Iodine has 53 electrons and a configuration of 2, 8, 18, 18, 7. Why does the fourth shell not reach maximum and therefore it picks up 7 electrons and begins a new shell? Write the noble gas electron configuration for tellurium. Step 1: Write the electron configuration of the atom in the following form: (1s) (2s, 2p) (3s, 3p) (3d) (4s, 4p) (4d) (4f) (5s, 5p) . . . Step 2: Identify the electron of interest, and ignore all electrons in higher groups (to the right in the list from Step 1). These do not shield electrons in lower groups. Step 3: Slater's Rules is now broken Give the electron configuration of iodine. In the formation of an ion, what is the expected behavior of an atom with the following electron configuration: 1s^2 2s^2 2p^6 3s^2 3p^6 4s^1 a) gain 1 electron b) lose 1 electron c) lose 2 electrons d) gain 2 electrons e) no correct response; Write the electron configuration of Ni. What element is represented by the electron configuration 1 s 2 2 s 2 2 p 6 3 s 2 3 p 6 4 s 2 3 d 10 4 p 6 5 s 2 4 d 10 5 p 2 ? An element has the electron configuration [Kr] 5s24d105p2. The element is: a. nonmetal b. transition element c. metal d. lanthanide e. actinide; What element is best described by the electron configuration 1s22s22p63s23p4? 2 Answers. Iodine is a halogen that is included in the group of halogens in the row 5 of the periodic table. The atomic number of iodine is 53 which shows that there are 53 protons and electrons in an iodine atom. When iodine gets an electron it is converted into Iodide and it's electrons get 54. The electronic configuration of iodine is, In writing the electron configuration for Magnesium the first two electrons will go in the 1s orbital. Since 1s can only hold two electrons the next 2 electrons for magnesium go in the 2s orbital. The nex six electrons will go in the 2p orbital. The p orbital can hold up to six electrons. We'll put six in the 2p orbital and then put the PIMuDo. 7. Move to the 3d orbital, which can hold a maximum of 10 electrons. Fill it with 10 electrons. 8. Next, move to the 4p orbital and fill it with 6 electrons. 9. Finally, move to the 5s orbital and fill it with 2 electrons. In total, iodine has 53 electrons, and its electron configuration is 1s2 2s2 2p6 3s2 3p6 4s2 3d10 4p6 5s2 4d10 5p5. I hope For example, the orbitals 2p 2 p and 3s 3 s do have the same value of n + l n + l. For 2p 2 p, the sum n + l n + l is 2 + 1 = 3 2 + 1 = 3. For 3s 3 s, the sum n + l n + l is 3 + 0 = 3 3 + 0 = 3. So, with your criteria, the 3s 3 s orbital should be more stable than the 2p 2 p, and should be filled before the 2p 2 p. It is not the case. For instance, to find the electron configuration for an oxygen atom, we start by recognizing that the atomic number of oxygen is 8, which implies that the oxygen atom has eight protons in its nucleus and eight electrons. The first two electrons occupy the 1s orbital, the next two fill the 2s, and the next four fill the 2p orbitals, leading to There is a 2-8-8 rule for these elements. The first shell is filled with 2 electrons, the second is filled with 8 electrons, and the third is filled with 8. You can see that sodium (Na) and magnesium (Mg) have a couple of extra electrons. They, like all atoms, want to be happy. They have two possibilities: they can try to get to eight electrons Step 1: Write the electron configuration of the atom in the following form: (1s) (2s, 2p) (3s, 3p) (3d) (4s, 4p) (4d) (4f) (5s, 5p) . . . Step 2: Identify the electron of interest, and ignore all electrons in higher groups (to the right in the list from Step 1). These do not shield electrons in lower groups. Step 3: Slater's Rules is now broken 8: 18: 18: 6: 53 I iodine : [Kr] 4d 10 5s 2 5p 5; 1s 2: 2s 2: 2p 6: 2003; Section 1, Basic Constants, Units, and Conversion Factors; Electron Configuration of

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