Structure of AtomClass 11 Chemistry Notes

Here are the comprehensive notes for the chapter Structure of Atom.

Introduction to the Atom

For centuries, thinkers believed that matter was made of fundamental, indivisible particles called atoms (from the Greek a-tomio, meaning 'uncut-able'). This idea was formally proposed as a scientific theory by John Dalton in 1808. Dalton's atomic theory successfully explained laws like the conservation of mass, but it couldn't explain phenomena like static electricity (why rubbing glass with silk generates a charge).

By the late 19th and early 20th centuries, experiments proved that atoms are, in fact, divisible. They are composed of even smaller, sub-atomic particles: electrons, protons, and neutrons. This discovery changed our entire understanding of matter and led to the development of new atomic models.

Discovery of Sub-Atomic Particles

The first clues about the atom's internal structure came from experiments involving the discharge of electricity through gases at very low pressures.

Discovery of the Electron

Experiments using cathode ray discharge tubes (glass tubes with metal electrodes, evacuated to a very low pressure) revealed the existence of the electron.

When a very high voltage is applied across the electrodes, a stream of particles moves from the negative electrode (cathode) to the positive electrode (anode). This stream is called cathode rays.

Properties of Cathode Rays (Electrons):

• They travel from the cathode to the anode.
• They are invisible, but their presence can be detected by the glow they produce on fluorescent or phosphorescent materials (like zinc sulphide). This is the principle behind old television picture tubes.
• They travel in straight lines in the absence of electric or magnetic fields.
• They are deflected by electric and magnetic fields in a way that indicates they are composed of negatively charged particles. These particles were named electrons.
• The characteristics of cathode rays do not change regardless of the material of the electrodes or the gas inside the tube. This proves that electrons are a fundamental constituent of all atoms.

Charge-to-Mass Ratio of the Electron

In 1897, J.J. Thomson conducted an experiment to measure the charge-to-mass ratio ($e/m_e$) of an electron. He applied electric and magnetic fields perpendicular to the path of the cathode rays.

He observed that the amount of deflection of the electrons depends on:

1. The magnitude of the negative charge: A greater charge leads to a greater deflection.
2. The mass of the particle: A lighter particle is deflected more easily.
3. The strength of the electric or magnetic field: A stronger field causes a greater deflection.

By carefully balancing the fields so that the cathode rays were not deflected, Thomson was able to determine the charge-to-mass ratio: $\frac{e}{m_e} = 1.758820 \times 10^{11} \text{ C kg}^{-1}$

Charge and Mass of the Electron

R.A. Millikan's Oil Drop Experiment (1906-14) was designed to determine the charge on a single electron. In this experiment, he observed tiny, charged oil droplets falling between two electrically charged plates. By adjusting the voltage, he could control the motion of the droplets.

Millikan concluded that the charge on any droplet was always an integral multiple of a fundamental unit of charge, which he identified as the charge of a single electron.

• Charge of an electron (e): The accepted value is $-1.602176 \times 10^{-19}$ C.
• Mass of an electron ($m_e$): Using Thomson's e/m ratio and Millikan's charge value, the mass of an electron was calculated. $m_e = \frac{e}{e/m_e} = \frac{1.602176 \times 10^{-19} \text{ C}}{1.758820 \times 10^{11} \text{ C kg}^{-1}} = 9.1094 \times 10^{-31} \text{ kg}$

Discovery of Protons and Neutrons

• Protons: Experiments with modified cathode ray tubes led to the discovery of canal rays, which were streams of positively charged particles. Unlike electrons, the properties of these particles depended on the gas inside the tube. The lightest and smallest positive ion was produced from hydrogen gas and was named the proton. It was characterized in 1919.

• Neutrons: The existence of a neutral particle was necessary to account for the total mass of an atom. In 1932, James Chadwick discovered the neutron by bombarding a thin sheet of beryllium with alpha-particles. This produced electrically neutral particles with a mass slightly greater than that of a proton.

