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GNDU Question Paper-2022
Bachelor of Computer Application (BCA) (Hons.)
1
st
Semester (Batch 2024-28) (CBGS)
CHEMISTRY
(Inorganic Chemistry-I)
Time Allowed: Three Hours Max. Marks:35
Note: Attempt Five questions in all, selecting at least One question from each section. The
Fifth question may be attempted from any section. All questions carry equal marks.
SECTION-A
1. (a) Write brief note on Hund's multiplicity rule.
(b) Give electronic configuration of Cr (Z=24) and Fe3+ (Z=26) .
(c) What do you understand by radial probability distribution curves? Draw radial
probability distribution curves for 3p and 3d orbitals. What information do these curves
provide?
2. (a) Name and draw various orbitals possible for n = 3 and/ = 1.
(b) What do you understand by quantum numbers ? Also discuss the physical significance
of different quantum numbers.
SECTION-B
3. (a) What are isoelectronic ions? Give one example.
(b) What is effective nuclear charge? Calculate effective nuclear charge for one of the
outer electrons (4s) of copper atom (At. No. = 29).
(c) Calculate the percentage ionic character in HBr molecule. Given electronegativity
values of H and Br are 2.1 and 2.8, respectively.
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4. (a) Define covalent radius of atoms. How it differs from van der Waals radius?
(b) What is ionization energy? Give its variation in a period and in a group in the periodic
table. Also discuss various factors in detail which affect ionization energy.
SECTION-C
5. (a) Discuss sp³d³ hybridization with one example.
(b) Draw MO diagram of CO molecule. Also predict its bond order.
6. (a) Discuss the shape of SF4 on the basis of VSEPR Theory.
(b) How will you prepare diborane? Draw its structure. Also discuss nature of bonding in
it.
SECTION-D
7. (a) Define coordination number. What is the coordination number of Na in NaCl?
(b) Draw and explain Born-Haber cycle for the formation of KCI.
8. (a) Write a brief note on radius ratio rules.
(b) What do you understand by defects in crystals? Briefly discuss the following types of
defects in crystals:
(i) Schottky defect
(ii) Frenkel defect.
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GNDU Answer Paper-2022
Bachelor of Computer Application (BCA) (Hons.)
1
st
Semester (Batch 2024-28) (CBGS)
CHEMISTRY
(Inorganic Chemistry-I)
Time Allowed: Three Hours Max. Marks:35
Note: Attempt Five questions in all, selecting at least One question from each section. The
Fifth question may be attempted from any section. All questions carry equal marks.
SECTION-A
1. (a) Write brief note on Hund's multiplicity rule.
(b) Give electronic configuration of Cr (Z=24) and Fe3+ (Z=26) .
(c) What do you understand by radial probability distribution curves? Draw radial
probability distribution curves for 3p and 3d orbitals. What information do these curves
provide?
Ans: 1. (a) Hund's Multiplicity Rule
Hund's Multiplicity Rule is a simple rule that helps us understand how electrons are
arranged in orbitals of the same energy (called degenerate orbitals). For example, the three
p orbitals (px, py, pz) have equal energy.
The rule states:
"Electrons first occupy all empty orbitals singly with parallel spins before any pairing takes
place."
This happens because electrons repel each other. By staying in separate orbitals first, they
remain farther apart, reducing repulsion and making the atom more stable.
Example
Suppose we have three electrons to fill in the p orbitals.
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Correct arrangement (Hund's Rule):
px py pz
↑ ↑ ↑
Each orbital gets one electron first.
Wrong arrangement:
px py pz
↑↓ ↑ _
Here, two electrons are paired before all orbitals are occupied. This is not allowed according
to Hund's Rule because it increases electron repulsion.
Why is this rule important?
It gives the most stable electronic arrangement.
It reduces electron-electron repulsion.
It helps explain the magnetic properties of atoms.
1. (b) Electronic Configuration of Cr (Z = 24) and Fe³⁺ (Z = 26)
(i) Chromium (Cr), Atomic Number = 24
Normally, we might expect the configuration to be:
1s² 2s² 2p⁶ 3s² 3p⁶ 3d⁴ 4s²
However, chromium is an exception.
Its actual electronic configuration is:
1s² 2s² 2p⁶ 3s² 3p⁶ 3d⁵ 4s¹
or
[Ar] 3d⁵ 4s¹
Why is Chromium an exception?
A half-filled d-subshell (3d⁵) is more stable than 3d⁴. Therefore, one electron from the 4s
orbital moves into the 3d orbital.
Normal:
3d : ↑ ↑ ↑ ↑
4s : ↑↓
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Actual:
3d : ↑ ↑ ↑ ↑ ↑
4s : ↑
This arrangement provides extra stability.
(ii) Fe³⁺ (Iron Ion)
Iron (Fe) has atomic number 26.
