
Question 9
Step-by-step solutions: a) Calcium (Ca): Calcium has an atomic
number of 20. The electron configuration for calcium can be deter-
mined by writing out the electron configuration for the preceding
noble gas, which is argon (Ar). Then, continuing with the electron
configuration for calcium.
The electron configuration for calcium is: 1s
²
2s
²
2p 3s
²
3p 4s
²
b) Manganese (Mn): Manganese has an atomic number of 25. The
electron configuration for manganese can be determined by writing
out the electron configuration for the preceding noble gas, which is
argon (Ar). Then, continuing with the electron configuration for
manganese.
The electron configuration for manganese is: 1s
²
2s
²
2p 3s
²
3p 4s
²
3d
c) Sulfur (S): Sulfur has an atomic number of 16. The electron
configuration for sulfur can be determined by writing out the electron
configuration for the preceding noble gas, which is neon (Ne). Then,
continuing with the electron configuration for sulfur.
The electron configuration for sulfur is: 1s
²
2s
²
2p 3s
²
3pQuestion
9: Write the electron configuration for the following elements: a)
Calcium (Ca) b) Manganese (Mn) c) Sulfur (S)
Step-by-step solutions: a) Calcium (Ca): Calcium has an atomic
number of 20. The electron configuration for calcium can be deter-
mined by writing out the electron configuration for the preceding
noble gas, which is argon (Ar). Then, continuing with the electron
configuration for calcium.
The electron configuration for calcium is: 1s
²
2s
²
2p 3s
²
3p 4s
²
b) Manganese (Mn): Manganese has an atomic number of 25. The
electron configuration for manganese can be determined by writing
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out the electron configuration for the preceding noble gas, which is
argon (Ar). Then, continuing with the electron configuration for
manganese.
The electron configuration for manganese is: 1s
²
2s
²
2p 3s
²
3p 4s
²
3d
c) Sulfur (S): Sulfur has an atomic number of 16. The electron
configuration for sulfur can be determined by writing out the electron
configuration for the preceding noble gas, which is neon (Ne). Then,
continuing with the electron configuration for sulfur.
The electron configuration for sulfur is: 1s
²
2s
²
2p 3s
²
3p
Question 10
Explain the VSEPR theory and apply it to predict the molecular
geometry of the following molecules:
a) CO2
b) NH3
c) SF6
Step-by-step Solutions:
a) CO2:
1. Determine the number of valence electrons in the molecule:
Carbon has 4 valence electrons, and each oxygen has 6 valence elec-
trons. Thus, the total number of valence electrons for CO2is 4+2(6) =
16.
2. Determine the central atom: In CO2, carbon is the central
atom.
3. Determine the electron pair geometry: Since CO2has 2 bonding
pairs and 0 lone pairs around the central atom, the electron pair
geometry is linear.
4. Determine the molecular geometry: With a linear electron pair
geometry, the molecular geometry for CO2is also linear.
b) NH3:
1. Determine the number of valence electrons in the molecule:
Nitrogen has 5 valence electrons, and each hydrogen has 1 valence
electron. Thus, the total number of valence electrons for NH3is
5 + 3(1) = 8.
2. Determine the central atom: In NH3, nitrogen is the central
atom.
3. Determine the electron pair geometry: With 3 bonding pairs
and 1 lone pair around the central atom, the electron pair geometry
of NH3is trigonal pyramidal.
4. Determine the molecular geometry: Based on the trigonal pyra-
midal electron pair geometry, the molecular geometry for NH3is also
trigonal pyramidal.
c) SF6:
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1. Determine the number of valence electrons in the molecule: Sul-
fur has 6 valence electrons, and each fluorine has 7 valence electrons.
Thus, the total number of valence electrons for SF6is 6 + 6(7) = 48.
2. Determine the central atom: In SF6, sulfur is the central atom.
3. Determine the electron pair geometry: With 6 bonding pairs
and 0 lone pairs around the central atom, the electron pair geometry
of SF6is octahedral.
4. Determine the molecular geometry: Based on the octahedral
electron pair geometry, the molecular geometry for SF6is also octa-
hedral.
These predictions are made using the VSEPR theory, which states
that electron pairs around a central atom will arrange themselves to
be as far apart as possible, resulting in specific molecular geome-
tries.Question 10:
Explain the VSEPR theory and apply it to predict the molecular
geometry of the following molecules:
a) CO2
b) NH3
c) SF6
Step-by-step Solutions:
a) CO2:
1. Determine the number of valence electrons in the molecule:
Carbon has 4 valence electrons, and each oxygen has 6 valence elec-
trons. Thus, the total number of valence electrons for CO2is 4+2(6) =
16.
2. Determine the central atom: In CO2, carbon is the central
atom.
3. Determine the electron pair geometry: Since CO2has 2 bonding
pairs and 0 lone pairs around the central atom, the electron pair
geometry is linear.
4. Determine the molecular geometry: With a linear electron pair
geometry, the molecular geometry for CO2is also linear.
b) NH3:
1. Determine the number of valence electrons in the molecule:
Nitrogen has 5 valence electrons, and each hydrogen has 1 valence
electron. Thus, the total number of valence electrons for NH3is
5 + 3(1) = 8.
2. Determine the central atom: In NH3, nitrogen is the central
atom.
3. Determine the electron pair geometry: With 3 bonding pairs
and 1 lone pair around the central atom, the electron pair geometry
of NH3is trigonal pyramidal.
4. Determine the molecular geometry: Based on the trigonal pyra-
midal electron pair geometry, the molecular geometry for NH3is also
trigonal pyramidal.
c) SF6:
12
1. Determine the number of valence electrons in the molecule: Sul-
fur has 6 valence electrons, and each fluorine has 7 valence electrons.
Thus, the total number of valence electrons for SF6is 6 + 6(7) = 48.
2. Determine the central atom: In SF6, sulfur is the central atom.
3. Determine the electron pair geometry: With 6 bonding pairs
and 0 lone pairs around the central atom, the electron pair geometry
of SF6is octahedral.
4. Determine the molecular geometry: Based on the octahedral
electron pair geometry, the molecular geometry for SF6is also octa-
hedral.
These predictions are made using the VSEPR theory, which states
that electron pairs around a central atom will arrange themselves to
be as far apart as possible, resulting in specific molecular geometries.
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