Properties of Fundamental Particles

Atomic Models

After the discovery of sub-atomic particles, scientists faced several major questions: How are these particles arranged? How is an atom stable? How does atomic structure explain the chemical properties of elements?

Thomson Model of the Atom (1898)

J.J. Thomson proposed that an atom is a sphere of uniformly distributed positive charge, with negatively charged electrons embedded in it, much like plums in a pudding or seeds in a watermelon.

• Key Features:

  • The atom is a sphere with a radius of approximately $10^{-10}$ m.
  • Positive charge is spread evenly throughout the sphere.
  • Electrons are embedded within the sphere to ensure the atom is electrically neutral.
  • The mass of the atom is assumed to be uniformly distributed.

• Limitation: While it explained the overall neutrality of an atom, this model was inconsistent with the results of later experiments, most notably Rutherford's scattering experiment.

Rutherford's Nuclear Model of the Atom

Ernest Rutherford, along with his students Hans Geiger and Ernest Marsden, conducted the famous $\alpha$-particle scattering experiment. They bombarded a very thin gold foil with a stream of high-energy, positively charged alpha ($\alpha$) particles.

Observations:

1. Most of the $\alpha$-particles passed straight through the gold foil without any deflection.
2. A small fraction of $\alpha$-particles were deflected by small angles.
3. A very few $\alpha$-particles (~1 in 20,000) bounced back, deflecting by nearly $180^\circ$.

Conclusions:

1. Most of the space in an atom is empty, which is why most $\alpha$-particles passed through undeflected.
2. The positive charge of the atom is not spread out but is concentrated in a very small, dense region, which he called the nucleus. This dense positive charge was responsible for repelling and deflecting the incoming positive $\alpha$-particles.
3. The volume of the nucleus is negligibly small compared to the total volume of the atom. The radius of an atom is about $10^{-10}$ m, while the nucleus is about $10^{-15}$ m.

Rutherford's Model (The Nuclear Model):

• The positive charge and most of the mass of the atom are concentrated in a tiny, dense nucleus.
• The nucleus is surrounded by electrons that orbit it at high speed in circular paths, similar to planets orbiting the sun.
• Electrons and the nucleus are held together by electrostatic forces of attraction.

Atomic Number and Mass Number

• Atomic Number (Z): The number of protons in the nucleus of an atom. In a neutral atom, the atomic number is also equal to the number of electrons.

  • Atomic number (Z) = number of protons = number of electrons (in a neutral atom)

• Mass Number (A): The total number of protons and neutrons in the nucleus. Protons and neutrons are collectively called nucleons.

  • Mass number (A) = number of protons (Z) + number of neutrons (n)

Isobars and Isotopes

We can represent an atom using the notation ${ }_{Z}^{A} X$, where X is the element symbol.

• Isotopes are atoms of the same element (identical atomic number, Z) but with different mass numbers (A). This difference is due to a different number of neutrons in the nucleus.

  • Example: Hydrogen has three isotopes: Protium (${ }_{1}^{1} \text{H}$), Deuterium (${ }_{1}^{2} \text{D}$), and Tritium (${ }_{1}^{3} \text{T}$).
  • Since chemical properties are determined by the number of electrons (which equals the number of protons), isotopes of an element have the same chemical behavior.

• Isobars are atoms of different elements that have the same mass number (A) but different atomic numbers (Z).

  • Example: Carbon-14 (${ }_{6}^{14} \text{C}$) and Nitrogen-14 (${ }_{7}^{14} \text{N}$).

Given

• Symbol: ${ }_{35}^{80} \text{Br}$
• Atomic Number, $Z = 35$
• Mass Number, $A = 80$
• The species is a neutral atom.

To Find

Number of protons, neutrons, and electrons.

Formula

• Number of protons = $Z$
• Number of electrons = $Z$ (for a neutral atom)
• Number of neutrons = $A - Z$

Solution

• Number of protons = $Z = 35$
• Number of electrons = $Z = 35$
• Number of neutrons = $80 - 35 = 45$

Final Answer The atom ${ }_{35}^{80} \text{Br}$ contains 35 protons, 35 electrons, and 45 neutrons.