Electronic configuration of neutral Fe:
[Ar] 3d⁶ 4s²
To form Fe³⁺, three electrons are removed.
Electrons are removed first from the 4s orbital, then from the 3d orbital.
Step 1:
Remove two electrons from 4s
[Ar] 3d⁶
Step 2:
Remove one electron from 3d
[Ar] 3d⁵
Therefore,
Electronic configuration of Fe³⁺ is
[Ar] 3d⁵
Again, the 3d⁵ half-filled configuration is highly stable.
1. (c) Radial Probability Distribution Curves
The radial probability distribution curve tells us where an electron is most likely to be
found at different distances from the nucleus.
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Instead of saying the exact position of an electron (which is impossible according to
quantum mechanics), scientists calculate the probability of finding it.
The graph shows:
X-axis: Distance from the nucleus
Y-axis: Probability of finding the electron
The higher the peak, the greater the chance of finding the electron at that distance.
Radial Probability Curve for 3p Orbital
Probability
^
| /\
| / \
| /\ / \
| / \____/ \____
|_____/________________________> Distance
Node
The 3p orbital has one radial node.
Therefore, it shows two probability peaks.
Radial Probability Curve for 3d Orbital
Probability
^
| /\
| / \
| / \
|_________________/______\________> Distance
The 3d orbital has no radial node.
It has one major probability peak.
What information do these curves provide?
The radial probability distribution curves give valuable information about electron behavior.
1. Most probable distance of the electron from the nucleus
o The highest peak shows where the electron is most likely to be found.
2. Presence of radial nodes
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o A radial node is a region where the probability of finding the electron is zero.
o The 3p orbital has one radial node, while the 3d orbital has none.
3. Electron penetration
o Orbitals that penetrate closer to the nucleus experience stronger nuclear
attraction.
o This affects the energy of orbitals.
4. Comparison of different orbitals
o These curves help compare how electrons are distributed in different orbitals
like s, p, d, and f.
5. Understanding atomic structure
o They explain why some orbitals are more stable than others and help predict
chemical behavior.
Conclusion
Hund's Multiplicity Rule explains that electrons fill orbitals singly with parallel spins before
pairing, making atoms more stable. Chromium and Fe³⁺ are important examples because
both finally achieve the stable half-filled 3d⁵ configuration. Radial probability distribution
curves show the regions where electrons are most likely to be found around the nucleus and
help us understand electron distribution, orbital penetration, and atomic stability. Together,
these concepts form the foundation of quantum chemistry and help explain the
arrangement and behavior of electrons inside atoms.
2. (a) Name and draw various orbitals possible for n = 3 and/ = 1.
(b) What do you understand by quantum numbers ? Also discuss the physical significance
of different quantum numbers.
Ans: 2. (a) Name and draw various orbitals possible for n = 3 and l = 1
To understand this question, imagine an atom as a multi-floor building where electrons live.
The principal quantum number (n) tells us the floor or energy level.
The azimuthal quantum number (l) tells us the type of room (orbital) on that floor.
Here:
n = 3 means the electron is in the third energy level (third shell).
l = 1 means it belongs to the p-subshell.
For l = 1, the magnetic quantum number (mₗ) can have three values:
mₗ = -1
mₗ = 0
mₗ = +1
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These three values represent three different p orbitals.
Therefore, when n = 3 and l = 1, the possible orbitals are:
3pₓ
3pᵧ
3p_z
Each orbital has the same shape but points in a different direction.
Shape of p-Orbitals
3px (Along X-axis)
( )
------●------
( )
3py (Along Y-axis)
( )
|
●
|
( )
3pz (Along Z-axis)
●
( = Coming out of the page)
( = Going into the page)
Important Points
All three 3p orbitals have equal energy in an isolated atom.
Each orbital can hold 2 electrons.
Therefore, the 3p subshell can accommodate a total of 6 electrons.
2. (b) What do you understand by Quantum Numbers? Discuss the physical significance of
different quantum numbers.
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Quantum numbers are a set of four numbers that completely describe the position and
behavior of an electron inside an atom.
Think of an electron like a student in a large school.
To identify one particular student, we need information such as:
Which building?
Which classroom?
Which seat?
Which direction is the student facing?
Similarly, scientists use four quantum numbers to identify every electron in an atom.
Without quantum numbers, it would be impossible to know exactly where an electron
belongs.
There are four quantum numbers.
1. Principal Quantum Number (n)
This tells us the main energy level (shell) in which the electron is present.
It also tells us:
Size of the orbital
Distance from the nucleus
Energy of the electron
Possible values:
n = 1, 2, 3, 4......
