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JF Basic Chemistry Tutorial : Atomic Theory

Shane Plunkett. plunkes@tcd.ie. JF Basic Chemistry Tutorial : Atomic Theory. Matter -anything that had mass and takes up space -classified into 3 types: elements, compounds and mixtures. Element -consists of only one kind of atom. So, an element is a pure substance.

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JF Basic Chemistry Tutorial : Atomic Theory

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  1. Shane Plunkett plunkes@tcd.ie JF Basic Chemistry Tutorial : Atomic Theory Matter -anything that had mass and takes up space -classified into 3 types: elements, compounds and mixtures Element -consists of only one kind of atom. So, an element is a pure substance. Atom -the smallest particle of an element that can exist. -made up of three subatomic particles protons, neutrons and electrons. How do we know this?

  2. Theories through time…. • 1890’s • J.J. Thompson - The cathode ray experiment • negative particles discharged were named electrons and denoted e- Conclusions: Can’t be a plum pudding structure! - central positively charged point and a large volume of empty space • Early 20th Century • The Plum Pudding model! • atoms made up of blobs of a positively charged jelly with electrons suspended in it • 1908 • Ernest Rutherford – The Gold Leaf Experiment • 2 students Geiger and Marsden shot α-particles (positively charged particles) at a thin piece of gold foil. Most passed through, some bounced back.

  3. 16.00 O 8 Mass Number or Atomic Mass (A) Element symbol Atomic Number (Z) Central point called the atomic nucleus and this is where all positively charged protons are located. Atoms do not have an overall charge. There must be the same number of electrons as there are protons. If we take a look at the periodic table, we see that it is made up of vertical blocks (these are called groups) and horizontal rows (these are called periods). Definition:The number of protons in an atomic nucleus is called the atomic number (Z) of the element. Note: the atomic number of an element is the smaller number e.g. look at Oxygen in the Periodic Table

  4. Questions Give the name, atomic number and group number of the element with the following Z value: (a) Z=16 Sulfur Group VI (b) Z=8 Oxygen Group VI (c) Z=19 Potassium Group I (d) Z=13 Aluminium Group III (e) Z=32 Germanium Group IV

  5. What about neutrons? A mass spectrometer is an instrument used by chemists to determine the mass of a given element. Not all atoms of a particular element have the same mass. Example:Neon (Ne) gives 3 types of atoms, each heavier than the last. Isotopes have the same atomic number but different atomic masses They must have the same number of electrons and protons but a different number of neutrons Explanation: There must be a third type of subatomic particle that contributes to this change in mass  Neutrons (n) have no electric charge, but have the same mass as protons. Definition: Atoms of a single element that differ from each other in mass are called isotopes

  6. Definition: The mass number or atomic mass (A) is the total number of protons and neutrons in a nucleus of an atom. Note: the mass number or atomic mass of an element is the larger number As we go across the rows of the periodic table, the mass of the nucleus increases. If we move onto the next element, Helium, we find: Atomic number = 2, i.e. 2 electrons and 2 protons in each He atom Mass number = 4.00 No. of neutrons in the helium atom = Mass no. – Atomic no. = 2 Each helium atom has 2 electrons, 2 protons and 2 neutrons Example Hydrogen: Atomic number = 1 This means a hydrogen atom possesses one electron and one proton Mass number = 1.0079 No. of neutrons in the hydrogen atom = Mass no. – Atomic no. = 1.0079 – 1  0 This means that the hydrogen atom has one electron, one proton and no neutrons.

  7. Sample Question Given Z (the atomic number), how many protons, neutrons, and electrons are in the following elements? (a) Z=6 (Carbon)  have 6 protons and 6 electrons number of neutrons = mass number – atomic number = 12 – 6 = 6 neutrons Now you try! Z = 30 Z=16 Z=35 (b) Z=12 (Magnesium)  have 12 protons and 12 electrons number of neutrons = 24 – 12 = 12 neutrons (c) Z=17 (Chlorine)  have 17 protons and 17 electrons number of neutrons = 35 – 17 = 18 neutrons Z = 30, Ans.: 30 electrons, 30 protons, 35 neutrons Z = 16, Ans.: 16 electrons, 16 protons, 16 neutrons Z = 35, Ans.: 35 electrons, 35 protons, 45 neutrons

  8. Sample Question The three naturally occuring isotopes of argon are 36Ar, 38Ar and 40Ar. How many protons, neutrons and electrons are present in each? Note that in isotopes, the mass numbers can change, but the atomic number always remains the same. It is only the number of neutrons that can change. Number of neutrons = mass number – atomic number 36Ar: no. of neutrons = 36 – 18 = 18 neutrons 38Ar: no. of neutrons = 38 – 18 = 20 neutrons 40Ar: no. of neutrons = 40 – 18 = 22 neutrons Mass numbers are 36, 38 and 40 respectively. From the Periodic Table, we know that the atomic number of argon is 18. This means that each isotope has 18 protons and 18 electrons.