Given

• Number of electrons = 18
• Number of protons = 16
• Number of neutrons = 16

To Find

The proper symbol for the species (${ }_{Z}^{A} X^{\text{charge}}$).

Formula

• Atomic Number, $Z$ = number of protons
• Mass Number, $A$ = number of protons + number of neutrons
• Charge = number of protons - number of electrons

Solution

First, identify the element from the atomic number.

• $Z$ = number of protons = 16. The element with $Z=16$ is Sulphur (S).

Next, calculate the mass number.

• $A$ = (number of protons) + (number of neutrons) = $16 + 16 = 32$.

Finally, determine if the species is an ion by calculating its charge.

• Charge = $16 - 18 = -2$.
• Since the species has 2 more electrons than protons, it is an anion with a 2- charge.

Combining these parts gives the symbol.

Final Answer The symbol for the species is ${ }_{16}^{32} \text{S}^{2-}$.

Drawbacks of Rutherford's Model

Despite its improvements, Rutherford's model had two major flaws:

1. It could not explain the stability of the atom. According to classical electromagnetic theory (by James Clerk Maxwell), an accelerating charged particle (like an electron orbiting a nucleus) must continuously emit electromagnetic radiation. By losing energy, the electron should spiral into the nucleus in just $10^{-8}$ seconds. This obviously does not happen, meaning the model was incomplete.
2. It said nothing about the arrangement or energies of the electrons. The model did not explain how electrons were distributed in their orbits or why different elements had different chemical properties.

Developments Leading to Bohr's Model of the Atom

To fix the problems with Rutherford's model, Niels Bohr incorporated two new revolutionary ideas:

1. The dual character of electromagnetic radiation (it behaves as both a wave and a particle).
2. Experimental results from atomic spectra, which showed that atoms emit or absorb energy only at specific, discrete values.

Wave Nature of Electromagnetic Radiation

James Maxwell (1870) proposed that when a charged particle accelerates, it produces and transmits oscillating electric and magnetic fields. These fields travel as waves called electromagnetic waves or electromagnetic radiation. Light is one form of this radiation.

Properties of Electromagnetic Waves:

• The oscillating electric and magnetic fields are perpendicular to each other and to the direction of wave propagation.
• Unlike sound waves, they do not require a medium and can travel through a vacuum.
• They travel in a vacuum at a constant speed, the speed of light (c), which is approximately $3.0 \times 10^8 \text{ m s}^{-1}$.

Electromagnetic radiation is characterized by several properties:

• Frequency ($\nu$): The number of waves that pass a point in one second. Its SI unit is Hertz (Hz), where $1 \text{ Hz} = 1 \text{ s}^{-1}$.
• Wavelength ($\lambda$): The distance between two consecutive crests or troughs of a wave. Its SI unit is the meter (m).
• Wavenumber ($\bar{\nu}$): The number of wavelengths per unit length. It is the reciprocal of wavelength ($\bar{\nu} = 1/\lambda$). Its common unit is $\text{cm}^{-1}$.

These properties are related by the equation: $c = \nu \lambda$

The electromagnetic spectrum is the range of all types of electromagnetic radiation, from long-wavelength radio waves to short-wavelength gamma rays. Visible light is only a tiny portion of this spectrum.

Given

• Wavelength of violet light, $\lambda_{\text{violet}} = 400 \text{ nm} = 400 \times 10^{-9} \text{ m}$
• Wavelength of red light, $\lambda_{\text{red}} = 750 \text{ nm} = 750 \times 10^{-9} \text{ m}$
• Speed of light, $c = 3.00 \times 10^8 \text{ m s}^{-1}$

To Find

The frequencies ($\nu$) for violet and red light.