Example:
n = 1 → First shell (K)
n = 2 → Second shell (L)
n = 3 → Third shell (M)
Higher value of n means:
Larger orbital
Electron farther from nucleus
Higher energy
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2. Azimuthal Quantum Number (l)
This tells us the shape of the orbital.
Its value depends upon n.
Possible values:
l = 0 to (n − 1)
For example:
If n = 3
l = 0,1,2
These represent:
l Value
Subshell
Shape
0
s
Spherical
1
p
Dumbbell
2
d
Clover leaf
3
f
Complex
Thus, l determines the type of orbital.
3. Magnetic Quantum Number (mₗ)
This tells us the orientation (direction) of an orbital in space.
Possible values:
mₗ = -l to +l
Example:
For l = 1
mₗ = -1,0,+1
Hence three orbitals are formed:
px
py
pz
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This is why the p-subshell has three orbitals.
4. Spin Quantum Number (mₛ)
This tells us the spin direction of an electron.
Every electron spins in one of two possible ways:
+½ (Clockwise)
−½ (Anticlockwise)
Only two electrons can occupy one orbital, and their spins must be opposite. This follows
the Pauli Exclusion Principle, which states that no two electrons in an atom can have the
same set of all four quantum numbers.
Example:
↑ ↓
One electron spins upward and the other spins downward.
Physical Significance of Quantum Numbers
Quantum Number
Symbol
Physical Significance
Principal Quantum
Number
n
Determines the shell, energy, size of the orbital and
distance from the nucleus.
Azimuthal
Quantum Number
l
Determines the subshell and the shape of the orbital (s, p,
d, f).
Magnetic Quantum
Number
mₗ
Determines the orientation of the orbital in three-
dimensional space.
Spin Quantum
Number
mₛ
Determines the direction of electron spin and explains
why only two electrons can occupy one orbital.
Conclusion
Quantum numbers are like the complete address of an electron inside an atom. The
principal quantum number (n) tells us the electron's energy level, the azimuthal quantum
number (l) describes the orbital shape, the magnetic quantum number (mₗ) specifies the
orbital's orientation, and the spin quantum number (mₛ) indicates the electron's spin
direction. For n = 3 and l = 1, the electron belongs to the 3p subshell, which contains three
orbitals3pₓ, 3pᵧ, and 3p_z. Together, these four quantum numbers uniquely identify
every electron in an atom and help explain the arrangement of electrons, the structure of
atoms, and the chemical behavior of elements.
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SECTION-B
3. (a) What are isoelectronic ions? Give one example.
(b) What is effective nuclear charge? Calculate effective nuclear charge for one of the
outer electrons (4s) of copper atom (At. No. = 29).
(c) Calculate the percentage ionic character in HBr molecule. Given electronegativity
values of H and Br are 2.1 and 2.8, respectively.
Ans: (a) Isoelectronic Ions
Imagine that electrons are students sitting in a classroom. Two ions may belong to different
elements, but if both have the same number of electrons, they are called isoelectronic ions.
Although the number of protons is different, the electron arrangement becomes identical.
Definition
Isoelectronic ions are ions that have the same number of electrons and the same
electronic configuration but different nuclear charges (different number of protons).
Example
Na⁺: Sodium has 11 electrons. After losing one electron, it has 10 electrons.
Mg²⁺: Magnesium has 12 electrons. After losing two electrons, it also has 10
electrons.
Na → Na⁺ = 10 electrons
Mg → Mg²⁺ = 10 electrons
Both have the same electronic configuration:
1s² 2s² 2p⁶
Therefore, Na⁺ and Mg²⁺ are isoelectronic ions.
(b) Effective Nuclear Charge (Zeff)
Every atom has a positively charged nucleus that attracts electrons. However, inner
electrons act like a shield, reducing the attraction felt by the outer electrons.
Think of it like this:
Nucleus (+29)
│
Inner Electrons
(Shielding Effect)
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│
4s Electron
The 4s electron does not feel the full +29 positive charge because the inner electrons block
part of the attraction.
Definition
Effective nuclear charge is the net positive charge experienced by an outer electron after
the shielding effect of inner electrons is taken into account.
Formula
eff
Where:
Z = Atomic number
S = Shielding constant
Copper Configuration
Copper (Z = 29)
1s² 2s² 2p⁶ 3s² 3p⁶ 3d¹⁰ 4s¹
Using Slater's Rules for the 4s electron:
Same shell (4s, 4p): 0 × 0.35 = 0
3s² 3p⁶ electrons: 8 × 0.85 = 6.8
3d¹⁰ electrons: 10 × 1.00 = 10
1s² 2s² 2p⁶ electrons: 10 × 1.00 = 10
Shielding constant:
Therefore,
eff
Answer
Effective nuclear charge on the 4s electron of copper = +2.2
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This means the outermost electron behaves as if it is attracted by only about +2.2 units of
positive charge, not the full +29.
