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Interpretation:
3s, 3p and 3d orbitals which have the same energy in a hydrogen atom but different energies in a many-electron atom should be explained by using the concept of quantum numbers and Aufbau principle.
Concept Introduction:
Energies of orbitals in the hydrogen atom
The energies of orbitals in the hydrogen atom depend only on the value of the principal quantum number (n). When n increases, energy also increases. For this reason, orbitals in the same shell have the same energy in spite of their subshell. The increasing order of energy of hydrogen orbitals is
1s < 2s = 2p < 3s = 3p = 3d < 4s = 4p = 4d = 4f
In the case of one 2s and three 2p orbitals in the second shell, they have the same energy. In the third shell, all nine orbitals (one 3s, three 3p and five 3d) have the same energy. All sixteen orbitals (one 4s, three 4p, five 4d and seven 4f) in the fourth shell have the same energy.
Figure 1
The energy levels of the different orbitals in hydrogen atom are easily explained by considering the given diagram. Here, each box represents one orbital. Orbitals with the same principal quantum number (n) have the same energy.
Energies of orbitals in a many electron atom
Energy of an orbital in a many electron atom depends on both the values of principle quantum number (n) and
Principal Quantum Number (n)
The principal quantum number (n) assigns the size of the orbital and specifies the energy of an electron. If the value of n is larger, then the average distance of an electron in the orbital from the nucleus will be greater. Therefore the size of the orbital is large. The principal quantum numbers have the integral values of 1, 2, 3 and so forth and it corresponds to the quantum number in
Angular Momentum Quantum Number (l)
The angular momentum quantum number (l) explains the shape of the atomic orbital. The values of l are integers which depend on the value of the principal quantum number, n. For a given value of n, the possible values of l range are from 0 to n − 1. If n = 1, there is only one possible value of l (l=0). If n = 2, there are two values of l: 0 and 1. If n = 3, there are three values of l: 0, 1, and 2. The value of l is selected by the letters s, p, d, and f. If l = 0, we have an s orbital; if l = 1, we have a p orbital; if l = 2, we have a d orbital and finally if l = 3, we have a f orbital. A collection of orbitals with the same value of n is called a shell. One or more orbitals with the same n and l values are referred to a subshell (sublevel). The value of l also has a slight effect on the energy of the subshell; the energy of the subshell increases with l (s < p < d < f).
Aufbau principle
In the ground state of an atom, an electron enters the orbital with lowest energy first and subsequent electrons are fed in the order of increasing energies. This is called Aufbau principle. The word 'aufbau' in German means 'building up'. Here, it refers to the filling up of orbitals with electrons. Aufbau principle explains the theoretical, sequential building up of the stepwise addition of protons and electrons for the elements in the periodic table.
Figure 2
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Chapter 7 Solutions
General Chemistry
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- Bookmarks Profiles Tab Window Help Chemical Formula - Aktiv Che X + → C 11 a app.aktiv.com Google Chrome isn't your default browser Set as default Question 12 of 16 Q Fri Feb 2 Verify it's you New Chrome availabl- Write the balanced molecular chemical equation for the reaction in aqueous solution for mercury(I) nitrate and chromium(VI) sulfate. If no reaction occurs, simply write only NR. Be sure to include the proper phases for all species within the reaction. 3 Hg(NO3)2(aq) + Cг2(SO4)3(aq) → 3 Hg₂SO (s) + 2 Cr(NO3), (aq) ean Ui mate co ence an climate bility inc ulnerabili women, main critic CLIMATE-INI ernational + 10 O 2 W FEB 1 + 4- 3- 2- 2 2 ( 3 4 NS 28 2 ty 56 + 2+ 3+ 4+ 7 8 9 0 5 (s) (1) Ch O 8 9 (g) (aq) Hg NR CI Cr x H₂O A 80 Q A DII A F2 F3 FA F5 F6 F7 F8 F9 #3 EA $ do 50 % 6 CO & 7 E R T Y U 8 ( 9 0 F10 34 F11 川 F12 Subr + delete 0 { P }arrow_forwardDeducing the reactants of a Diels-Alder reaction n the molecule on the right-hand side of this organic reaction be made in good yield from no more than two reactants, in one step, by moderately heating the reactants? ? Δ • If your answer is yes, then draw the reactant or reactants in the drawing area below. You can draw the reactants in any arrangement you like. • If your answer is no, check the box under the drawing area instead. Explanation Check Click and drag to start drawing a structure. >arrow_forwardPredict the major products of the following organic reaction: + Some important notes: A ? • Draw the major product, or products, of the reaction in the drawing area below. • If there aren't any products, because no reaction will take place, check the box below the drawing area instead. • Be sure to use wedge and dash bonds when necessary, for example to distinguish between major products that are enantiomers. Explanation Check Click and drag to start drawing a structure.arrow_forward
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