Formula

$c = \nu \lambda \quad \implies \quad \nu = \frac{c}{\lambda}$

Solution

For violet light: $\nu_{\text{violet}} = \frac{3.00 \times 10^8 \text{ m s}^{-1}}{400 \times 10^{-9} \text{ m}} = 7.50 \times 10^{14} \text{ Hz}$

For red light: $\nu_{\text{red}} = \frac{3.00 \times 10^8 \text{ m s}^{-1}}{750 \times 10^{-9} \text{ m}} = 4.00 \times 10^{14} \text{ Hz}$

Final Answer The frequency range of the visible spectrum is from $4.0 \times 10^{14}$ Hz (red) to $7.5 \times 10^{14}$ Hz (violet).

Particle Nature of Electromagnetic Radiation: Planck's Quantum Theory

Classical wave theory could not explain several phenomena, including:

• Black-body radiation: The nature of radiation emitted by hot objects.
• Photoelectric effect: The ejection of electrons from a metal surface when light shines on it.

These observations suggested that energy is not continuous but is exchanged in discrete packets.

Black-Body Radiation An ideal object that absorbs and emits all frequencies of radiation is called a black body. When a black body is heated, it emits radiation. The intensity and wavelength of this radiation depend only on its temperature. Classical physics could not explain the observed distribution of emitted wavelengths.

In 1900, Max Planck proposed a revolutionary idea. He suggested that atoms and molecules could emit or absorb energy only in discrete quantities, which he called a quantum. The energy of a single quantum is directly proportional to the frequency of the radiation.

This relationship is described by Planck's equation: $E = h\nu$ Where:

• $E$ is the energy of one quantum.
• $\nu$ is the frequency of the radiation.
• $h$ is Planck's constant, with a value of $6.626 \times 10^{-34} \text{ J s}$.

The Photoelectric Effect In 1887, H. Hertz observed that when light shines on certain metals, electrons are ejected. This is the photoelectric effect.

Key Observations:

1. Electrons are ejected instantly, with no time delay.
2. The number of electrons ejected is proportional to the intensity (brightness) of the light.
3. For each metal, there is a characteristic minimum frequency, called the threshold frequency ($\nu_0$), below which no electrons are ejected, no matter how intense the light is.
4. Above the threshold frequency, the kinetic energy of the ejected electrons increases with the frequency of the light.

In 1905, Albert Einstein explained the photoelectric effect using Planck's quantum theory. He proposed that light itself is made of a stream of particles called photons, with each photon having an energy of $E = h\nu$.

Einstein's Explanation:

• When a photon strikes an electron, it transfers its entire energy to the electron instantaneously.
• A certain amount of energy, called the work function ($W_0$), is required to remove the electron from the metal. The work function is related to the threshold frequency by $W_0 = h\nu_0$.
• If the photon's energy ($h\nu$) is greater than the work function, the electron is ejected. The excess energy becomes the kinetic energy of the electron.

This is summarized by the photoelectric effect equation: $h\nu = h\nu_0 + \frac{1}{2}m_e v^2$ or $h\nu = W_0 + K.E.$

Dual Behaviour of Electromagnetic Radiation

The photoelectric effect and black-body radiation demonstrate the particle-like nature of light. However, phenomena like interference and diffraction demonstrate its wave-like nature. This led to the conclusion that light has a dual behaviour, acting as both a wave and a particle depending on the experiment.

Given

• Threshold frequency, $\nu_0 = 7.0 \times 10^{14} \text{ s}^{-1}$
• Frequency of incident radiation, $\nu = 1.0 \times 10^{15} \text{ s}^{-1}$
• Planck's constant, $h = 6.626 \times 10^{-34} \text{ J s}$

To Find

The kinetic energy (K.E.) of the emitted electron.