(c) Percentage Ionic Character of HBr
A chemical bond can be covalent (sharing electrons) or ionic (transfer of electrons). Most
bonds lie somewhere between these two extremes.
The greater the difference in electronegativity, the greater the ionic character.
Step 1: Find the electronegativity difference
H = 2.1
Br = 2.8
Δχ = 2.8 − 2.1 = 0.7
Step 2: Use the formula
Ionic Character
Substitute the values:
Answer
Percentage ionic character of HBr = approximately 11.5%.
Quick Revision
Concept
Easy Meaning
Isoelectronic ions
Different ions having the same number of electrons. Example:
Na⁺ and Mg²⁺.
Effective nuclear
charge (Zeff)
The actual positive pull felt by an outer electron after inner
electrons reduce (shield) the nucleus's attraction.
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Shielding effect
Inner electrons block part of the nucleus's attraction, so outer
electrons feel a smaller pull.
Electronegativity
The ability of an atom to attract shared electrons in a bond.
Percentage ionic
character
Indicates how ionic a bond is. A larger electronegativity
difference means a higher percentage ionic character.
Final Answers
(a) Isoelectronic ions are ions having the same number of electrons. Example: Na⁺
and Mg²⁺.
(b) Effective nuclear charge on the 4s electron of Cu = +2.2.
(c) Percentage ionic character of HBr ≈ 11.5%.
By understanding these three ideas together, you can see that atoms become stable by
arranging electrons, outer electrons do not feel the full pull of the nucleus because of
shielding, and differences in electronegativity determine how ionic or covalent a chemical
bond will be. These concepts are closely connected and form the foundation for
understanding atomic structure and chemical bonding.
4. (a) Define covalent radius of atoms. How it differs from van der Waals radius?
(b) What is ionization energy? Give its variation in a period and in a group in the periodic
table. Also discuss various factors in detail which affect ionization energy.
Ans: Atoms are extremely tiny particles, so their exact size cannot be measured directly.
Scientists determine the size of an atom by measuring the distance between the nuclei of
two nearby atoms. Two important terms used to describe atomic size are covalent radius
and van der Waals radius.
Covalent Radius
The covalent radius is half of the distance between the nuclei of two identical atoms that
are joined together by a single covalent bond.
For example, in a chlorine molecule (Cl₂), two chlorine atoms are connected by a covalent
bond. If the distance between the two nuclei is 198 pm (picometers), then:
Covalent Radius = 198 ÷ 2 = 99 pm
This radius tells us how large an atom is when it forms a chemical bond.
van der Waals Radius
The van der Waals radius is half of the distance between the nuclei of two atoms that are
close to each other but are not chemically bonded. These atoms attract each other only
through weak intermolecular forces called van der Waals forces.
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Since there is no strong bond pulling the atoms together, they remain farther apart.
Difference between Covalent Radius and van der Waals Radius
Covalent Radius
van der Waals Radius
Measured between bonded atoms.
Measured between non-bonded atoms.
Distance is smaller because atoms are
bonded.
Distance is larger because atoms are only weakly
attracted.
Used for molecules having covalent
bonds.
Used for noble gases and molecules without
covalent bonding.
Diagram
1. Covalent Radius
H -------- H
<----74 pm---->
Covalent Radius = 74/2 = 37 pm
2. van der Waals Radius
He He
<--- Larger Distance --->
(No chemical bond)
Conclusion:
The covalent radius measures the size of an atom when it forms a covalent bond, whereas
the van der Waals radius measures the size of an atom when it is simply near another atom
without forming a bond. Therefore, the van der Waals radius is always larger than the
covalent radius.
4. (b) What is Ionization Energy? Explain its Variation in a Period and a Group. Also discuss
the Factors Affecting Ionization Energy.
Imagine that an atom is like a house and the electrons are family members living inside it.
The nucleus is like the head of the family holding everyone together. If you want to remove
one family member from the house, you need to use some effort. Similarly, removing an
electron from an atom also requires energy. This required energy is called Ionization
Energy.
Definition of Ionization Energy
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Ionization Energy is the minimum amount of energy required to remove the outermost
(most loosely held) electron from an isolated gaseous atom to form a positive ion.
Example:
Na(g) + Energy → Na⁺(g) + e⁻
The energy supplied removes one electron from the sodium atom.
Variation of Ionization Energy in the Periodic Table
1. Across a Period (Left → Right)
Ionization energy generally increases.
Why?
Nuclear charge increases.
Atomic size becomes smaller.
Electrons are attracted more strongly by the nucleus.
More energy is needed to remove an electron.
Example:
Li Be B C N O F Ne
↑------------------------------↑
Ionization Energy Increases
So, Lithium has lower ionization energy, while Neon has the highest ionization energy in
that period.
2. Down a Group (Top → Bottom)
Ionization energy generally decreases.
