Show In Atomic Structure we need to be familiar with a few ionisation energy trends. In this video let's discuss the first ionisation energy trend across Period 3. First ionisation energy is the energy required to remove 1 mole of electrons from 1 mole of gaseous atoms according to the following equation: M(g) -> M+(g) + e- The first ionisation energy (IE) trend across Period 3 is given in the sketch below.
There are 3 observations: 1. General increase in first IEThe concept used to explain the general trend is involving effective nuclear charge. Across the period proton number increases hence nuclear charge increases. Electrons are added to the same principal quantum shell so shielding effect or screening effect stays constant. Effective nuclear charge increases hence attraction between nucleus and valence electron is stronger. More energy is required to remove the valence electron hence first IE increases across Period 3.
2. Slight decrease in IE from Mg to AlThe first anomaly is the decrease in IE from Mg to Al. We need to write down the electron configuration for Mg and Al and compare where the electrons are removed.
Electron is removed from 3s subshell for Mg while electron is removed from 3p subshell for Al, which is further away from the nucleus or has a higher energy level. The attraction between nucleus and that electron is weaker hence less energy is required to remove it. Therefore Al has a lower first IE. 3. Slight decrease in IE from P to SThe second anomaly between P and S requires us to draw the electron in box diagram as electrons are removed from 3p subshell for both elements.
For sulfur, electron is removed from a fully filled 3p orbital hence it experiences interelectronic repulsion between the electron pair in the same 3p orbital. This makes it easier to remove that electron, therefore S has a lower first IE. For the detailed step-by-step discussion on the first ionisation energy trend for Period 3 elements, check out this video! Topic: Atomic Structure, Physical Chemistry, A Level Chemistry, Singapore Back to other previous Physical Chemistry Video Lessons. Found this A Level Chemistry video useful? Please LIKE this video and SHARE it with your friends! Join my 2000+ subscribers on my YouTube Channel for new A Level Chemistry video lessons every week. Check out other A Level Chemistry Video Lessons here! Need an experienced tutor to make Chemistry simpler for you? Do consider signing up for my A Level H2 Chemistry Tuition classes at Bishan or online tuition classes! The ionization energy is the quantity of energy that an isolated, gaseous atom in the ground electronic state must absorb to discharge an electron, resulting in a cation. \[\ce{ H(g) \rightarrow H^{+}(g) + e^{-}} \] This energy is usually expressed in kJ/mol, or the amount of energy it takes for all the atoms in a mole to lose one electron each. When considering an initially neutral atom, expelling the first electron will require less energy than expelling the second, the second will require less energy than the third, and so on. Each successive electron requires more energy to be released. This is because after the first electron is lost, the overall charge of the atom becomes positive, and the negative forces of the electron will be attracted to the positive charge of the newly formed ion. The more electrons that are lost, the more positive this ion will be, the harder it is to separate the electrons from the atom. In general, the further away an electron is from the nucleus, the easier it is for it to be expelled. In other words, ionization energy is a function of atomic radius; the larger the radius, the smaller the amount of energy required to remove the electron from the outer most orbital. For example, it would be far easier to take electrons away from the larger element of Ca (Calcium) than it would be from one where the electrons are held tighter to the nucleus, like Cl (Chlorine). In a chemical reaction, understanding ionization energy is important in order to understand the behavior of whether various atoms make covalent or ionic bonds with each other. For instance, the ionization energy of Sodium (alkali metal) is 496KJ/mol (1) whereas Chlorine's first ionization energy is 1251.1 KJ/mol (2). Due to this difference in their ionization energy, when they chemically combine they make an ionic bond. Elements that reside close to each other in the periodic table or elements that do not have much of a difference in ionization energy make polar covalent or covalent bonds. For example, carbon and oxygen make CO2 (Carbon dioxide) reside close to each other on a periodic table; they, therefore, form a covalent bond. Carbon and chlorine make CCl4 (Carbon tetrachloride) another molecule that is covalently bonded.
As described above, ionization energies are dependent upon the atomic radius. Since going from right to left on the periodic table, the atomic radius increases, and the ionization energy increases from left to right in the periods and up the groups. Exceptions to this trend is observed for alkaline earth metals (group 2) and nitrogen group elements (group 15). Typically, group 2 elements have ionization energy greater than group 13 elements and group 15 elements have greater ionization energy than group 16 elements. Groups 2 and 15 have completely and half-filled electronic configuration respectively, thus, it requires more energy to remove an electron from completely filled orbitals than incompletely filled orbitals. Alkali metals (IA group) have small ionization energies, especially when compared to halogens or VII A group (see diagram 1). In addition to the radius (distance between nucleus and the electrons in outermost orbital), the number of electrons between the nucleus and the electron(s) you're looking at in the outermost shell have an effect on the ionization energy as well. This effect, where the full positive charge of the nucleus is not felt by outer electrons due to the negative charges of inner electrons partially canceling out the positive charge, is called shielding. The more electrons shielding the outer electron shell from the nucleus, the less energy required to expel an electron from said atom. The higher the shielding effect the lower the ionization energy (see diagram 2). It is because of the shielding effect that the ionization energy decreases from top to bottom within a group. From this trend, Cesium is said to have the lowest ionization energy and Fluorine is said to have the highest ionization energy (with the exception of Helium and Neon).