Formula

$K.E. = h(\nu - \nu_0)$

Solution

First, let's express both frequencies in the same power of 10. $\nu = 1.0 \times 10^{15} \text{ s}^{-1} = 10.0 \times 10^{14} \text{ s}^{-1}$

Now substitute the values into the formula. $K.E. = (6.626 \times 10^{-34} \text{ J s}) \times (10.0 \times 10^{14} \text{ s}^{-1} - 7.0 \times 10^{14} \text{ s}^{-1})$ $K.E. = (6.626 \times 10^{-34} \text{ J s}) \times (3.0 \times 10^{14} \text{ s}^{-1})$ $K.E. = 1.988 \times 10^{-19} \text{ J}$

Final Answer The kinetic energy of the emitted electron is $1.988 \times 10^{-19}$ J.

Atomic Spectra

When white light passes through a prism, it splits into a continuous rainbow of colors, called a continuous spectrum. However, when atoms in the gas phase are excited (by heating or electric discharge), they emit light only at specific, discrete wavelengths. This produces a line spectrum or atomic spectrum.

• Emission Spectrum: The spectrum of radiation emitted by a substance that has absorbed energy. It appears as a series of bright lines against a dark background.
• Absorption Spectrum: The spectrum produced when a continuous source of radiation is passed through a sample. The sample absorbs light at the same specific wavelengths it would emit, leaving dark lines in the continuous spectrum.

Each element has a unique line spectrum, which can be used to identify it, much like a fingerprint.

Line Spectrum of Hydrogen The hydrogen atom has the simplest line spectrum. The spectral lines of hydrogen are grouped into several series:

• Lyman Series: (UV region) Transitions ending at $n_1=1$.
• Balmer Series: (Visible region) Transitions ending at $n_1=2$.
• Paschen Series: (Infrared region) Transitions ending at $n_1=3$.
• Brackett Series: (Infrared region) Transitions ending at $n_1=4$.
• Pfund Series: (Infrared region) Transitions ending at $n_1=5$.

Johannes Rydberg developed a general formula to describe all the series in the hydrogen spectrum: $\bar{\nu} = 109,677 \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} \right) \text{ cm}^{-1}$ where $109,677 \text{ cm}^{-1}$ is the Rydberg constant for hydrogen, and $n_2 > n_1$.

Bohr's Model for the Hydrogen Atom

In 1913, Niels Bohr proposed a model for the hydrogen atom that explained its stability and line spectrum by incorporating the idea of quantization.

Bohr's Postulates:

1. An electron in a hydrogen atom moves around the nucleus in a fixed circular path of constant radius and energy. These paths are called orbits or stationary states.
2. The energy of an electron in an orbit does not change with time. An electron can move to a higher energy orbit by absorbing energy or move to a lower energy orbit by emitting energy. This energy change is not continuous.
3. The frequency of radiation absorbed or emitted during a transition between two states with energies $E_1$ and $E_2$ is given by: $\nu = \frac{\Delta E}{h} = \frac{E_2 - E_1}{h}$
4. The angular momentum of an electron in an orbit is quantized. It can only have values that are an integral multiple of $h/2\pi$. $m_e v r = n \frac{h}{2\pi}$ where $n$ is an integer ($1, 2, 3, ...$) called the principal quantum number.

Key Results of Bohr's Theory for Hydrogen:

• Radii of Stationary States: The radius of the n-th orbit is given by: $r_n = n^2 a_0$ where $a_0 = 52.9 \text{ pm}$ is the radius of the first orbit (Bohr orbit).
• Energy of Stationary States: The energy of the n-th orbit is given by: $E_n = -R_H \left( \frac{1}{n^2} \right)$ where $R_H = 2.18 \times 10^{-18} \text{ J}$ is the Rydberg constant (in terms of energy).

Explanation of the Hydrogen Line Spectrum Bohr's model explains the line spectrum as resulting from an electron transitioning between two allowed energy levels. The energy difference ($\Delta E$) between an initial orbit ($n_i$) and a final orbit ($n_f$) is: $\Delta E = R_H \left( \frac{1}{n_i^2} - \frac{1}{n_f^2} \right)$ When an electron drops from a higher orbit ($n_i$) to a lower one ($n_f$), energy is emitted as a photon of a specific frequency, creating an emission line.