Reason:
Atomic size increases.
More electron shells are added.
Outer electrons are farther from the nucleus.
Shielding effect increases.
Less energy is needed to remove the outermost electron.
Example:
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Li
↓
Na
↓
K
↓
Rb
↓
Cs
Ionization Energy Decreases
Factors Affecting Ionization Energy
1. Atomic Size
The larger the atom, the farther the outer electron is from the nucleus.
Large atom → Low ionization energy
Small atom → High ionization energy
Example:
Cesium has a lower ionization energy than Lithium because its outer electron is much
farther away.
2. Nuclear Charge
The nucleus contains positively charged protons.
More protons mean stronger attraction for electrons.
Stronger attraction means more energy is required to remove an electron.
Therefore:
Higher nuclear charge = Higher ionization energy
3. Shielding (Screening) Effect
Inner-shell electrons block some of the nucleus's attractive force from reaching the outer
electrons.
More inner electrons mean:
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Greater shielding
Weaker attraction on outer electrons
Lower ionization energy
4. Distance of the Outer Electron
If the outermost electron is close to the nucleus, it is held tightly.
Shorter distance → Higher ionization energy
Greater distance → Lower ionization energy
5. Electronic Configuration
Atoms with completely filled or half-filled orbitals are especially stable.
Removing an electron from these stable arrangements requires extra energy.
Examples:
Noble gases (Ne, Ar) have very high ionization energies.
Nitrogen has a higher ionization energy than oxygen because its half-filled p-orbitals
are more stable.
6. Penetration Effect
Electrons in different orbitals are not equally close to the nucleus.
The order of penetration is:
s > p > d > f
Since s-electrons stay closest to the nucleus, they are held most strongly and require the
highest energy to remove.
Simple Trend Diagram
Increasing Ionization Energy
←------------------------------------→
Li Be B C N O F Ne
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Top
Li
Na
K
Rb
Cs
Bottom
Ionization Energy Decreases
Key Points to Remember
Covalent radius is half the distance between two bonded atoms.
van der Waals radius is half the distance between two non-bonded atoms and is
always larger than the covalent radius.
Ionization energy is the minimum energy needed to remove the outermost electron
from a gaseous atom.
Across a period, ionization energy increases because atoms become smaller and
nuclear attraction becomes stronger.
Down a group, ionization energy decreases because atomic size and shielding
increase.
The main factors affecting ionization energy are atomic size, nuclear charge,
shielding effect, distance of the outer electron, electronic configuration, and
penetration effect.
SECTION-C
5. (a) Discuss sp³d³ hybridization with one example.
(b) Draw MO diagram of CO molecule. Also predict its bond order.
Ans: Hybridization is a process in which the atomic orbitals of an atom mix together to form
new orbitals called hybrid orbitals. These hybrid orbitals have the same energy and help the
atom form stable chemical bonds. Think of it like mixing different colors of paint to create a
new color. Similarly, different orbitals mix to create new orbitals suitable for bonding.
What is sp³d³ Hybridization?
In sp³d³ hybridization, one s orbital, three p orbitals, and three d orbitals combine
together.
1 s orbital
3 p orbitals
3 d orbitals
Total orbitals mixed = 7
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So, 7 identical sp³d³ hybrid orbitals are formed.
These seven orbitals arrange themselves in a shape called the Pentagonal Bipyramidal
Geometry.
This geometry contains:
5 atoms in one pentagonal plane.
2 atoms above and below the plane.
Simple Diagram
X
|
X - A - X
/ | \
X X X
\ /
X-----X
|
X
(A = Central Atom, X = Surrounding Atoms)
The actual shape is Pentagonal Bipyramidal.
Example: IF₇ (Iodine Heptafluoride)
The best example of sp³d³ hybridization is IF₇.
Central atom = Iodine (I)
Surrounding atoms = 7 Fluorine (F) atoms
Iodine has 7 valence electrons, and it forms 7 sigma (σ) bonds with seven fluorine atoms.
To accommodate these seven bonds, iodine undergoes sp³d³ hybridization, producing 7
hybrid orbitals, each forming one σ bond with a fluorine atom.
Thus, IF₇ has a Pentagonal Bipyramidal shape.
Key Points
One s, three p, and three d orbitals combine.
Total hybrid orbitals formed = 7
Geometry = Pentagonal Bipyramidal
Bond angle ≈ 72° (in pentagonal plane) and 90° (between axial and equatorial
atoms).
Example = IF₇
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(b) Draw the Molecular Orbital (MO) Diagram of CO Molecule and Predict its Bond Order
The Carbon Monoxide (CO) molecule consists of one Carbon atom (C) and one Oxygen
atom (O).
Carbon contributes 6 electrons.
Oxygen contributes 8 electrons.
