2.1.1 Sub-Atomic Particles
Atoms consist of three fundamental sub-atomic particles: Electrons, Protons, and Neutrons. Their properties are summarized below:
1. Electron
- Charge: Negatively charged with a value of $1.6022 \times 10^{-19} \text{ C}$.
- Relative Charge: $-1$.
- Mass: $9.1095 \times 10^{-31} \text{ kg}$.
- Discovery: Discovered by J.J. Thomson.
- Behavior: Deflected towards the positive pole of an electric field.
2. Proton
- Charge: Positively charged, equal in magnitude to an electron ($1.6022 \times 10^{-19} \text{ C}$).
- Relative Charge: $+1$.
- Mass: $1.6727 \times 10^{-27} \text{ kg}$ (about 1836 times heavier than an electron).
- Behavior: Deflected towards the negative pole of an electric field.
3. Neutron
- Charge: Neutral (no charge).
- Relative Charge: $0$.
- Mass: $1.6750 \times 10^{-27} \text{ kg}$ (slightly more massive than a proton).
- Behavior: Passes through electric and magnetic fields un-deflected.
Summary Table: Properties of Sub-atomic Particles
| Particle | Mass (kg) | Charge (C) | Relative Mass | Relative Charge |
|---|---|---|---|---|
| Proton | $1.672 \times 10^{-27}$ | $+1.602 \times 10^{-19}$ | $1$ | $+1$ |
| Neutron | $1.675 \times 10^{-27}$ | $0$ | $1$ | $0$ |
| Electron | $9.109 \times 10^{-31}$ | $-1.602 \times 10^{-19}$ | $\frac{1}{1836}$ | $-1$ |
2.1.2 Behaviour in Electric Field
When sub-atomic particles pass through an electric field, they behave differently based on their charge and mass:
- Electrons: Move towards the positive pole. Because they are much lighter than protons, they show a more significant curve/deflection.
- Protons: Move towards the negative pole.
- Neutrons: Exhibit no deviation and move in a straight line because they are electrically neutral.
Relevant Questions & Answers
Q1: Who discovered the electron and what is its mass?
A: The electron was discovered by J.J. Thomson. Its mass is $9.1095 \times 10^{-31} \text{ kg}$.
Q2: How does the mass of a proton compare to that of an electron?
A: A proton is significantly heavier than an electron; it is approximately 1836 times the mass of an electron.
Q3: Why do neutrons pass through electric fields without any deflection?
A: Neutrons carry no electrical charge (they are neutral), so they do not experience electrostatic forces when passing through an electric field.
Q4: In an electric field, which direction will a proton be deflected?
A: Since a proton is positively charged, it will be deflected towards the negative pole of the electric field.
Q5: Which sub-atomic particle is the most massive?
A: The neutron is the most massive sub-atomic particle, with a mass of $1.6750 \times 10^{-27} \text{ kg}$, which is slightly more than the mass of a proton.
2.1.3 Atomic Number and Mass Number
1. Atomic Number (Z)
- Definition: Atomic number is the number of protons in the nucleus of an atom.
- Neutral Atoms: Since an atom is a neutral particle, the number of electrons revolving around the nucleus is equal to the number of protons.
- Representation: It is represented by the symbol $Z$.
2. Mass Number (A)
- Definition: Mass number (or nucleon number) represents the total number of protons and neutrons in the nucleus.
- Representation: It is represented by the symbol $A$.
- Relationship: Mass number is approximately equal to the atomic mass of an atom.
Key Formulas
- Mass Number: $\text{Atomic number} + \text{Number of neutrons} = \text{Mass number}$ ($P + N = A$).
- Neutral Atom: $\text{Number of Protons} = \text{Number of electrons}$.
- Cations (Positive Ions): $\text{Number of electrons} = \text{Atomic number} - \text{magnitude of charge on cation}$.
- Anions (Negative Ions): $\text{Number of electrons} = \text{Atomic number} + \text{magnitude of charge on anion}$.
Concept Assessment Exercise 2.1
Question: How many electrons are present in the following species: $Mg^{+2}$, $Al^{+3}$, $S^{-2}$?
Solutions:
Note: To solve these, we use the atomic numbers ($Z$) for Magnesium (12), Aluminum (13), and Sulfur (16).
-
$Mg^{+2}$ (Magnesium Ion):
- Atomic Number ($Z$) = 12.
- Charge = $+2$.
- Electrons = $12 - 2 = \mathbf{10} \text{ electrons}$.
-
$Al^{+3}$ (Aluminum Ion):
- Atomic Number ($Z$) = 13.
