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Bicol University (BS Food technology) College of Science LABORATORY ACTIVITIES DEPARTMENT OF CHEMISTRY Sem 1 2015-2016 Chem 1–General Chemistry ELECTRONIC STRUCTURE OF THE ATOM Activity No. ____ The present model of the atom tells us that the electrons in the atom move in threedimensional space around the nucleus but NOT in an orbit that has a definite radius. The position of the electron CANNOT be defined exactly but the PROBABILITY of finding it at a given position can be determined. The region in space where the electron is most likely to be found is called an ORBITAL. The orbital describes a region in space around the nucleus where an electron moves. It is an energy state, thus, an orbital has a characteristic energy. An orbital is characterized by four QUANTUM NUMBERS. THE PRINCIPAL QUANTUM NUMBER, n The PRINCIPAL QUANTUM NUMBER, n, refers to main energy levels or shells of an orbital. It is related to the average distance of an electron from the nucleus, and the energy of the orbital. The higher the n, the farther the electron is from the nucleus and the higher the energy of the orbital. The principal quantum number also gives us an idea of the size of an atom. The higher the n value, the farther the electron is from the nucleus and so the larger the atom is. The principal quantum number can only have positive integral values (1, 2, 3, 4 . . .) THE AZIMUTHAL QUANTUM NUMBER, l (angular momentum quantum number) The AZIMUTHAL QUANTUM NUMBER, l, represents energy sublevel or subshells and take integral values from 0 to n-1. It is related to the shape of the orbital; the different shapes have been assigned a corresponding letter. l name of orbital n l 1 0 0 1 0 1 2 2 3 [Q1] 0 1 2 3 4 5 s p d f g h orbital symbol/ designation 1s 2s 2p 3s 3p 3d Give the orbital symbol for the orbitals in the fourth main energy level. THE MAGNETIC QUANTUM NUMBER, ml The MAGNETIC QUANTUM NUMBER, ml, relates to the spatial orientation of the orbital in space relative to the other orbital's orientation. It can have integral values from -l through 0 to +l. 2l + 1 38 Bicol University (BS Food technology) College of Science LABORATORY ACTIVITIES DEPARTMENT OF CHEMISTRY Sem 1 2015-2016 Chem 1–General Chemistry For example, for l = 0, there is only one possible value for ml. Thus, there is only one type of s orbital. On the other hand, for l = 1, ml can have three possible values; -1 0, +1. There are three types of p orbitals, each different in the way they are oriented in space. l = 0 2 (0) + 1 l = 1 2 (1) + 1 1 orientation 0 3 orientations (-1, 0, +1) [Q2] Based on the ml value for the d and f orbitals, determine how many types of d and f orbitals there are. THE SPIN QUANTUM NUMBER. ms The SPIN QUANTUM NUMBER, ms, represents electron spin. Since there are only two possible spins —- clockwise and counterclockwise — for an electron, ms can have two values: ─½ or +½. The spin quantum number led to the PAULI'S EXCLUSION PRINCIPLE. In a given atom, no two electrons can have the same set of four quantum numbers. This implies that an orbital can accommodate a maximum of two electrons. The electron configuration of an atom describes the arrangements of electrons in the orbitals of the atom. This arrangement follows some rules/principles. THE AUFBAU PRINCIPLE AUFBAU is a German term that means "building up." As protons are added one by one to the nucleus to build up the elements, electrons are likewise added to the orbital. For instance, a hydrogen atom has one proton and one electron, and this electron occupies a certain orbital. As another proton is added to the hydrogen nucleus to give the helium atom, an electron is likewise added. But which orbital would this electron occupy? A more general question would be, as electrons are added to the orbitals, in what order will the orbitals be filled? Do you start with the orbital with the highest quantum number, n, that is, the orbital with the highest energy? Or do you start with the lowest? THE n + l RULE This rule will aid you in determining which orbital is