Total electrons in CO = 14
According to Molecular Orbital (MO) Theory, these electrons are filled into molecular
orbitals from lower energy to higher energy, following the Aufbau Principle.
MO Energy Order
For CO, the molecular orbitals are filled in the following order:
Higher Energy
σ*2p
----------
π*2p
-------------
π2p
↑↓ ↑↓
----------
σ2p
↑↓
----------
π2p
----------
σ2s*
↑↓
----------
σ2s
↑↓
Lower Energy
Electronic Configuration of CO
The electronic configuration is:
(σ1s)² (σ1s)² (σ2s)² (σ2s)² (π2p)⁴ (σ2p)²
Only the valence molecular orbitals are considered when discussing bonding.
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Bond Order Formula
The bond order is calculated using:
Bond Order
Bonding Electrons Antibonding Electrons
Calculation
Bonding electrons = 10
Antibonding electrons = 4
So,
Bond Order
Final Answer
Bond Order = 3
This means the carbon and oxygen atoms are connected by a triple bond (C≡O).
Why is the Bond Order Important?
A higher bond order means:
Stronger chemical bond.
Shorter bond length.
Greater stability.
Since CO has a bond order of 3, it has one of the strongest bonds among diatomic
molecules.
6. (a) Discuss the shape of SF4 on the basis of VSEPR Theory.
(b) How will you prepare diborane? Draw its structure. Also discuss nature of bonding in
it.
Ans: The shape of SF₄ (Sulphur Tetrafluoride) can be easily understood with the help of
VSEPR Theory. VSEPR stands for Valence Shell Electron Pair Repulsion Theory. This theory
says that electron pairs present around the central atom repel each other and arrange
themselves as far apart as possible to reduce repulsion. Because of this arrangement,
molecules get different shapes.
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In the SF₄ molecule, Sulphur (S) is the central atom. Sulphur has 6 valence electrons. It
forms four covalent bonds with four fluorine (F) atoms by using four of its electrons. After
making these four bonds, one lone pair of electrons remains on the sulphur atom.
So, around the sulphur atom there are:
4 bond pairs
1 lone pair
This means there are 5 electron pairs in total. According to VSEPR Theory, five electron pairs
arrange themselves in a trigonal bipyramidal geometry. However, the lone pair occupies an
equatorial position because it experiences less repulsion there than at the axial positions.
As a result, the actual shape of the SF₄ molecule becomes see-saw (or distorted
tetrahedral) instead of trigonal bipyramidal. The lone pair pushes the bonded fluorine
atoms slightly closer together, causing the bond angles to become smaller than the ideal
values.
Diagram of SF₄
F (axial)
|
S
.. | \
F(eq) | F(eq)
(Lone Pair)
|
F (axial)
(.. = Lone Pair)
Key Points
Central atom = Sulphur (S)
Bond pairs = 4
Lone pairs = 1
Electron pair geometry = Trigonal Bipyramidal
Molecular shape = See-Saw
Reason = Lone pair repels bonding pairs more strongly.
(b) How will you prepare Diborane? Draw its structure. Also discuss the nature of bonding
in it.
Diborane (B₂H₆) is one of the most interesting compounds of boron because its bonding is
different from ordinary molecules.
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Preparation of Diborane
Diborane is commonly prepared by reacting Sodium Borohydride (NaBH₄) with Iodine (I₂).
Chemical Equation:
In the laboratory, diborane can also be prepared by reducing boron trifluoride (BF₃) with
lithium aluminium hydride (LiAlH₄).
Structure of Diborane
A diborane molecule contains:
2 Boron atoms
6 Hydrogen atoms
Out of the six hydrogen atoms:
Four are terminal hydrogen atoms, each attached to one boron atom by a normal
covalent bond.
Two are bridge hydrogen atoms, which connect both boron atoms.
Structure
H
|
H B H
\ /
H H
/ \
H B H
|
H
(The two middle H atoms are bridge hydrogen atoms.)
Nature of Bonding in Diborane
The bonding in diborane is unusual because boron has only three valence electrons and
cannot form enough normal covalent bonds to complete its octet.
There are two types of bonds:
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1. Four normal covalent (2-center, 2-electron) bonds
o These are formed between boron and the four terminal hydrogen atoms.
2. Two bridge bonds (3-center, 2-electron bonds)
o Each bridge hydrogen is shared between two boron atoms.
o These bonds involve three atoms (BHB) but only two electrons.
o Therefore, they are called three-center two-electron (3c2e) bonds, also
known as banana bonds because of their curved shape.
These bridge bonds help both boron atoms achieve greater stability despite having fewer
electrons.