- Charge = $+3$.
- Electrons = $13 - 3 = \mathbf{10} \text{ electrons}$.
-
$S^{-2}$ (Sulfide Ion):
- Atomic Number ($Z$) = 16.
- Charge = $-2$.
- Electrons = $16 + 2 = \mathbf{18} \text{ electrons}$.
2.1.3 Atomic Number and Mass Number
- Atomic Number (Z): The number of protons in an atom's nucleus. In a neutral atom, this also equals the number of electrons.
- Mass Number (A): The total number of protons and neutrons in the nucleus. It is approximately equal to the atomic mass.
- Key Formula: $P + N = A$.
Calculating Electrons in Ions
- Cations: $\text{Electrons} = Z - \text{magnitude of positive charge}$.
- Anions: $\text{Electrons} = Z + \text{magnitude of negative charge}$.
2.1.4 Atomic and Ionic Radius
Atomic radius is the average distance from the nucleus to the outermost electrons.
Periodic Trends
| Trend Direction | Atomic Radius Behavior | Reasoning |
|---|---|---|
| Across a Period (Left to Right) | Decreases | Increased nuclear charge while shielding remains constant. |
| Down a Group (Top to Bottom) | Increases | Increasing number of shells and shielding effect from intervening electrons. |
Ionic Radius Specifics
- Cations (+): Always smaller than their parent atoms because of increased effective nuclear charge and decreased electron-electron repulsion.
- Anions (-): Generally larger than their parent atoms (often formed by nonmetals).
- Trends: Ionic radius increases down a group. Across a period, the trend matches atomic radius (decreasing) if the ions compared have the same charge.
Concept Assessment Exercises
Exercise 2.1: Electron Counts
(Using Atomic Numbers: Mg=12, Al=13, S=16)
- $Mg^{+2}$: $12 - 2 = \mathbf{10}$ electrons.
- $Al^{+3}$: $13 - 3 = \mathbf{10}$ electrons.
- $S^{-2}$: $16 + 2 = \mathbf{18}$ electrons.
Exercise 2.2: Periodic Trends
- As you move from left to right across a period, the atomic radius generally decreases.
- Down a group in the periodic table, the atomic radius tends to increase.
- When an atom loses electrons to become a cation, its ionic radius decreases (is smaller) compared to its atomic radius.
- Among elements, from top to bottom within a group, the ionic radius of cations tends to increase, while the ionic radius of anions tends to increase.
- In general, nonmetals tend to form negative (anion) ions with larger ionic radii compared to their atomic radii.
2.2 Quantum Numbers
Quantum numbers are a set of four numerals used to specify the exact location and behavior of an electron in an atom. They describe the energy, shape, orientation, and spin of orbitals.
1. Principal Quantum Number ($n$)
- Definition: Represents Bohr’s shell in quantum form.
- Values: $n = 1, 2, 3, \dots$.
- Significance: As $n$ increases, the energy and size of the shell increase. It also corresponds to the period number in the periodic table.
2. Azimuthal Quantum Number ($l$)
- Definition: Describes the shapes of orbitals within a shell (subshells).
- Values: Ranges from $0$ to $n - 1$.
- Notation: $l$ values correspond to subshells: $0=s$, $1=p$, $2=d$, $3=f$.
- Significance: It determines the orbital's angular momentum.
3. Magnetic Quantum Number ($m$)
- Definition: Describes the orientation of an orbital in space around the nucleus.
- Values: Ranges from $-l$ to $+l$.
- Orbital Splitting:
- s-orbital ($l=0$): $m=0$; only one orientation (spherically symmetrical).
- p-orbital ($l=1$): $m = +1, 0, -1$; three degenerate orbitals ($p_x, p_y, p_z$).
- d-orbital ($l=2$): $m = -2, -1, 0, +1, +2$; five orbitals ($d_{xy}, d_{yz}, d_{xz}, d_{x^2-y^2}, d_{z^2}$).
- f-orbital ($l=3$): Seven orientations ($m$ from $-3$ to $+3$).
4. Spin Quantum Number ($s$)
- Definition: Describes the self-rotation (spin) of an electron about its own axis.
- Values: $+1/2$ (clockwise, $\uparrow$) and $-1/2$ (anti-clockwise, $\downarrow$).
- Rule: A single orbital can hold a maximum of two electrons, which must have opposite spins to cancel their magnetic fields.
2.2.1 Quantum Numbers and Electronic Configuration
Electronic configuration is the distribution of electrons around the nucleus in orbits and orbitals. A complete configuration requires all four quantum numbers to specify each electron's location.