Limitations of Bohr's Model:

• It could not explain the finer details of the hydrogen spectrum (like lines that were actually closely spaced doublets).
• It failed to explain the spectra of atoms with more than one electron (multi-electron atoms).
• It could not explain the splitting of spectral lines in the presence of a magnetic field (Zeeman effect) or an electric field (Stark effect).
• It could not explain how atoms form molecules through chemical bonds.

The fundamental flaw was that Bohr's model still treated the electron as a particle moving in a well-defined path, ignoring its wave-like nature.

Given

• Initial state, $n_i = 5$
• Final state, $n_f = 2$
• Rydberg constant, $R_H = 2.18 \times 10^{-18} \text{ J}$
• Planck's constant, $h = 6.626 \times 10^{-34} \text{ J s}$
• Speed of light, $c = 3.0 \times 10^8 \text{ m s}^{-1}$

To Find

(i) Frequency of the emitted photon, $\nu$ (ii) Wavelength of the emitted photon, $\lambda$

Formula

$\Delta E = R_H \left( \frac{1}{n_i^2} - \frac{1}{n_f^2} \right)$ $\nu = \frac{|\Delta E|}{h}$ $\lambda = \frac{c}{\nu}$

Solution

(i) Calculate the energy difference

$\Delta E = 2.18 \times 10^{-18} \text{ J} \left( \frac{1}{5^2} - \frac{1}{2^2} \right) = 2.18 \times 10^{-18} \text{ J} \left( \frac{1}{25} - \frac{1}{4} \right)$ $\Delta E = 2.18 \times 10^{-18} \text{ J} (0.04 - 0.25) = 2.18 \times 10^{-18} \text{ J} (-0.21)$ $\Delta E = -4.58 \times 10^{-19} \text{ J}$ The negative sign indicates that energy is emitted. For calculating frequency, we use the magnitude of $\Delta E$.

(ii) Calculate the frequency

$\nu = \frac{4.58 \times 10^{-19} \text{ J}}{6.626 \times 10^{-34} \text{ J s}} = 0.691 \times 10^{15} \text{ Hz} = 6.91 \times 10^{14} \text{ Hz}$ Answer for frequency = $6.91 \times 10^{14}$ Hz

(iii) Calculate the wavelength

$\lambda = \frac{3.0 \times 10^8 \text{ m s}^{-1}}{6.91 \times 10^{14} \text{ Hz}} = 0.434 \times 10^{-6} \text{ m} = 434 \times 10^{-9} \text{ m} = 434 \text{ nm}$ Answer for wavelength = $434$ nm

Towards the Quantum Mechanical Model of the Atom

Two key developments paved the way for a more accurate model of the atom:

Dual Behaviour of Matter

In 1924, Louis de Broglie proposed that if radiation can have both wave and particle properties, then matter (like electrons) should also exhibit dual behaviour. He proposed a relationship between the wavelength and momentum of a particle.

de Broglie's Equation: $\lambda = \frac{h}{mv} = \frac{h}{p}$ Where:

• $\lambda$ is the de Broglie wavelength.
• $h$ is Planck's constant.
• $m$ is the mass of the particle.
• $v$ is the velocity of the particle.
• $p$ is the momentum of the particle.

This was experimentally confirmed when it was shown that a beam of electrons can be diffracted, a property characteristic of waves.

Given

• Mass of electron, $m = 9.1 \times 10^{-31}$ kg
• Kinetic Energy, K.E. = $3.0 \times 10^{-25}$ J
• Planck's constant, $h = 6.626 \times 10^{-34}$ J s

To Find

The de Broglie wavelength, $\lambda$.

