Structure of the Atom — NCERT Solutions, AI Tutor & Practice

An atom is made up of three fundamental particles:

| Particle | Symbol | Charge | Relative Mass (u) | Location | |----------|--------|--------|-------------------|----------| | Electron | $e^-$ | $-1$ (negative) | $\frac{1}{1836}$ (negligible) | Outside nucleus, in shells | | Proton | $p^+$ | $+1$ (positive) | $1$ | Inside nucleus | | Neutron | $n^0$ | $0$ (neutral) | $1$ | Inside nucleus |

Key points:

• The mass of a proton and neutron is nearly equal ($\approx 1\,\text{u}$).
• The electron is about $1836$ times lighter than a proton, so its mass is usually ignored.
• An atom is electrically neutral because the number of protons equals the number of electrons.

Thomson's Model (Plum Pudding Model):

J.J. Thomson proposed that:

1. An atom is a sphere of positive charge in which electrons (negatively charged) are embedded, like seeds in a watermelon or plums in a pudding.
2. The negative and positive charges are equal in magnitude, so the atom as a whole is electrically neutral.

Analysis:

• This model successfully explained the electrical neutrality of the atom.
• However, it did not have a nucleus and could not explain the results of Rutherford's later scattering experiment.

Conclusion: Thomson's model was an important first step, but it was soon replaced because it wrongly assumed positive charge was spread throughout the atom.

Experiment: Rutherford bombarded a thin gold foil with fast-moving, positively charged $\alpha$-particles.

Observations:

1. Most of the $\alpha$-particles passed straight through the foil undeflected.
2. A few were deflected by small angles.
3. A very few (about 1 in 12000) bounced straight back.

Conclusions:

1. Since most particles passed straight through → most of the atom is empty space.
2. A few were deflected → the positive charge occupies very little space.
3. Very few bounced back → all the positive charge and nearly all the mass is concentrated in a very small, dense centre called the nucleus.

Nuclear Model:

• The nucleus is positively charged and lies at the centre.
• Electrons revolve around the nucleus.
• The size of the nucleus is very small compared to the atom.

Drawback of Rutherford's model:

• According to classical physics, an electron revolving around the nucleus in a circular orbit is under acceleration.
• An accelerating charged particle radiates (loses) energy continuously.
• Losing energy, the electron would spiral inward and fall into the nucleus, making the atom unstable.
• But atoms are stable in reality — this contradiction was the main drawback.

Bohr's Model: Niels Bohr overcame this by proposing:

1. Electrons revolve only in certain discrete permitted orbits (called energy levels or shells).
2. While revolving in these special orbits, electrons do not radiate energy.
3. These orbits/shells are represented by $K, L, M, N,\dots$ or $n = 1, 2, 3, 4,\dots$

Thus Bohr's model explained the stability of the atom.

Rules for filling electrons (Bohr–Bury scheme):

1. Maximum number of electrons in a shell is given by $2n^2$, where $n$ is the shell number.
   • $K (n=1): 2 \times 1^2 = 2$
   • $L (n=2): 2 \times 2^2 = 8$
   • $M (n=3): 2 \times 3^2 = 18$
   • $N (n=4): 2 \times 4^2 = 32$
2. The outermost shell can hold a maximum of 8 electrons.
3. Electrons are filled in a stepwise manner — inner shells fill before outer shells (a new shell begins only after the previous one is filled as per the rules).

Electronic configuration of Magnesium ($Z = 12$):

• Total electrons $= 12$
• $K$ shell: $2$
• $L$ shell: $8$
• $M$ shell: $2$

$\text{Mg} = 2, 8, 2$

Valency: It is the combining capacity of an atom. It is the number of electrons an atom loses, gains, or shares to achieve a stable (octet/duplet) outer shell.

Nitrogen ($Z = 7$):

• Configuration: $2, 5$
• Outermost electrons = $5$
• To complete octet, it needs to gain $8 - 5 = 3$ electrons.
• Valency = 3

Calcium ($Z = 20$):

• Configuration: $2, 8, 8, 2$
• Outermost electrons = $2$
• It is easier to lose 2 electrons than to gain 6.
• Valency = 2

Rule: If outermost electrons $\leq 4$, valency = number of outer electrons (lost). If $> 4$, valency = $8 -$ outer electrons (gained).