Relevant Questions & Answers
Q1: What is the maximum number of electrons that can reside in a single orbital?
A: A maximum of two electrons can reside in one orbital, provided they have opposite spins.
Q2: If $n = 3$, what are the possible values for the azimuthal quantum number ($l$)?
A: The values range from $0$ to $n-1$. For $n=3$, the possible values for $l$ are $0, 1,$ and $2$.
Q3: Which quantum number determines the orientation of an orbital in space?
A: The Magnetic Quantum Number ($m$) determines the orientation.
Q4: Explain why the s-orbital is spherically symmetrical.
A: For an s-orbital, the value of $l$ is $0$, meaning the magnetic quantum number $m$ can only be $0$. Having only one orientation makes it spherically symmetrical.
Q5: What do the $+1/2$ and $-1/2$ values represent in the spin quantum number?
A: $+1/2$ represents a clockwise spin ($\uparrow$), while $-1/2$ represents an anti-clockwise spin ($\downarrow$).
2.3 Rules of Electronic Configuration
Electronic configuration refers to the distribution of electrons in orbitals or subshells around the nucleus. To determine the configuration, three fundamental principles and one mathematical rule are followed:
1. Auf Bau Principle
- Rule: Electrons are distributed in orbitals/subshells in order of increasing energy.
- Process: Low-energy orbitals are filled first, followed by higher-energy orbitals.
- Energy Order: $1s \lt 2s \lt 2p \lt 3s \lt 3p \lt 4s \lt 3d \dots$
2. n + l Rule
- Rule: The energy of an orbital is determined by the sum of its principal ($n$) and azimuthal ($l$) quantum numbers.
- Principle: Higher $n + l$ value indicates higher energy.
- Example: For a $4s$ orbital, $n+l = 4+0=4$; for a $3d$ orbital, $n+l = 3+2=5$. Therefore, $3d$ has greater energy than $4s$.
3. Pauli Exclusion Principle
- Rule: No two electrons in an atom can have the same set of all four quantum numbers.
- Significance: Even if $n$, $l$, and $m$ are identical (meaning electrons are in the same orbital), the spin quantum number ($s$) must be different.
- Example: In a $3p_x$ orbital, one electron will have $s = +1/2$ ($\uparrow$) and the other $s = -1/2$ ($\downarrow$).
4. Hund’s Rule
- Rule: If degenerate orbitals (orbitals with equal energy) are available, electrons prefer to reside separately with the same spin rather than pairing up with opposite spins.
- Example: For three electrons in $p$ orbitals, the configuration is $p_x^1, p_y^1, p_z^1$ rather than pairing them in one orbital first.
- Stability: Half-filled and completely filled orbitals are more stable (e.g., Chromium and Copper).
Concept Assessment Exercises
Exercise 2.4
- In which orbital of the atom will the $11^{th}$ electron be configured?
A: Following the order $1s^2, 2s^2, 2p^6, 3s^1$, the $11^{th}$ electron will be in the $3s$ orbital. - Place these orbitals in increasing energy order: $5s, 4p, 4s, 3d$.
- A: Using the $n+l$ rule and Aufbau principle: $4s \lt 3d \lt 4p \lt 5s$.
Exercise 2.5: Which orbital has greater energy?
- $5f$ or $6p$:
$5f$ ($n+l = 5+3=8$) vs $6p$ ($n+l = 6+1=7$). $5f$ has greater energy. - $5s$ or $4d$:
$5s$ ($n+l = 5+0=5$) vs $4d$ ($n+l = 4+2=6$). $4d$ has greater energy.
Quick Summary Table of Quantum Numbers for Electrons
| Electron | $n$ | $l$ | $m$ | $s$ |
|---|---|---|---|---|
| Electron A | 3 | 1 | -1 | $+1/2$ |
| Electron B | 3 | 1 | -1 | $-1/2$ |
2.6 Ionization Energy
Ionization energy (I.E.) is defined as the minimum energy required to remove the outermost, most loosely bonded electron from an isolated gaseous atom. To accurately measure this energy without the influence of external factors like heat of fusion or vaporization, the atom must be in a gaseous state.
Examples:
- Sodium: $Na(g) \rightarrow Na^{+}(g) + 1e^{-}$ ($I.E. = 496 \text{ kJ/mol}$).
- Magnesium: $Mg(g) \rightarrow Mg^{+}(g) + 1e^{-}$ ($I.E. = 738 \text{ kJ/mol}$).
