Structure of the Atom

The properties of the three sub-atomic particles are compared below:

| Property | Electron | Proton | Neutron |
|---|---|---|---|
| Discovery | J.J. Thomson (1897) | E. Goldstein (1886) | J. Chadwick (1932) |
| Symbol | $e^-$ | $p^+$ | $n$ |
| Charge | $-1$ (negative) | $+1$ (positive) | $0$ (neutral) |
| Absolute charge | $1.6 \times 10^{-19}$ C | $1.6 \times 10^{-19}$ C | Zero |
| Mass | $9.1 \times 10^{-31}$ kg ($\approx \frac{1}{2000}$ u) | $1.673 \times 10^{-27}$ kg ($\approx 1$ u) | $1.675 \times 10^{-27}$ kg ($\approx 1$ u) |
| Location in atom | Outside nucleus (in shells) | Inside nucleus | Inside nucleus |

Key points:
- Electrons are negatively charged and revolve around the nucleus in fixed shells.
- Protons are positively charged and are present in the nucleus.
- Neutrons are neutral (no charge) and are also present in the nucleus.
- The mass of an electron is negligible compared to protons and neutrons.

J.J. Thomson's Model (Plum Pudding Model): Thomson proposed that an atom is a sphere of positive charge with electrons embedded in it like plums in a pudding.

Limitations:
1. Thomson's model could not explain the results of Rutherford's alpha-particle scattering experiment. If the positive charge were spread uniformly throughout the atom, the alpha particles should have passed through with very little deflection. However, Rutherford observed that some alpha particles were deflected at large angles and some even bounced back, which Thomson's model could not account for.
2. The model did not mention the existence of a nucleus — a dense, positively charged centre of the atom.
3. It could not explain the stability of the atom in terms of arrangement of electrons.

Conclusion: Thomson's model was an early attempt but was soon replaced by Rutherford's nuclear model.

Rutherford's Nuclear Model: Rutherford proposed that the atom has a tiny, dense, positively charged nucleus at the centre, and electrons revolve around it in circular orbits.

Limitations:
1. Instability of the atom: According to classical electromagnetic theory, a charged particle (electron) moving in a circular orbit continuously loses energy by emitting radiation. As it loses energy, it should spiral inward and eventually fall into the nucleus. This would make the atom unstable and collapse within $10^{-8}$ seconds. But atoms are known to be stable — this contradiction could not be explained by Rutherford's model.
2. Could not explain the atomic spectrum: If electrons lose energy continuously, they should emit a continuous spectrum of radiation. However, atoms emit a line spectrum (discrete lines), which Rutherford's model failed to explain.
3. No information about electronic arrangement: The model did not say anything about how electrons are arranged or distributed around the nucleus.

Conclusion: Rutherford's model was an improvement over Thomson's model but had serious drawbacks that were later addressed by Bohr's model.

Bohr's Model of the Atom (1913):

Neils Bohr proposed an improved model of the atom to overcome the limitations of Rutherford's model. The main postulates are:

1. Nucleus: The atom has a small, dense, positively charged nucleus at its centre (retained from Rutherford's model).

2. Fixed orbits (shells): Electrons revolve around the nucleus in fixed circular paths called orbits or shells. These shells are designated as K, L, M, N, ... (or numbered 1, 2, 3, 4, ...).

3. Discrete energy levels: Each orbit has a fixed amount of energy associated with it. As long as an electron remains in a particular orbit, it does not radiate energy. Hence the atom is stable.

4. Energy change during transition: When an electron jumps from a lower energy orbit to a higher energy orbit, it absorbs energy. When it falls from a higher to a lower orbit, it emits energy in the form of radiation. This explains the line spectrum of atoms.

Diagram (description): The nucleus is at the centre; K shell is closest (lowest energy), followed by L, M, N shells at increasing distances and energies.

Significance: Bohr's model successfully explained the stability of the atom and the line spectrum of hydrogen.