Formula

$K.E. = \frac{1}{2}mv^2 \implies v = \sqrt{\frac{2 \times K.E.}{m}}$ $\lambda = \frac{h}{mv}$

Solution

First, calculate the velocity ($v$) of the electron. $v = \sqrt{\frac{2 \times (3.0 \times 10^{-25} \text{ kg m}^2 \text{s}^{-2})}{9.1 \times 10^{-31} \text{ kg}}} = \sqrt{0.659 \times 10^6 \text{ m}^2 \text{s}^{-2}} = 812 \text{ m s}^{-1}$

Now, use the velocity to calculate the wavelength ($\lambda$). $\lambda = \frac{6.626 \times 10^{-34} \text{ J s}}{(9.1 \times 10^{-31} \text{ kg})(812 \text{ m s}^{-1})} = \frac{6.626 \times 10^{-34}}{7390 \times 10^{-31}} \text{ m}$ $\lambda = 0.0008967 \times 10^{-3} \text{ m} = 8.967 \times 10^{-7} \text{ m} = 896.7 \text{ nm}$

Final Answer The wavelength of the electron is $896.7$ nm.

Heisenberg's Uncertainty Principle

In 1927, Werner Heisenberg stated that it is impossible to determine simultaneously and with perfect accuracy both the exact position and the exact momentum (or velocity) of a microscopic particle like an electron.

Heisenberg's Uncertainty Principle Equation: $\Delta x \times \Delta p_x \geq \frac{h}{4\pi}$ Where:

• $\Delta x$ is the uncertainty in position.
• $\Delta p_x$ is the uncertainty in momentum.

This means that the more precisely we know the position of an electron ($\Delta x$ is small), the less precisely we know its momentum ($\Delta p_x$ is large), and vice versa.

Significance: This principle rules out the idea of electrons having definite paths or orbits as described by Bohr. If we cannot know both position and velocity at the same time, we cannot predict the trajectory. This fundamentally changed how we view the electron in an atom, shifting from a model of certainty to one of probability.

The Quantum Mechanical Model of the Atom

This model, developed by Erwin Schrödinger and Werner Heisenberg in 1926, incorporates the dual behaviour of matter and the uncertainty principle. It describes the behavior of electrons in atoms not in terms of definite orbits, but in terms of probabilities.

Key Features of the Quantum Mechanical Model:

1. Quantized Energy: The energy of electrons in atoms is quantized, meaning they can only have specific, allowed energy values.
2. Wave Function ($\Psi$): The Schrödinger equation provides solutions called wave functions ($\Psi$). A wave function is a mathematical function that describes the electron. An atomic orbital is defined as the wave function for an electron in an atom.
3. Probability, not Certainty: The exact path of an electron cannot be known. Instead, we can determine the probability of finding an electron in a certain region of space.
4. Probability Density ($|\Psi|^2$): The square of the wave function, $|\Psi|^2$, at any point in space gives the probability density of finding the electron at that point. Regions of high probability density are where the electron is most likely to be found.

Orbitals and Quantum Numbers

Each orbital in an atom is described by a set of three quantum numbers ($n, l, m_l$), which arise from the solution of the Schrödinger equation. A fourth quantum number ($m_s$) describes the electron itself.

1. Principal Quantum Number (n):

   • Describes the main energy level or shell.
   • Values: Positive integers ($n = 1, 2, 3, ...$).
   • Determines the size and, to a large extent, the energy of the orbital. Higher $n$ means a larger orbital and higher energy.
   • Shells are also designated by letters: K ($n=1$), L ($n=2$), M ($n=3$), etc.

2. Azimuthal Quantum Number (l):

   • Also known as the orbital angular momentum or subsidiary quantum number.
   • Describes the shape of the orbital and defines the subshell.
   • Values: Integers from $0$ to $n-1$.
   • Subshells are designated by letters:
     • $l=0 \rightarrow$ s subshell (spherical shape)
     • $l=1 \rightarrow$ p subshell (dumbbell shape)
     • $l=2 \rightarrow$ d subshell (complex shapes)
     • $l=3 \rightarrow$ f subshell (even more complex shapes)

3. Magnetic Quantum Number ($m_l$):

   • Describes the spatial orientation of the orbital.
   • Values: Integers from $-l$ to $+l$, including $0$.
   • The number of possible $m_l$ values ($2l+1$) gives the number of orbitals in a subshell.
     • s subshell ($l=0$): one orbital ($m_l=0$)
     • p subshell ($l=1$): three orbitals ($m_l = -1, 0, +1$)
     • d subshell ($l=2$): five orbitals ($m_l = -2, -1, 0, +1, +2$)

4. Electron Spin Quantum Number ($m_s$):

   • Describes the intrinsic angular momentum of an electron, which is its "spin".
   • An electron can spin in one of two directions, represented by two possible values.
   • Values: $+1/2$ (spin up, $\uparrow$) or $-1/2$ (spin down, $\downarrow$).