Important Points to Remember
SF₄
o Central atom: Sulphur
o Bond pairs: 4
o Lone pair: 1
o Electron geometry: Trigonal bipyramidal
o Molecular shape: See-Saw
Diborane (B₂H₆)
o Prepared from NaBH₄ and I₂ or BF₃ and LiAlH₄
o Contains 4 terminal hydrogen atoms and 2 bridge hydrogen atoms
o Has four normal covalent bonds and two three-center two-electron
(banana) bonds
Thus, SF₄ demonstrates how lone pair repulsion changes molecular shape according to
VSEPR Theory, while diborane is a classic example of electron-deficient bonding, where
special three-center two-electron bonds stabilize the molecule. These two compounds are
important examples in inorganic chemistry because they show that molecular shape and
bonding are not always explained by ordinary covalent bonding alone.
SECTION-D
7. (a) Define coordination number. What is the coordination number of Na in NaCl?
(b) Draw and explain Born-Haber cycle for the formation of KCI.
Ans: The coordination number is the number of nearest neighboring ions or atoms that
surround a particular ion in a crystal structure.
In simple words, imagine you are standing in the middle of a group of friends. The number
of friends standing closest to you is like your coordination number. Similarly, in an ionic
crystal, every ion is surrounded by a fixed number of oppositely charged ions.
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For example, in Sodium Chloride (NaCl), each sodium ion (Na⁺) is surrounded by six
chloride ions (Cl⁻). At the same time, each chloride ion (Cl⁻) is also surrounded by six
sodium ions (Na⁺). This arrangement gives the crystal stability because every positive ion is
attracted equally by six negative ions.
Structure of NaCl
Cl⁻
|
Cl⁻ Na⁺ Cl⁻
|
Cl⁻
Front: Cl⁻
Back : Cl⁻
Here, the sodium ion has 6 nearest chloride ions around it.
Answer
Coordination Number: The number of nearest neighboring ions surrounding an ion
in a crystal.
Coordination Number of Na⁺ in NaCl = 6
7. (b) Draw and Explain the BornHaber Cycle for the Formation of KCl
The BornHaber Cycle is a step-by-step energy diagram used to explain how an ionic
compound is formed from its elements. Instead of showing the reaction in one single step,
it divides the process into several smaller energy changes.
Think of it like building a house. A house is not built in one step. First, bricks are made,
cement is prepared, walls are built, and finally the roof is added. Similarly, potassium
chloride (KCl) is formed through several energy steps.
Overall Reaction
The formation of KCl involves the following steps:
Step 1: Sublimation of Potassium
Solid potassium is converted into gaseous potassium atoms.
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K(s) → K(g)
This requires energy because atoms must separate from the solid.
Step 2: Ionization of Potassium
A gaseous potassium atom loses one electron to form a positive ion.
K(g) → K⁺(g) + e⁻
Energy is absorbed because removing an electron is difficult.
Step 3: Dissociation of Chlorine
A chlorine molecule splits into chlorine atoms.
½Cl₂(g) → Cl(g)
Energy is needed to break the ClCl bond.
Step 4: Electron Affinity of Chlorine
A chlorine atom gains an electron to become a chloride ion.
Cl(g) + e⁻ → Cl⁻(g)
Energy is released because chlorine attracts electrons strongly.
Step 5: Lattice Formation
The gaseous potassium ion and chloride ion combine to form solid potassium chloride.
K⁺(g) + Cl⁻(g) → KCl(s)
A large amount of energy is released, making the crystal stable. This released energy is
called lattice energy.
BornHaber Cycle Diagram
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K(g) + Cl(g)
|
Ionization | Electron Affinity
|
K(g) + Cl(g)
|
Lattice Energy
↓
KCl(s)
↑
Heat of Formation (ΔHf)
K(s) + ½Cl(g)
|
Sublimation & Bond Dissociation
Why is the BornHaber Cycle Important?
The BornHaber Cycle helps us understand:
How ionic compounds are formed step by step.
Which steps absorb energy and which release energy.
Why ionic compounds like KCl are stable.
How lattice energy can be calculated using Hess's Law.
Key Points to Remember
Coordination Number: Number of nearest neighboring ions around an ion.
Coordination Number of Na⁺ in NaCl = 6.
BornHaber Cycle: A thermochemical cycle showing all energy changes during the
formation of an ionic compound.
Energy is absorbed during sublimation, ionization, and bond dissociation.
Energy is released during electron affinity and lattice formation.
The large lattice energy released is the main reason why KCl is a stable ionic
compound.
8. (a) Write a brief note on radius ratio rules.
(b) What do you understand by defects in crystals? Briefly discuss the following types of
defects in crystals:
(i) Schottky defect
(ii) Frenkel defect.
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Ans: The Radius Ratio Rule is a simple rule used in chemistry to predict how ions are
arranged in an ionic crystal. It tells us which type of hole (space) a positive ion (cation) can
fit into among negative ions (anions).