The three atomic models discussed in the chapter are compared below:

| Feature | Thomson's Model | Rutherford's Model | Bohr's Model |
|---|---|---|---|
| Year | 1898 | 1911 | 1913 |
| Shape of atom | Sphere of uniform positive charge | Mostly empty space with a nucleus | Mostly empty space with a nucleus |
| Nucleus | Not proposed | Tiny, dense, positively charged nucleus | Tiny, dense, positively charged nucleus |
| Electrons | Embedded in positive sphere (like plums in pudding) | Revolve around nucleus in circular orbits | Revolve in fixed shells (K, L, M, N...) with discrete energies |
| Energy of electrons | Not specified | Continuous — electrons lose energy and spiral in | Discrete — electrons in fixed energy levels, stable orbits |
| Stability of atom | Could not explain | Could not explain (electrons should collapse) | Explained — electrons in fixed orbits do not radiate energy |
| Atomic spectrum | Could not explain | Could not explain | Explained (line spectrum due to electron transitions) |
| Limitation | No nucleus; disproved by Rutherford's experiment | Atom should collapse; no line spectrum explanation | Does not explain multi-electron atoms fully |

Conclusion: Each model was an improvement over the previous one. Bohr's model was the most successful among the three in explaining atomic stability and hydrogen spectrum.

Rules for Distribution of Electrons in Shells (Bohr-Bury Rules):

1. Maximum electrons in a shell: The maximum number of electrons that can be accommodated in a shell is given by the formula $2n^2$, where $n$ is the shell number.
- K shell ($n=1$): $2 \times 1^2 = 2$ electrons
- L shell ($n=2$): $2 \times 2^2 = 8$ electrons
- M shell ($n=3$): $2 \times 3^2 = 18$ electrons (but for first 18 elements, max 8 are filled)

2. Outermost shell rule: The outermost shell cannot have more than 8 electrons, even if the formula allows more.

3. Penultimate shell rule: The second last (penultimate) shell cannot have more than 18 electrons.

4. Sequential filling: Shells are filled in order — a new shell is started only when the inner shell is completely filled.

5. Valence shell: The last shell (outermost) is called the valence shell and the electrons in it are called valence electrons.

Example — Electronic configurations of first 18 elements:

| Element | Atomic No. | K | L | M |
|---|---|---|---|---|
| H | 1 | 1 | — | — |
| He | 2 | 2 | — | — |
| Li | 3 | 2 | 1 | — |
| Be | 4 | 2 | 2 | — |
| B | 5 | 2 | 3 | — |
| C | 6 | 2 | 4 | — |
| N | 7 | 2 | 5 | — |
| O | 8 | 2 | 6 | — |
| F | 9 | 2 | 7 | — |
| Ne | 10 | 2 | 8 | — |
| Na | 11 | 2 | 8 | 1 |
| Mg | 12 | 2 | 8 | 2 |
| Al | 13 | 2 | 8 | 3 |
| Si | 14 | 2 | 8 | 4 |
| P | 15 | 2 | 8 | 5 |
| S | 16 | 2 | 8 | 6 |
| Cl | 17 | 2 | 8 | 7 |
| Ar | 18 | 2 | 8 | 8 |

Definition of Valency:
Valency is the combining capacity of an atom of an element. It is determined by the number of electrons in the outermost shell (valence electrons) of an atom.

- If the number of valence electrons is 4 or less, the valency = number of valence electrons.
- If the number of valence electrons is more than 4, the valency = $8 -$ (number of valence electrons).

Example 1 — Silicon (Si):
- Atomic number of Si = 14
- Electronic configuration: K = 2, L = 8, M = 4
- Valence electrons = 4
- Since valence electrons $\leq 4$, Valency of Si = 4
- Silicon can form 4 bonds (e.g., $\text{SiCl}_4$).

Example 2 — Oxygen (O):
- Atomic number of O = 8
- Electronic configuration: K = 2, L = 6
- Valence electrons = 6
- Since valence electrons > 4, Valency of O = $8 - 6 =$ 2
- Oxygen can form 2 bonds (e.g., $\text{H}_2\text{O}$).

Conclusion: Valency tells us how many bonds an atom can form with other atoms.

