When comparing the first ionization energies of elements vithin the same period, it is noted that boron first ionization energy than beryllium. This is due to the greater shielding has a of the electrons from the subshell to an area closer to the nucleus, which is not possible for boron the electron present in the atom. As a result, this electron is penetration shielded, making it to remove. 2p harder beryllium smaller ineffectively easier effectively 25

When comparing the first ionization energies of elements vithin the same period, it is noted that boron first ionization energy than beryllium. This is due to the greater shielding has a of the electrons from the subshell to an area closer to the nucleus, which is not possible for boron the electron present in the atom. As a result, this electron is penetration shielded, making it to remove. 2p harder beryllium smaller ineffectively easier effectively 25

Chemistry

10th Edition

ISBN:9781305957404

Author:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste

Publisher:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste

Chapter1: Chemical Foundations

Section: Chapter Questions

Problem 1RQ: Define and explain the differences between the following terms. a. law and theory b. theory and...

See similar textbooks

Similar questions

A. If an element with the valence configuration 4s'3d10 loses 2 electron(s), these electron(s) would be removed from the |subshell(s). B. An element with the valence electron configuration 4s²4p* would form a monatomic ion with a charge of In order to form this ion, the element will| |electron(s) from/into the| | subshell(s).
Use noble gas notation to depict electron configurations for the following metals of the third transition series. (Express your answer using the noble gas notation. For example, the electron configuration of Li would be entered as [He]2s¹.) a. Tungsten, W. Tungsten is mainly used in the production of hard materials based on tungsten carbide. b. Gold, Au. Gold and its alloys are used in monetary exchange, jewelry, and investment.
Some patterns of electron configuration are listed below. In each case, [G] stands for a noble-gas core and n, m, or o stand for integer numbers like 1, 2, 3 and so on. Note: assume G is lighter than Rn. Think about the chemical elements made of atoms with an electron configuration that matches each pattern. In what block or blocks of the Periodic Table would these elements lie? Would they be metals, nonmetals, metalloids, some of each? Use your answers to these questions to complete the table. electron configuration of atoms 6 [G]ns²np [G]ns¹ 2 [G]ns²mf³ 3 blocks 00 P S 00 P 00 S location of elements (check all that apply) S C р р f р f metals or nonmetals? 000 000 000 metals nonmetals metalloids metals nonmetals metalloids metals nonmetals metalloids
D. Give the quantum numbers for the last electron of each element given in Part II following the table below. Element 1 subshell N - 7 1s22s22p3 70 Ga,17⁰ Ar 1840 Sb,1122 112 AU79197 n main energy level 2 s=0 p=1 d=2 f=3 1 m₁ orbital -1 to 0 to +1 -1, 0, +1 +1 m₂ magnetic spin (+1/2, -1/2) +1/2
In Universe L, recently discovered by an intrepid team of chemists who also happen to have studied interdimensional travel, quantum mechanics works just as it does in our universe, except that there are four d orbitals instead of the usual number we observe here. Use these facts to write the ground-state electron configurations of the seventh and eighth elements in the first transition series in Universe L. Note: you may use [X] to stand for the electron configuration of the noble gas at the end of the row before the first transition series. seventh transition metal: eighth transition metal:
Answer number 1 to 5
Part A: How does ionization energy change down a group? increases from left to right decreases from left to right increases from top to bottom decreases from top to bottom Part B: Why is Ionization energy smaller for larger atoms? Sheilding. The nucleus cannot hold onto the electrons as well and loses them easier Shielding. The nucleus holds onto the electrons stronger and does not lose them easily There is no explanation Part C: According to the trend, which of these 4 elements has the largest ionization energy? Be (atomic# 4) Sr (atomic# 38) F (atomic# 9) I (atomic# 53) Part D: According to the trend, which of these 4 elements has the smallest ionization energy? Be (atomic# 4) Sr (atomic# 38) F (atomic# 9)…