The symbol \(I_1\) stands for the first ionization energy (energy required to take away an electron from a neutral atom) and the symbol \(I_2\) stands for the second ionization energy (energy required to take away an electron from an atom with a +1 charge. Each succeeding ionization energy is larger than the preceding energy. This means that \(I_1 < I_2 < I_3 < ... < I_n\) will always be true. Example of how ionization energy increases as succeeding electrons are taken away. \[ Mg \,(g) \rightarrow Mg^+\,(g) + e^- \;\;\; I_1= 738\, kJ/mol \] \[Mg^+ \,(g) \rightarrow Mg^{2+}\, (g) + e^- \;\;\; I_2= 1451\, kJ/mol\] See first, second, and third ionization energies of elements/ions in Table 3.
Electron orbitals are separated into various shells which have strong impacts on the ionization energies of the various electrons. For instance, let us look at aluminum. Aluminum is the first element of its period with electrons in the 3p shell. This makes the first ionization energy comparably low to the other elements in the same period, because it only has to get rid of one electron to make a stable 3s shell, the new valence electron shell. However, once you've moved past the first ionization energy into the second ionization energy, there is a large jump in the amount of energy required to expel another electron. This is because you now are trying to take an electron from a fairly stable and full 3s electron shell. Electron shells are also responsible for the shielding that was explained above.
Both ionization energy and electron affinity have similar trend in the periodic table. For example, just as ionization energy increases along the periods, electron affinity also increases. Likewise, electron affinity decreases from top to bottom due to the same factor, i.e., shielding effect. Halogens can capture an electron easily as compared to elements in the first and second group. This tendency to capture an electron in a gaseous state is termed as electronegativity. This tendency also determines one of the chemical differences between Non metallic and metallic elements. Diagram 3: showing increasing trend of electron affinity from left to right (9).
Diagram 4: showing decreasing pattern of electron affinities of elements from top to bottom (9)
As indicated above, the elements to the right side of periodic table (diagram 3) have tendency to receive the electron while the one at the left are more electropositive. Also, from left to right, the metallic characteristics of elements decrease (4).
The difference of electronegativity or ionization energies between two reacting elements determine the fate of the type of bond. For example, there is a big difference of ionization energies and electronegativity between Na and. Cl. Therefore, sodium completely removes the electron from its outermost orbital and chlorine completely accepts the electron, and as a result we have an ionic bond (4). However, in cases where there is no difference in electronegativity, the sharing of electrons produces a covalent bond. For example, electronegativity of Hydrogen is 2.1 and the combination of two Hydrogen atoms will definitely make a covalent bond (by sharing of electrons). The combination of Hydrogen and Fluorine (electronegativity=3.96) will produce a polar covalent bond because they have small differences between electronegativity (5).
1) By looking at following electronic configuration of elements, can you predict which element has the lowest first ionization energy?
2) The ionization energy of \(Na^{+3}\) ion is one of the following (7)
3) Ionization energies and electron affinities are
4) Ionization energy is the ability to capture an electron: 5) The second ionization energy of Mg is greater than second ionization energy of Al: 6) Which group would generally have the lowest first ionization energy?
7) Sulfur has a first ionization energy of 999.6 kJ/mol. Rubidium has a first ionization energy of 403 kJ/mol. What bond do they form when chemically combined?
8) Ionization energy, when supplied to an atom, results in a(n)
9) Low first ionization energy is considered a property of 10) Gallium has a first ionization energy of 578.8 kJ/mol, and calcium has a first ionization energy of 589.8 kJ/mol . According to periodic trends, one would assume that calcium, being to the left of gallium, would have the lower ionization energy. Explain, in terms of orbitals, why these numbers make sense.
1) Element D, 2) D, 3) B, 4) A, 5) B, 6) B, 7) C, 8) C, 9) A 10) Gallium has one electron in the 4p orbital, which can be expelled to reveal a more stable and full 4s orbital. Calcium, however, has a fully stable 4s orbital as its valence orbital, which you would have to disrupt to take an electron away from. References
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Why does first ionisation energy decrease?
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