(i) Atomic Number (Z):
The atomic number of an element is defined as the number of protons present in the nucleus of its atom. Since an atom is electrically neutral, the number of protons equals the number of electrons.
$$Z = \text{Number of protons} = \text{Number of electrons}$$
Example: Carbon (C) has 6 protons, so its atomic number $Z = 6$.

(ii) Mass Number (A):
The mass number of an atom is defined as the total number of protons and neutrons (nucleons) present in the nucleus.
$$A = \text{Number of protons} + \text{Number of neutrons}$$
Example: Carbon has 6 protons and 6 neutrons, so its mass number $A = 6 + 6 = 12$. It is written as $^{12}_6\text{C}$.

(iii) Isotopes:
Isotopes are atoms of the same element that have the same atomic number but different mass numbers (i.e., same number of protons but different number of neutrons).
Example: Hydrogen has three isotopes:
- Protium: $^1_1\text{H}$ (1 proton, 0 neutrons)
- Deuterium: $^2_1\text{H}$ (1 proton, 1 neutron)
- Tritium: $^3_1\text{H}$ (1 proton, 2 neutrons)

All have the same atomic number (1) but different mass numbers.

(iv) Isobars:
Isobars are atoms of different elements that have the same mass number but different atomic numbers.
Example: Calcium ($^{40}_{20}\text{Ca}$) and Argon ($^{40}_{18}\text{Ar}$) are isobars — both have mass number 40 but different atomic numbers (20 and 18 respectively).

Two Uses of Isotopes:
1. An isotope of uranium ($^{235}\text{U}$) is used as fuel in nuclear reactors to generate electricity.
2. An isotope of iodine ($^{131}\text{I}$) is used in the treatment of thyroid cancer (goitre).
3. *(Bonus)* An isotope of cobalt ($^{60}\text{Co}$) is used in the treatment of cancer (radiotherapy).

Given: $\text{Na}^+$ ion has completely filled K and L shells.

Step 1: Electronic configuration of Na atom
- Atomic number of Na = 11
- Number of electrons in Na = 11
- Electronic configuration: K = 2, L = 8, M = 1

Step 2: Formation of Na⁺ ion
- $\text{Na}^+$ is formed when a sodium atom loses 1 electron from its outermost shell (M shell).
$$\text{Na} \rightarrow \text{Na}^+ + e^-$$
- Number of electrons in $\text{Na}^+$ = $11 - 1 = 10$

Step 3: Electronic configuration of Na⁺
- 10 electrons are distributed as: K = 2, L = 8
- K shell capacity = 2 (completely filled ✓)
- L shell capacity = 8 (completely filled ✓)
- M shell has 0 electrons.

Conclusion: Since $\text{Na}^+$ has only 10 electrons, they completely fill the K shell (2 electrons) and L shell (8 electrons), with no electrons in the M shell. Hence, $\text{Na}^+$ has completely filled K and L shells.

Given:
- Isotope 1: $^{79}\text{Br}$, abundance = 49.7%
- Isotope 2: $^{81}\text{Br}$, abundance = 50.3%

Formula:
$$\text{Average atomic mass} = \frac{(\text{mass of isotope 1} \times \% \text{ abundance}_1) + (\text{mass of isotope 2} \times \% \text{ abundance}_2)}{100}$$

Calculation:
$$\text{Average atomic mass} = \frac{(79 \times 49.7) + (81 \times 50.3)}{100}$$

$$= \frac{3926.3 + 4074.3}{100}$$

$$= \frac{8000.6}{100}$$

$$= 80.006 \text{ u}$$

$$\boxed{\text{Average atomic mass of Bromine} \approx 80 \text{ u}}$$

Given:
- Average atomic mass of element X = 16.2 u
- Two isotopes: $^{16}\text{X}$ (mass = 16 u) and $^{18}\text{X}$ (mass = 18 u)

Let the percentage of $^{16}\text{X}$ = $a$%
Then the percentage of $^{18}\text{X}$ = $(100 - a)$%