Atomic Number ($Z$): The number of protons present in the nucleus of an atom. $Z = \text{number of protons}$

Mass Number ($A$): The total number of protons and neutrons (nucleons) in the nucleus. $A = \text{protons} + \text{neutrons}$

Given: protons $= 17$, neutrons $= 18$

Atomic number: $Z = 17$

Mass number: $A = 17 + 18 = 35$

Identification: The element with atomic number $17$ is Chlorine (Cl).

Symbol representation: $^{35}_{17}\text{Cl}$

Isotopes: Atoms of the same element having the same atomic number (same number of protons) but different mass numbers (different number of neutrons).

Examples:

1. Hydrogen has three isotopes: $^{1}_{1}\text{H}$ (protium), $^{2}_{1}\text{H}$ (deuterium), $^{3}_{1}\text{H}$ (tritium).
2. Carbon has isotopes: $^{12}_{6}\text{C}$ and $^{14}_{6}\text{C}$.
3. Chlorine: $^{35}_{17}\text{Cl}$ and $^{37}_{17}\text{Cl}$.

Uses of isotopes:

1. An isotope of uranium ($^{235}\text{U}$) is used as fuel in nuclear reactors.
2. An isotope of cobalt ($^{60}\text{Co}$) is used in the treatment of cancer.
3. An isotope of iodine ($^{131}\text{I}$) is used in the treatment of goitre (thyroid disorder).

(Any two uses are sufficient.)

Given:

• Isotopes: $^{35}\text{Cl}$ and $^{37}\text{Cl}$
• Ratio $= 3 : 1$

Step 1: Convert ratio to percentages.

• $^{35}\text{Cl} = \frac{3}{4} \times 100 = 75\%$
• $^{37}\text{Cl} = \frac{1}{4} \times 100 = 25\%$

Step 2: Calculate weighted average. $\text{Average mass} = \frac{(35 \times 75) + (37 \times 25)}{100}$ $= \frac{2625 + 925}{100} = \frac{3550}{100}$ $= 35.5\,\text{u}$

Conclusion: The average atomic mass of chlorine is $\mathbf{35.5\,u}$.

Isobars: Atoms of different elements having different atomic numbers but the same mass number.

Example:

• Calcium: $^{40}_{20}\text{Ca}$ (20 protons, 20 neutrons)
• Argon: $^{40}_{18}\text{Ar}$ (18 protons, 22 neutrons)

Both have mass number $= 40$ but different atomic numbers.

Difference between Isotopes and Isobars:

| Feature | Isotopes | Isobars | |---------|----------|---------| | Element | Same element | Different elements | | Atomic number | Same | Different | | Mass number | Different | Same | | Neutrons | Different | Different |

Conclusion: Isotopes share the atomic number; isobars share the mass number.

Given: $O^{2-}$, atomic number $Z = 8$, mass number $A = 16$.

Step 1: Protons Protons $= Z = 8$ (the number of protons never changes when an ion forms).

Step 2: Neutrons $\text{Neutrons} = A - Z = 16 - 8 = 8$

Step 3: Electrons

• A neutral oxygen atom has $8$ electrons.
• The $2-$ charge means the atom has gained 2 extra electrons. $\text{Electrons} = 8 + 2 = 10$

Answer:

• Protons $= 8$
• Neutrons $= 8$
• Electrons $= 10$

Understand: The nucleus contains two types of particles:

• Protons — positively charged ($+1$)
• Neutrons — no charge (neutral, $0$)

Analyze:

• Neutrons contribute mass to the nucleus but do not carry any charge.
• Therefore, the only charged particles inside the nucleus are the protons.
• Since all protons are positive and there is nothing negative inside the nucleus (electrons are outside), the total charge is positive.

Conclude: The nucleus is positively charged because it contains protons (positive) and neutrons (neutral); the presence of neutral neutrons does not reduce or cancel the positive charge of the protons.