2.6.2 Factors Influencing Ionization Energy
Several physical and electronic factors determine the magnitude of ionization energy:
- Nuclear Charge: As nuclear charge increases across a period (left to right), the attraction for outermost electrons increases, which in turn increases the ionization energy.
- Atomic Radii: Smaller radii strengthen the nucleus's pull on electrons, increasing I.E. Conversely, as size increases down a group, I.E. decreases.
- Shielding Effect: Inner electron shells act as a shield, reducing the effective nuclear charge felt by outer electrons. This makes it easier to remove electrons, leading to a decrease in I.E. down a group.
- Spin Pair Repulsion: When electrons are paired in an orbital, repulsion occurs, increasing the electron's energy and making it easier to remove. This explains slight anomalies, such as the decrease in I.E. from Nitrogen to Oxygen.
2.6.1 Periodic Trends of Ionization Energy
- Across a Period (Left to Right): Generally increases due to increasing nuclear charge and decreasing atomic radius, which brings electrons closer to the nucleus.
- Down a Group (Top to Bottom): Generally decreases as atoms become larger and the shielding effect increases, making it easier to remove the valence electron.
- Anomalies:
- Group IIIA: Lower I.E. is observed because there is only one electron in the p-orbital (unstable configuration).
- Group VA: Higher I.E. is observed due to the stability of half-filled p-orbitals.
Relevant Questions & Answers
Q1: Why must an atom be in the gaseous state when measuring ionization energy?
A: It must be in the gaseous state to avoid the influence of other energy factors such as heat of fusion, bond dissociation energy, and heat of vaporization.
Q2: How does the shielding effect change across a period compared to down a group?
A: From left to right across a period, the shielding effect remains constant. However, moving down a group, the shielding effect increases as the number of electron shells increases.
Q3: Why does Nitrogen have a higher ionization energy than Oxygen, despite being to its left?
A: Nitrogen has a stable half-filled p-orbital. In Oxygen, electron pair repulsion in the p-orbital makes it slightly easier to remove an electron, causing a slight decrease in I.E..
Q4: What is the relationship between atomic radius and ionization energy?
A: They are inversely related; as the atomic radius decreases, the ionization energy typically increases because the nucleus has a stronger grip on the electrons.
2.6.4 Electronic Configuration and Position in Periodic Table by Using Successive Ionization Energy Data
Successive ionization energy values serve as an index for the valence of an atom. Large gaps between successive ionization energies indicate that the next electron is being removed from a lower (inner) shell, which is much closer to the nucleus and more strongly held.
Case Study 1: Sodium ($Na$)
- First Ionization Energy: $495 \text{ kJ/mol}$.
- Second Ionization Energy: $4560 \text{ kJ/mol}$.
- Observation: There is a massive "jump" after the first electron is removed.
- Conclusion: This indicates there is only one electron in the valence shell; therefore, Sodium must be a part of Group 1.
Case Study 2: Magnesium ($Mg$)
- First Ionization Energy: $735 \text{ kJ/mol}$.
- Second Ionization Energy: $1445 \text{ kJ/mol}$.
- Third Ionization Energy: $7730 \text{ kJ/mol}$.
- Observation: The gap between the first and second values is relatively small, but the gap between the second and third values is very large.
- Conclusion: This indicates that the third electron is being removed from an inner shell. Since two electrons are removed easily, Magnesium belongs to Group 2.
Concept Assessment Exercise 2.6
1. How does the trend in ionization energy across a period reflect changes in atomic structure and nuclear charge?
Answer: Across a period, the atomic number (number of protons) increases, which strengthens the nuclear charge. Simultaneously, the atomic radius decreases, bringing valence electrons closer to the nucleus. These factors increase the electrostatic attraction, making it harder to remove an electron and thus increasing the ionization energy.
2. What factors contribute to the general decrease in ionization energy down a group?
Answer: Two main factors contribute:
- Increasing Atomic Size: As the number of shells increases, the valence electrons move further away from the nucleus.
- Shielding Effect: The number of inner electron shells increases, acting as a "shield" that reduces the effective nuclear pull on the outermost electrons.
3. In terms of ionization energy, why do noble gases possess the highest values among their respective periods, and how does this relate to their chemical reactivity?
Answer: Noble gases have completely filled valence shells, which is a highly stable electronic configuration. They also have the smallest atomic radii and highest effective nuclear charges in their respective periods. This results in extremely high ionization energy, meaning they do not easily lose electrons, which explains their very low chemical reactivity (inertness).
2.7 Electron Affinity
Definition: Electron affinity is the amount of energy released when an electron is added to the outermost shell of an isolated gaseous atom to form a negative ion (anion). It is the opposite process of ionization energy.