Using the formula for average atomic mass:
$$\text{Average atomic mass} = \frac{(16 \times a) + (18 \times (100 - a))}{100}$$

Setting up the equation:
$$16.2 = \frac{16a + 1800 - 18a}{100}$$

$$16.2 \times 100 = 1800 - 2a$$

$$1620 = 1800 - 2a$$

$$2a = 1800 - 1620 = 180$$

$$a = 90$$

Therefore:
- Percentage of $^{16}\text{X}$ = $\boxed{90\%}$
- Percentage of $^{18}\text{X}$ = $100 - 90 = \boxed{10\%}$

Verification: $\frac{(16 \times 90) + (18 \times 10)}{100} = \frac{1440 + 180}{100} = \frac{1620}{100} = 16.2$ u ✓

Given: Atomic number $Z = 3$

Step 1: Find the electronic configuration
- Number of electrons = 3 (since $Z = 3$)
- Distribution: K = 2, L = 1

Step 2: Find the valency
- Valence electrons (electrons in outermost shell) = 1
- Since valence electrons $\leq 4$, Valency = number of valence electrons = 1

Step 3: Name the element
- The element with atomic number 3 is Lithium (Li).

Answer: The valency of the element is 1 and the element is Lithium (Li).

Given:
- Species X: Protons = 6, Neutrons = 6
- Species Y: Protons = 6, Neutrons = 8

Step 1: Calculate Mass Numbers
$$\text{Mass number} = \text{Number of protons} + \text{Number of neutrons}$$

- Mass number of X = $6 + 6 = \mathbf{12}$, so X is $^{12}_6\text{C}$
- Mass number of Y = $6 + 8 = \mathbf{14}$, so Y is $^{14}_6\text{C}$

Step 2: Identify the relation
- Both X and Y have the same number of protons = 6 (same atomic number)
- But they have different mass numbers (12 and 14)

Conclusion: X and Y are Isotopes of each other. They are both isotopes of Carbon — Carbon-12 ($^{12}\text{C}$) and Carbon-14 ($^{14}\text{C}$). Isotopes are atoms of the same element with the same atomic number but different mass numbers.

(a) J.J. Thomson proposed that the nucleus of an atom contains only nucleons.
Answer: F (False)
J.J. Thomson did not propose the concept of a nucleus at all. He proposed the 'plum pudding model' where the atom is a sphere of uniform positive charge with electrons embedded in it. The nucleus was discovered by Rutherford.

(b) A neutron is formed by an electron and a proton combining together. Therefore, it is neutral.
Answer: F (False)
A neutron is an independent sub-atomic particle discovered by J. Chadwick. It is not formed by the combination of an electron and a proton. It is a fundamental particle present in the nucleus.

(c) The mass of an electron is about $\frac{1}{2000}$ times that of proton.
Answer: T (True)
The mass of an electron is approximately $9.1 \times 10^{-31}$ kg, while the mass of a proton is approximately $1.67 \times 10^{-27}$ kg. The ratio is approximately $\frac{1}{1837} \approx \frac{1}{2000}$. So this statement is true.

(d) An isotope of iodine is used for making tincture iodine, which is used as a medicine.
Answer: F (False)
Tincture iodine is a solution of normal iodine ($\text{I}_2$) in alcohol — it does not use a radioactive isotope of iodine. The isotope $^{131}\text{I}$ is used for the treatment of thyroid disorders, not for making tincture iodine.

Correct Answer: (a) Atomic Nucleus ✓

Justification: Rutherford's gold foil experiment (alpha-particle scattering experiment) showed that most alpha particles passed straight through, but a few were deflected at large angles and some bounced back. This led Rutherford to conclude that the atom has a tiny, dense, positively charged centre called the nucleus. Electrons were discovered by J.J. Thomson, and neutrons by J. Chadwick.

(b) Electron ✗, (c) Proton ✗, (d) Neutron ✗

Correct Answer: (c) different number of neutrons ✓

Justification: Isotopes are atoms of the same element with the same atomic number (same number of protons) but different mass numbers. Since mass number = protons + neutrons, and protons are the same, the difference lies in the number of neutrons. Isotopes have the same chemical properties (same valence electrons) but different physical properties.