Imagine you have large footballs arranged together. Now, you want to place a small tennis
ball in the empty space between them. If the tennis ball is too small, it will not touch all the
footballs properly and the arrangement becomes unstable. If it is too large, it will not fit.
Similarly, in ionic crystals, the size of the positive ion compared to the negative ion
determines where it can fit.
The radius ratio is calculated as:
Radius Ratio = Radius of Cation / Radius of Anion
Depending on this ratio, the cation occupies different positions in the crystal, known as the
coordination number.
Radius Ratio
Coordination Number
Shape
Less than 0.155
2
Linear
0.155 0.225
3
Triangular
0.225 0.414
4
Tetrahedral
0.414 0.732
6
Octahedral
Greater than 0.732
8
Cubic
Simple Diagram
Large ions (O)
O
O ● O
O
● = Small positive ion (Cation)
O = Large negative ions (Anions)
The position of the ● (cation) depends on its size. A larger cation can touch more
surrounding anions, resulting in a higher coordination number.
Importance of Radius Ratio Rule
Predicts the crystal structure of ionic compounds.
Explains why different salts have different arrangements.
Helps determine the stability of ionic crystals.
Widely used in solid-state chemistry and material science.
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(b) Defects in Crystals
A crystal is an orderly arrangement of atoms, ions, or molecules. In an ideal crystal, every
particle occupies its correct position. However, in reality, no crystal is perfectly arranged.
Sometimes particles are missing, displaced, or occupy the wrong position. These
irregularities are called crystal defects.
Think of a classroom where every student has an assigned seat. If one student is absent or
someone sits in another student's seat, the seating arrangement is disturbed. This is similar
to defects in crystals.
Crystal defects affect many properties such as:
Density
Electrical conductivity
Strength
Color
Diffusion of ions
The two most common point defects in ionic crystals are Schottky defect and Frenkel
defect.
(i) Schottky Defect
A Schottky defect occurs when equal numbers of positive ions (cations) and negative ions
(anions) are missing from their normal lattice positions.
Since both ions are missing equally, the crystal remains electrically neutral, but the total
number of particles decreases.
Example
Sodium Chloride (NaCl)
Potassium Chloride (KCl)
Cesium Chloride (CsCl)
Diagram
Normal Crystal
Na+ Cl− Na+
Cl− Na+ Cl−
Na+ Cl− Na+
Schottky Defect
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Na+ Cl− Na+
Cl− □ Cl−
Na+ □ Na+
□ = Missing Na+ and Cl− ions
Characteristics
Equal number of ions are missing.
Electrical neutrality is maintained.
Density of the crystal decreases because some ions are absent.
Common in ionic compounds where cations and anions are of similar size.
(ii) Frenkel Defect
A Frenkel defect occurs when a small positive ion leaves its normal position and moves
into an empty space (interstitial site) within the crystal.
Here, no ion leaves the crystal, so the total number of ions remains the same.
Imagine a student leaving their assigned chair and sitting in an empty corner of the
classroom. The student is still inside the classroom, but the seating arrangement is
disturbed.
Example
Silver Chloride (AgCl)
Silver Bromide (AgBr)
Zinc Sulphide (ZnS)
Diagram
Before Defect
Ag+ Cl− Ag+
Cl− Ag+ Cl−
Ag+ Cl− Ag+
After Frenkel Defect
Ag+ Cl− Ag+
Cl− □ Cl−
Ag+ Cl− Ag+
Ag+
(Interstitial Position)
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Characteristics
Only the cation moves.
Electrical neutrality is maintained.
Density remains almost unchanged because no ions are lost.
Usually found when the cation is much smaller than the anion.
Difference Between Schottky and Frenkel Defects
Schottky Defect
Frenkel Defect
Equal numbers of cations and anions
are missing.
A cation leaves its position and occupies an
interstitial site.
Density decreases.
Density remains almost the same.
Number of ions decreases.
Number of ions remains the same.
Common in NaCl, KCl, CsCl.
Common in AgCl, AgBr, ZnS.
Conclusion
The Radius Ratio Rule helps us predict how ions arrange themselves in an ionic crystal
based on the relative sizes of the cation and anion. A proper radius ratio leads to a stable
crystal structure with an appropriate coordination number.
Real crystals, however, are never perfectly arranged. They contain crystal defects, which are
small imperfections in the crystal lattice. The two important types are the Schottky defect,
where equal numbers of positive and negative ions are missing (reducing the crystal's
density), and the Frenkel defect, where a small positive ion moves to an interstitial position
without leaving the crystal (keeping the density nearly unchanged). Understanding these
concepts is important because crystal defects strongly influence the physical properties and
practical applications of materials such as salts, semiconductors, and electronic devices.
This paper has been carefully prepared for educational purposes. If you notice any mistakes or
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