  9. We have so far built up a picture of the atom that looks like this: Atomic Nucleus -contains the protons and neutrons Empty space Electrons are found somewhere in here

  10. We know that electrons orbit the atomic nucleus. Can we be more definite about where electrons can be found? Four types of orbitals: s, p, d and f. We will look at the first 3. YES! There is a probabilityassociated with finding an electron in a particular location around the nucleus. Definition:An orbital is a region of space where the probability of finding an electron is large. 

  11. Quantum Numbers of Atomic Orbitals Atomic orbitals have 4 quantum numbers associated with them 1. The Prinicipal Quantum Number (n) -positive, whole number (n=1, 2, 3…) -tells you about the size of the orbital, i.e., the distance from the nucleus -tells you about the energy of the orbital; the bigger the number, the higher the energy level -the orbitals form a series of shells (like the layers of an onion). Shells of higher n surround shells of lower n. 2. The Angular Momentum Quantum Number (l) -can be a whole number with values from 0 to n-1 -tells you about the shape of the orbital, i.e. if it is an s, p or d orbital -you can get its values when you know n, e.g. if n=1, then l can only equal 0, so you have an s-orbital if n=2, then l can equal 0 or 1, so you have s- and p-orbitals if n=3, then l can equal 0, 1 or 2, so can have s-, p- and d-orbitals

  12. 3. The Magnetic Quantum Number (ml) -whole number from –lthrough 0 to +l - labels the orbitals and tells you how many of each orbital type you have or: anticlockwise represented as an arrow pointing downwards (↓) and is given the value -½ 4. The Spin Magnetic Quantum Number (ms) -electrons behave like spinning spheres and this behaviour is called spin • spin can be: clockwise (usually represented as an arrow pointing upwards (↑) and is given the value +½

  13. Sample Question Determine the total number of orbitals in a shell with a principle quantum number n = 3. Step 1: Determine the angular momentum quantum numbers associated with n = 3 To find out how many of each orbital we have, use the magnetic quantum number, ml Angular momentum quantum number, l = 0, 1, 2, …, n -1 For n =3, there are l = 0, 1, 2 subshells For l =0, we have an s-orbital l =1, we have a p-orbital l =2, we have a d-orbital

  14. Step 2: Use the Magnetic Quantum Number to determine the number of each type of orbital we have Magnetic quantum number, ml= l, l -1, l -2, …, -l For l=0; ml=0 This means we have one s-orbital Now you try! How many orbitals are there in the shell with n=2? Ans.: 4 orbitals (1s-orbital and 3 p-orbitals) For l=1; ml=1, 1-1, -1 =1, 0, -1 This means we have threep-orbitals For l=2: ml= 2, 2-1, 2-2, 2-3, -2 =2, 1, 0, -1, -2 This means we have fived-orbitals So, in total, we have 1+3+5 = 9 orbitals in the shell with n=3.

  15. What do these orbitals look like? s-orbitals -spherical in shape p-orbitals -a cloud with two lobes on opposite sides of the nucleus. -the two lobes are separated by a planar region called a nodal plane -there are 3 p-orbitals which point along the x, y, and z axes respectively px pz py

  16. d-orbitals -there are five d-orbitals named after the directions in which they point dx2 + y2 dxz dxy dz2 dyz

  17. There are rules for how electrons will accommodate these orbitals. Rule 1: Electrons will occupy the orbital of lowest available energy. The energy of orbitals increases in the order s<p<d,f Rule 2: Each orbital can accommodate a maximum of two electrons. The two electrons in an orbital must have opposite spins (i.e. one must be clockwise and the other must be anticlockwise). This is called the Pauli Exclusion Principle.

  18. Example: Helium, He Atomic number = 2Lowest energy orbital available is first s-orbital, denoted the 1s orbital Because we have 2 electrons and each orbital can have 2 electrons, we can place both electrons of He into the 1s orbital: Nucleus containing 2 protons and 2 neutrons  Electron  Electron 1s orbital

  19. The name we give to the build-up of these electrons in orbitals is the electronic configuration So, the electronic configuration of the He atom is 1s2, where the superscript 2 indicates the number of electrons present in the orbital. Question: How many electrons are present in the hydrogen atom and how would you write its electronic configuration? Answer: Hydrogen has one electron and an electronic configuration of 1s1 After the 1s orbital, next comes the 2s orbital. Again this can hold 2 electrons Question: How many electrons are present in the lithium atom and how would you write its electronic configuration? Answer: Lithium has 3 electrons and an electronic configuration of 1s2 2s1

  20. Once the 2s orbital has been filled, the 2p orbital is now filled First, the 1s and 2s orbitals are filled. This accounts for 4 out of the 5 electrons. Example: Boron, B Atomic number = 5 Because the 3s orbital is too high in energy and we can only place two electrons in each orbital, the final electron goes into the 2p orbital Hence, the electronic configuration for the ground state boron atom is 1s2 2s2 2p1 The third rule we need to know is called Hund’s Ruleand states that electrons will occupy degenerate orbitals (those of the same energy) singly before pairing (happens in buses).