(a) ✗ — Isotopes have different physical properties (e.g., different densities, melting points).
(b) ✗ — Isotopes have the same chemical properties.
(d) ✗ — Isotopes have the same atomic number.

Correct Answer: (b) 8 ✓

Justification:
- Atomic number of Cl = 17; electronic configuration of Cl atom: K=2, L=8, M=7 (valence electrons = 7)
- $\text{Cl}^-$ is formed when Cl gains 1 electron: total electrons = $17 + 1 = 18$
- Electronic configuration of $\text{Cl}^-$: K=2, L=8, M=8
- Valence electrons (outermost shell M) = 8

(a) 16 ✗, (c) 17 ✗, (d) 18 ✗ (18 is total electrons, not valence electrons)

Correct Answer: (d) 2,8,1 ✓

Justification:
- Atomic number of Sodium (Na) = 11
- Number of electrons = 11
- Filling shells in order: K shell = 2, L shell = 8, M shell = 1
- Electronic configuration = 2, 8, 1

Shells must be filled in order (K first, then L, then M), so options (b) and (c) are incorrect arrangements. Option (a) accounts for only 10 electrons.

(a) ✗, (b) ✗, (c) ✗

Key relations used:
- Atomic number ($Z$) = Number of protons = Number of electrons (for neutral atom)
- Mass number ($A$) = Number of protons + Number of neutrons
- Number of neutrons = $A - Z$

Row 1: Atomic number = 9
- Protons = 9, Electrons = 9
- Neutrons = 10 (given)
- Mass number = $9 + 10 = 19$
- Element with $Z = 9$ is Fluorine (F)
- Entry: $Z=9$, $A=19$, Neutrons=10, Protons=9, Electrons=9, Fluorine

Row 2: $Z = 16$, $A = 32$, Name = Sulphur
- Protons = 16, Electrons = 16
- Neutrons = $32 - 16 = 16$
- Entry: $Z=16$, $A=32$, Neutrons=16, Protons=16, Electrons=16, Sulphur

Row 3: $A = 24$, Protons = 12
- Atomic number $Z = 12$, Electrons = 12
- Neutrons = $24 - 12 = 12$
- Element with $Z = 12$ is Magnesium (Mg)
- Entry: $Z=12$, $A=24$, Neutrons=12, Protons=12, Electrons=12, Magnesium

Row 4: $A = 2$, Protons = 1
- Atomic number $Z = 1$, Electrons = 1
- Neutrons = $2 - 1 = 1$
- Element with $Z = 1$ and $A = 2$ is Deuterium ($^2_1\text{H}$) — an isotope of Hydrogen
- Entry: $Z=1$, $A=2$, Neutrons=1, Protons=1, Electrons=1, Deuterium (Hydrogen)

Row 5: $A = 1$, Neutrons = 0, Protons = 1, Electrons = 0
- Atomic number $Z = 1$
- This is a proton / Hydrogen ion (H⁺) or Protium nucleus
- Since electrons = 0, it is the $\text{H}^+$ ion (proton), but as an atomic species with $Z=1$, $A=1$, it is Protium (H) — the most common isotope of Hydrogen. (Note: electrons = 0 indicates it is the $\text{H}^+$ ion/proton.)
- Entry: $Z=1$, $A=1$, Neutrons=0, Protons=1, Electrons=0, Hydrogen (Protium / H⁺)

Completed Table:

| Atomic Number | Mass Number | Number of Neutrons | Number of Protons | Number of Electrons | Name |
|---|---|---|---|---|---|
| 9 | 19 | 10 | 9 | 9 | Fluorine |
| 16 | 32 | 16 | 16 | 16 | Sulphur |
| 12 | 24 | 12 | 12 | 12 | Magnesium |
| 1 | 2 | 1 | 1 | 1 | Deuterium (Hydrogen) |
| 1 | 1 | 0 | 1 | 0 | Hydrogen (H⁺ / Protium) |