first filled up as electrons are added to the orbitals in an atom. Each added electron enters the orbital s in increasing energy. That is, each added electron enters the orbital with the lowest energy. The lower the (n + l) value the lower is the energy of the orbital. If the (n + l) values are equal, the one with the lower n value has the lower energy. Consider the orbitals 1s, 2s, 3d, 4p, and 4f. Orbital 1s 2s 3d 4p 4f n 1 2 3 4 4 l 0 0 2 1 3 (n + l) 1 2 5 5 7 From the (n + l) rule, we say that of the five orbitals above, 1s has the lowest energy, followed by 2s, then 3d, 4p and 4f. Thus 1s orbital will be filled up first and 4f last. [Q3] For each pair of orbitals, determine which has the lower energy. 38 Bicol University (BS Food technology) College of Science LABORATORY ACTIVITIES DEPARTMENT OF CHEMISTRY Sem 1 2015-2016 1. 2s and 2p 2. 3d and 4f 3. 4d and 3p Chem 1–General Chemistry 4. 4p and 3d 5. 6s and 5d A memory device, which would aid you in determining the order by which orbitals are filled up, is given below. This might be a familiar figure to you since it is often printed in commercial periodic tables. You have to realize at this point that this memory device was constructed on the basis of the (n + l) rule. 1s 2s 3s 4s 5s 6s 7s 8s 2p 3p 4p 5p 6p 7p 8p 3d 4d 5d 6d 7d …. 4f 5f 6f 7f 5g …. …. For example: Writing the electron configuration of the sodium atom, 11 Na. The sodium atom, Na, is atom number 11. Therefore, it has 11 electrons. The first two will occupy the 1s orbital (an s orbital can accommodate a maximum of two electrons). The 2s orbital will be occupied next, followed by 2p (the 2p orbitals can accommodate a maximum of 6 electrons since there are three p orbitals, each orbital accommodating a maximum of 2 electrons). The remaining electron will occupy the 3s orbital. Thus the electron configuration of the 11Na atom is 1s2 2s2 2p6 3s1 [Q4] Write the electron configuration of the following atoms: Mg, P, V HUND'S RULE Consider the three 2p orbitals. These three orbitals have the same energy (same n + l value). The three 2p orbitals are DEGENERATE, that is, they have the same energy in the absence of a magnetic field. The same thing is true with the 3p orbitals, as well as the 4p, 5p, etc. The five 3d orbitals are degenerate, and so are the seven 4f orbitals. We have to address a problem here: how do we fill up degenerate orbitals? For instance, how do we place four electrons in the 2p orbitals? This question can be resolved using the Hund's Rule: Every orbital in a subshell is singly occupied with one electron before any orbital is doubly occupied, and all electrons in singly occupied orbitals have the same spin. Consider the carbon atom. It has 6 electrons. From our memory device, the first 2 electrons must occupy the 1s orbital, the next two electrons will occupy the 2s, and the last will occupy the 2p orbitals. Thus: 1s2 2s2 2p2. Since there are three p orbitals, the electron configuration can be written as 1s2 2s2 2p1 2p1 2p0. Note that there are two electrons in the 2p orbitals and that they occupy two different p orbitals singly. We can use the ORBITAL DIAGRAM to illustrate Hund's Rule. For the electrons in the C atom: 1s ↓ ↑ 2s ↓ ↑ 2p ↓ 2p ↓ 2p 38 Bicol University (BS Food technology) College of Science LABORATORY ACTIVITIES DEPARTMENT OF CHEMISTRY Sem 1 2015-2016 [Q5] Chem 1–General Chemistry Write the electron configuration of the vanadium atom illustrating Hund's Rule. ELECTRONIC CONFIGURATION OF IONS An ION results from a neutral atom that has gained (anion) or lost (cation) one or more of its electrons. The Na+ ion is a Na atom (which has 11 electrons) that has lost one electron from its 3s orbital. (Note: The electron lost is one in the highest energy level.). The F - ion is an F atom (which has 9 electrons) that has gained one electron. 11 9 Na 1s2 2s2 2p6 3s1 Neutral atom 11 p+ 11 e- F → 1s2 2s2 2p5 Neutral atom 9 p+ 9 e- → 9 F- Na+ 1s2 2s2 2p6 Cation: 11 p+ 10 e11 1s2 2s2 2p6 Anion: 9 p+ 10 e- Consider the ions Fe2+ and Fe3+. 