Example:
- Fluorine: $F(g) + 1e^{-} \rightarrow F^{-}(g)$ ($\Delta H = -328 \text{ kJ/mol}$).
2.7.1 Periodic Trends of Electron Affinity
The trend for electron affinity generally follows the same pattern as ionization energy:
- Across a Period (Left to Right): Electron affinity generally increases. This is because the nuclear charge increases and the atomic size decreases, making the nucleus more effective at attracting an incoming electron.
- Down a Group (Top to Bottom): Electron affinity generally decreases. As the atom gets larger and the shielding effect increases, the nucleus has a weaker attraction for an added electron.
2.8 Electronegativity
Definition: Electronegativity is the relative ability of an atom in a molecule to attract the shared pair of electrons towards itself. Unlike ionization energy and electron affinity, it is a property of an atom in a chemical bond.
Periodic Trends of Electronegativity
- Across a Period (Left to Right): Electronegativity increases. The increase in nuclear charge and decrease in atomic radius allow the nucleus to pull shared electrons more strongly.
- Down a Group (Top to Bottom): Electronegativity decreases. Increased atomic size and shielding reduce the nucleus's ability to attract shared electrons.
Relevant Questions & Answers
Q1: What is the main difference between electron affinity and electronegativity?
A: Electron affinity refers to the energy change when a single atom gains an electron in the gaseous state. Electronegativity is the ability of an atom to attract a shared pair of electrons within a chemical bond.
Q2: Why does electron affinity increase across a period?
A: It increases because the nuclear charge increases and the atomic radius decreases, which strengthens the pull of the nucleus on incoming electrons.
Q3: Which element is typically considered the most electronegative in the periodic table?
A: Based on the trends, Fluorine is the most electronegative element because it is located at the top right (excluding noble gases) where nuclear pull is strongest and size is smallest.
Q4: Why does electronegativity decrease down a group?
A: As you move down a group, the number of shells increases, making the atom larger. This increased distance and the shielding effect of inner electrons weaken the nucleus's hold on shared electron pairs.
2.11 Metallic and Non-Metallic Character
Elements are broadly classified into two categories based on their physical and chemical properties: Metals and Non-metals.
1. Metallic Character (Electropositivity)
- Definition: The ability of an atom to lose its valence electrons and form a positive ion (cation).
- Nature: Metals are generally solid, malleable, ductile, and good conductors of heat and electricity.
- Trend: Metallic character is closely related to Ionization Energy. Atoms with low ionization energy lose electrons more easily and thus exhibit stronger metallic character.
2. Non-Metallic Character (Electronegativity)
- Definition: The ability of an atom to gain electrons and form a negative ion (anion).
- Nature: Non-metals are usually gases or brittle solids and are poor conductors.
- Trend: This character is related to Electron Affinity and Electronegativity. Atoms that attract electrons strongly exhibit higher non-metallic character.
Periodic Trends
| Direction | Metallic Character | Non-Metallic Character |
|---|---|---|
| Across a Period (Left to Right) | Decreases | Increases |
| Down a Group (Top to Bottom) | Increases | Decreases |
2.12 Melting and Boiling Points
Melting and boiling points reflect the strength of the forces holding the atoms or molecules together in a substance.
Periodic Trends
- Across a Period:
- For metals (Groups I to IV), melting points increase as the number of binding electrons increases.
- Carbon (Group IV) typically has the highest melting point due to its giant covalent structure.
- From Group V to VII, melting points decrease because these elements exist as simple molecules held by weak intermolecular forces.
- Down a Group:
- Group I & II (Metals): Melting points decrease because the atomic size increases, weakening the metallic bond.
- Group VII & VIII (Non-metals): Melting points increase because larger atoms/molecules have stronger London dispersion forces.
Relevant Questions & Answers
Q1: Why does metallic character increase down a group?
A: As you move down a group, the atomic size increases and the shielding effect becomes stronger. This lowers the ionization energy, making it easier for the atom to lose its valence electrons.
Q2: Why do melting points of alkali metals decrease down the group?
A: As the atoms get larger, the distance between the nucleus and the sea of delocalized electrons increases, which weakens the metallic bond.
Q3: Which group in the periodic table generally contains the elements with the highest melting points?
A: Group IV elements, like Carbon and Silicon, often have the highest melting points because they form giant covalent lattice structures.
Q4: Explain the trend of non-metallic character across a period.
A: Across a period, nuclear charge increases and atomic size decreases. This makes the atoms better at attracting and gaining electrons, thus increasing non-metallic character.