electronic configuration practice- use periodic table with electron configuration table\nin the space below…

electronic configuration practice- use periodic table with electron configuration table\nin the space below, write the unabbreviated electron configurations of the following elements:\n1. sodium\n2. magnesium\n3. iron\n4. potassium\n5. selenium\nin the space below, write the abbreviated electron configurations of the following elements:\n6. cobalt\n7. silver\n8. tellurium\n9. radium\n10.iodine\ndetermine what elements are denoted by the following electron configurations:\n11.1s² 2s² 2p⁶ 3s² 3p⁴\n12.1s² 2s² 2p⁶ 3s² 3p⁶ 4s² 3d¹⁰4p⁶ 5s¹\n13.kr 5s² 4d¹⁰5p³\nexplain what is wrong with the following electron configurations:\n14.1s² 2s² 2p⁶ 3s² 3p⁶ 4s² 4d¹⁰4p⁶\n15.1s² 2s² 2p⁶ 3s³ 3d⁵\nidentify the element and determine the valence electrons.\n| electron configuration | element symbol | valence electrons |\n| 1s² 2s² 2p⁶ 3s² 3p⁶ 4s² 3d¹⁰4p⁶ 5s¹ | | |\n| 1s²2s²2p⁶3s²3p⁶4s²3d⁶ | | |\n| 1s²2s²2p⁶3s²3p⁶4s²3d¹⁰4p⁵ | | |\n| 1s²2s²2p⁶3s²3p⁶4s²3d¹⁰4p⁵ | | |\n| 1s² 2s² 2p⁶ 3s² 3p⁵ | | |
Answer
Explanation:
Step1: Recall electron - filling rules
Electrons fill orbitals in order of increasing energy (Aufbau principle), with s - orbitals holding a maximum of 2 electrons, p - orbitals holding 6, d - orbitals holding 10, and f - orbitals holding 14.
Step2: Unabbreviated electron configurations
Sodium (Na, atomic number = 11)
$1s^{2}2s^{2}2p^{6}3s^{1}$
Magnesium (Mg, atomic number = 12)
$1s^{2}2s^{2}2p^{6}3s^{2}$
Iron (Fe, atomic number = 26)
$1s^{2}2s^{2}2p^{6}3s^{2}3p^{6}4s^{2}3d^{6}$
Potassium (K, atomic number = 19)
$1s^{2}2s^{2}2p^{6}3s^{2}3p^{6}4s^{1}$
Selenium (Se, atomic number = 34)
$1s^{2}2s^{2}2p^{6}3s^{2}3p^{6}4s^{2}3d^{10}4p^{4}$
Step3: Abbreviated electron configurations
Cobalt (Co, atomic number = 27)
$[Ar]4s^{2}3d^{7}$
Silver (Ag, atomic number = 47)
$[Kr]4d^{10}5s^{1}$
Tellurium (Te, atomic number = 52)
$[Kr]5s^{2}4d^{10}5p^{4}$
Radium (Ra, atomic number = 88)
$[Rn]7s^{2}$
Iodine (I, atomic number = 53)
$[Kr]5s^{2}4d^{10}5p^{5}$
Step4: Identify elements from electron configurations
$1s^{2}2s^{2}2p^{6}3s^{2}3p^{4}$
Sulfur (S) since the total number of electrons is 16.
$1s^{2}2s^{2}2p^{6}3s^{2}3p^{6}4s^{2}3d^{10}4p^{6}5s^{1}$
Rubidium (Rb) as the total number of electrons is 37.
$[Kr]5s^{2}4d^{10}5p^{3}$
Antimony (Sb) as adding the electrons from $[Kr]$ (36) and the additional 15 electrons gives 51 electrons.
Step5: Find errors in electron configurations
$1s^{2}2s^{2}2p^{6}3s^{2}3p^{6}4s^{2}4d^{10}4p^{6}$
The 4d - subshell should not fill before the 5s - subshell. The correct order is $1s^{2}2s^{2}2p^{6}3s^{2}3p^{6}4s^{2}3d^{10}4p^{6}5s^{2}4d^{10}$
$1s^{2}2s^{2}2p^{6}3s^{3}3d^{5}$
The 3s - orbital can only hold 2 electrons, so it should be $1s^{2}2s^{2}2p^{6}3s^{2}3p^{6}4s^{1}3d^{5}$ (following Hund's rule and Aufbau principle).
Step6: Identify elements and valence electrons
$1s^{2}2s^{2}2p^{6}3s^{2}3p^{6}4s^{2}3d^{10}4p^{6}5s^{1}$
Element: Rubidium (Rb), Valence electrons: 1
$1s^{2}2s^{2}2p^{6}3s^{2}3p^{6}4s^{2}3d^{6}$
Element: Iron (Fe), Valence electrons: 8 (2 in 4s and 6 in 3d)
$1s^{2}2s^{2}2p^{6}3s^{2}3p^{6}4s^{2}3d^{10}4p^{5}$
Element: Bromine (Br), Valence electrons: 7
$1s^{2}2s^{2}2p^{6}3s^{2}3p^{6}4s^{2}3d^{10}4p^{5}$
Element: Bromine (Br), Valence electrons: 7
$1s^{2}2s^{2}2p^{6}3s^{2}3p^{5}$
Element: Chlorine (Cl), Valence electrons: 7
Answer:
- $1s^{2}2s^{2}2p^{6}3s^{1}$
- $1s^{2}2s^{2}2p^{6}3s^{2}$
- $1s^{2}2s^{2}2p^{6}3s^{2}3p^{6}4s^{2}3d^{6}$
- $1s^{2}2s^{2}2p^{6}3s^{2}3p^{6}4s^{1}$
- $1s^{2}2s^{2}2p^{6}3s^{2}3p^{6}4s^{2}3d^{10}4p^{4}$
- $[Ar]4s^{2}3d^{7}$
- $[Kr]4d^{10}5s^{1}$
- $[Kr]5s^{2}4d^{10}5p^{4}$
- $[Rn]7s^{2}$
- $[Kr]5s^{2}4d^{10}5p^{5}$
- Sulfur (S)
- Rubidium (Rb)
- Antimony (Sb)
- 4d - subshell filled before 5s - subshell; correct is $1s^{2}2s^{2}2p^{6}3s^{2}3p^{6}4s^{2}3d^{10}4p^{6}5s^{2}4d^{10}$
- 3s - orbital has 3 electrons instead of 2; correct is $1s^{2}2s^{2}2p^{6}3s^{2}3p^{6}4s^{1}3d^{5}$
- Rubidium (Rb), 1
- Iron (Fe), 8
- Bromine (Br), 7
- Bromine (Br), 7
- Chlorine (Cl), 7