26 Fe 1s2 2s2 2p6 3s2 3p6 4s2 3d6 or 1s2 2s2 2p6 3s2 3p6 4s2 3d2 3d1 3d1 3d1 1 3d It could lose the two electrons from the 4s orbital to give the Fe 2+ ion or lose the two electrons in the 4s orbital and one in the 3d orbital to give the Fe 3+ ion. [Q6] Consider the S2- ion. 1. How many electrons does it have? 2. Write its electron configuration. ISOELECTRONIC ATOMS AND IONS Complete the table below. Atom/Ion 1. 2. 3. 4. Number of Electrons Electron Configuration Na+ Ne S2Ar Note that Na+ and Ne has the same number of electrons and the same electron configuration. Na+ and Ne are said to be ISOELECRONIC. S2- and Ar are also isoelectronic. [Q7] Give two anions and two cations that are isoelectronic with Kr. VALENCE ELECTRONS AND VALENCE ELECTRON CONFIGURATION The electron configuration of Na (11 electrons) is 1s2 2s2 2p6 3s1. The CORE ELECTRONS in the Na atom are those that have the principal quantum numbers (n) 1 and 2, that is, the electrons in the 1s, 2s, and 2p orbitals. Na has 10 core electrons, with the following configuration: 1s2 2s2 2p6. There is only one electron in the orbital with the highest principal quantum number (highest n: highest main energy level), the 3s orbital. This is the VALENCE ELECTRON of the Na atom. It is the 38 Bicol University (BS Food technology) College of Science LABORATORY ACTIVITIES DEPARTMENT OF CHEMISTRY Sem 1 2015-2016 one involved in chemical bonding CONFIGURATION of the Na atom is 3s1. Chem 1–General Chemistry and chemical reactions. The VALENCE ELECTRON The S atom has the electron configuration: 1s2 2s2 2p6 3s2 3p4. S has 6 valence electrons, and its valence electron configuration is 3s2 3p4. [Q8] Examine the electron configuration of Mg and P in Q4. Determine the number of valence electrons and the valence electron configuration of each. Activity No. ____ Electronic Structure of the Atom Name ______________________________________ Course/Year/Section __________ Date ______________ Q1] Give the orbital symbol for the orbitals in the fourth main energy level. orbital symbol/ n l designation 4 [Q2] are. Based on the ml value for the d and f orbitals, determine how many types of d and f orbitals there l = 2 l = 3 [Q3] For each pair of orbitals, determine which has the lower energy. Orbital n l (n + l) Orbital n s s 2s 4p 2p 3d 3d 6s 4f 5d 4d l (n + l) 38 Bicol University (BS Food technology) College of Science LABORATORY ACTIVITIES DEPARTMENT OF CHEMISTRY Sem 1 2015-2016 Chem 1–General Chemistry 3p [Q4] Write the electron configuration of the following atoms: 12 Mg 15 P 23 V [Q5] Write the electron configuration of the vanadium atom illustrating Hund's Rule. 1s 2s 2p 2p 2p 3s 3p 3p 3p 4s 3d 3d 3d 3d [Q6] Consider the S2- ion. a) How many electrons does it have? no. of e2S b) Mg, P, V Write its electron configuration. electron configuration [Q7] Give two anions and two cations that are isoelectronic with Kr. Anions Cations [Q8] Examine the electron configuration of Mg and P in Q4. electrons and the valence electron configuration of each. Atom Number of Valence Electrons 3d Determine the number of valence Valence Electron Configuration Mg P EXERCISES ON QUANTUM NUMBERS Name ______________________________________ Course/Year/Section __________ Date ______________ A. Tell what type of orbital (such as 1s, 3d, and so on) is described by each of the following sets of quantum numbers. a. b. c. d. e. n n n n n =1, = 3, = 6, = 5, = 7, l=0 l=2 l=1 l=3 l=5 f. g. h. i. j. n n n n n = 2, = 3, =6, = 5, = 7, l=0 l=1 l=3 l=2 l=6 B. Tell how many electrons there can be, in a given atoms, with the following sets of quantum numbers. a. b. n=3 n = 2, l = 1 f. g. n=2 n = 3, l = 1 38 Bicol University (BS Food technology) College of Science LABORATORY ACTIVITIES DEPARTMENT OF CHEMISTRY Sem 1 2015-2016 c. d. +½ e. C. n = 4, l = 2, ml = 0 n = 3, l = 2, ml = 1, ms = n=7 Chem 1–General Chemistry h. i. —½ j. n = 3, l = 2, ml = 1 n = 4, l = 1, ml = 0, ms = n=8 Tell what is wrong with each of the following sets of quantum numbers. a. b. c. d. e. 2, 2, 0, +½ 3, 1, 2, ─½ 2, 0, 0, 0 3, ½, 0, ─½ ─3, 2, 0, +½ f. g. 3, 4, 2, ─½ 3, 2, 1, 0 h. i. j. 4, 2, 3, +½ 2, -1, 0, ─½ 3, 1, +½, ─½ 38