  21. Example:The atomic number of carbon is 6 and the electronic configuration is 1s2 2s2 2p2 The 2p orbital fills up as follows:      1s2 2s2 2p2 Note! The 2p orbitals have been filled up singly!     1s2 2s2 2p2

  22. QuestionWhat are the electronic configurations of the following elements (a) Nitrogen, (b) Magnesium, (c) Chlorine, (d) Silicon, (e) Potassium? The elements we have dealt with so far have only contained s and p orbitals These elements make up the s and p blocks of the periodic table The first two groups are called the s-blockand the last six groups are called the p-block In between these, we find the d-block or the transition metal elements The d-block elements are so called because this is where we begin to fill the d-orbitals 5 d-orbitals  can accommodate 10 electrons

  23. QuestionWhat are the electronic configurations of the following elements (a) Nitrogen, (b) Magnesium, (c) Chlorine, (d) Silicon, (e) Potassium? Answer: (a) Nitrogen has 7 electrons and an electronic configuration of 1s2 2s2 2p3 (b) Magnesium has 12 electrons and an electronic configuration of 1s2 2s2 2p6 3s2 (c) Chlorine has 17 electrons and an electronic configuration of 1s2 2s2 2p6 3s2 3p5 (d) Silicon has 14 electrons and an electronic configuration of 1s2 2s2 2p6 3s2 3p2 (e) Potassium has 19 electrons and an electronic configuration of 1s2 2s2 2p6 3s2 3p6 4s1

  24. ExampleThe atomic number of Iron is 26 we have 26 electrons          We begin to fill up the orbitals as follows: 2s2 2p6 3s2 3p6 1s2 We have filled 18 out of 26 electrons Now, we face a problem! As n increases, the sublevel energies get closer together. This results in the overlap of some sublevels and here we have such a case. The 4s sublevel is slightly lower in energy that the 3d orbital, so it is filled first.

  25.       4s2 3d6 Question What is the electronic configuration of (a) Vanadium, (b) Cobalt, (c) Zinc? Answer: (a) Vanadium has 23 electrons and an electronic configuration of 1s2 2s2 2p6 3s2 3p6 4s2 3d3 (b) Cobalt has 27 electrons and an electronic configuration of 1s2 2s2 2p6 3s2 3p6 4s2 3d7 (c) Zinc has 30 electrons and an electronic configuration of 1s2 2s2 2p6 3s2 3p6 4s2 3d10

  26. The nature of light and Atomic Spectra Light is emitted in discrete or definite packets called quanta or phonons ΔE = hν where h is Planck’s constant, 6.63 × 10-34Js Frequency of light, ν, increases proportionally with increase in energy, E: Electrons possess both kinetic and potential energy Electrons can behave like rocks on a cliff – when the rock falls it gives up potential energy  If electron falls towards nucleus, they give up potential energy When excited electrons in atoms fall from a high energy level to a low energy state, light is emitted with a specific frequency or colour Electrons must occupy certain energy levels – like steps on a ladder If an electron absorbs a photon or quantum of light, it is elevated to a higher energy level – an excited state When the electron falls to lower energy state, it gives up this energy in specific quanta

  27. Atoms with all electrons in their lowest energy levels are said to be in their ground state ΔE = Efinal – Einitial= 2.18 × 10-18J x The energy difference between any two levels may be found by using Rydberg’s constant: where 2.18 × 10-18J is the Rydberg constant and n is a positive integer representing the shell number We can convert the units of the Rydberg constant from Joules to metres-1 ν = c/λ ΔE = hν ΔE = 2.18 × 10-18 J x Combine these to give: h = Planck’s constant ΔE = hc = 2.18 × 10-18 J λ x c = Speed of light λ= wavelength

  28.  1 = 2.18 × 10-18Jλhc x = (2.18 × 10-18J) . (6.626 × 10-34Js)(3.00 × 108 ms-1) x = 1.0967 × 107 m-1 x Example What is the wavelength of a photon emitted during a transition from the ninitial = 5 state to the nfinal = 2 state in the hydrogen atom? ΔE = R = 2.18 × 10-18 J (1/4 – 1/25) = 4.58 × 10-19 J

  29. To calculate the wavelength of the photon: = (3.00 × 108ms-1)(6.63 × 10-34Js) 4.58 × 10-19 J λ= c/ν= ch/ΔE = 4.34 × 10-7m For wavelengths, must convert to nanometres: = 4.34 × 10-7 m × (1 × 109 nm)/(1 m) = 434 nm Question: What is the wavelength of a photon emitted during a transition from ninitial = 4 to nfinal= 2 state in the H atom? Answer: 487 nm

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