Shapes of Atomic Orbitals

The shape of an orbital is represented by a boundary surface diagram, which encloses the region where there is a 90% probability of finding the electron.

• s Orbitals ($l=0$): Are spherically symmetric. The size of the s orbital increases as the principal quantum number $n$ increases ($1s < 2s < 3s, ...$).
• p Orbitals ($l=1$): Have a dumbbell shape with two lobes on either side of the nucleus. There are three p orbitals in each p subshell, oriented along the x, y, and z axes ($p_x, p_y, p_z$).
• d Orbitals ($l=2$): Have more complex shapes. Four of the five d orbitals have a cloverleaf shape ($d_{xy}, d_{yz}, d_{xz}, d_{x^2-y^2}$), and the fifth ($d_{z^2}$) has a dumbbell shape with a donut-like ring around the middle.

Nodes are regions where the probability of finding an electron is zero.

• Radial nodes are spherical surfaces. The number of radial nodes for an orbital is $n-l-1$.
• Angular nodes are planes or cones. The number of angular nodes is equal to $l$.
• Total number of nodes = $n-1$.

Energies of Orbitals

• In a hydrogen atom: The energy of an orbital depends only on the principal quantum number, $n$. Orbitals within the same shell are degenerate (have the same energy).

  • Energy order: $1s < (2s = 2p) < (3s = 3p = 3d) < ...$

• In multi-electron atoms: The energy of an orbital depends on both $n$ and $l$. This is because of electron-electron repulsions and the shielding effect, where inner-shell electrons partially block the attraction of the nucleus from outer-shell electrons.

  • Within a shell, the energy of subshells increases in the order: $s < p < d < f$.
  • This splitting of energy levels leads to a more complex energy order for filling orbitals.

The (n+l) Rule helps predict the energy order of orbitals in multi-electron atoms:

1. The lower the value of $(n+l)$, the lower the energy of the orbital.
2. If two orbitals have the same $(n+l)$ value, the one with the lower $n$ value has lower energy.
   • Example: For 4s, $n+l = 4+0=4$. For 3d, $n+l = 3+2=5$. Therefore, 4s has lower energy than 3d.

The general order for filling orbitals is: $1s, 2s, 2p, 3s, 3p, 4s, 3d, 4p, 5s, 4d, 5p, 6s, 4f, 5d, 6p, 7s, ...$

Filling of Orbitals in Atoms

The distribution of electrons into the orbitals of an atom is called its electronic configuration. This is governed by three rules:

1. Aufbau Principle: (German for 'building up') Electrons fill the lowest energy orbitals available first before moving to higher energy orbitals.
2. Pauli Exclusion Principle: No two electrons in an atom can have the same set of four quantum numbers. This means an orbital can hold a maximum of two electrons, and they must have opposite spins ($\uparrow\downarrow$).
3. Hund's Rule of Maximum Multiplicity: When filling degenerate orbitals (orbitals of the same energy, like the three p orbitals), electrons will fill each orbital singly with parallel spins ($\uparrow$) before any orbital gets a second electron.

Electronic Configuration of Atoms

We can represent electronic configurations in two ways:

1. Notation: $1s^2 2s^2 2p^6 ...$ where the superscript indicates the number of electrons in the subshell.
2. Orbital Diagram: Uses boxes for orbitals and arrows for electrons.
   • Example: Nitrogen (Z=7): $1s^2 2s^2 2p^3$
     • Orbital Diagram: