Question

electron configuration practice worksheet
in the space below, write the unabbreviated electron configurations of the following elements:

1. sodium
2. iron
3. bromine
4. barium
5. neptunium

in the space below, write the abbreviated electron configurations of the following elements:

1. cobalt
2. silver
3. tellurium
4. radium
5. lawrencium

determine what elements are denoted by the following electron configurations:

1. 1s²2s²2p⁶3s²3p⁴
2. 1s²2s²2p⁶3s²3p⁶4s²3d¹⁰4p⁶5s¹
3. kr 5s²4d¹⁰5p³
4. xe 6s²4f¹⁴5d⁸
5. rn 7s²5f¹¹

determine which of the following electron configurations are not valid:

1. 1s²2s²2p⁶3s²3p⁶4s²4d¹⁰4p⁵
2. 1s²2s²2p⁶3s²3d⁵
3. ra 7s²5f⁸
4. kr 5s²4d¹⁰5p⁶
5. xe

Explanation:

Step1: Recall electron - filling rules

Use the Aufbau principle, Pauli - exclusion principle, and Hund's rule to write electron configurations.

Step2: Determine atomic numbers

Find the atomic numbers of the elements from the periodic table. The atomic number is equal to the number of electrons for a neutral atom.

Step3: Write unabbreviated electron configurations

Sodium (Na), atomic number = 11

$1s^{2}2s^{2}2p^{6}3s^{1}$

Iron (Fe), atomic number = 26

$1s^{2}2s^{2}2p^{6}3s^{2}3p^{6}4s^{2}3d^{6}$

Bromine (Br), atomic number = 35

$1s^{2}2s^{2}2p^{6}3s^{2}3p^{6}4s^{2}3d^{10}4p^{5}$

Barium (Ba), atomic number = 56

$1s^{2}2s^{2}2p^{6}3s^{2}3p^{6}4s^{2}3d^{10}4p^{6}5s^{2}4d^{10}5p^{6}6s^{2}$

Neptunium (Np), atomic number = 93

$1s^{2}2s^{2}2p^{6}3s^{2}3p^{6}4s^{2}3d^{10}4p^{6}5s^{2}4d^{10}5p^{6}6s^{2}4f^{14}5d^{10}6p^{6}7s^{2}5f^{4}6d^{1}$

Step4: Write abbreviated electron configurations

Cobalt (Co), atomic number = 27

$[Ar]4s^{2}3d^{7}$

Silver (Ag), atomic number = 47

$[Kr]5s^{1}4d^{10}$

Tellurium (Te), atomic number = 52

$[Kr]5s^{2}4d^{10}5p^{4}$

Radium (Ra), atomic number = 88

$[Rn]7s^{2}$

Lawrencium (Lr), atomic number = 103

$[Rn]7s^{2}5f^{14}6d^{1}$

Step5: Identify elements from electron configurations

$1s^{2}2s^{2}2p^{6}3s^{2}3p^{4}$

The total number of electrons is 2 + 2+6 + 2+4=16. The element is sulfur (S).

$1s^{2}2s^{2}2p^{6}3s^{2}3p^{6}4s^{2}3d^{10}4p^{6}5s^{1}$

The total number of electrons is 2+2 + 6+2+6+2+10+6 + 1=37. The element is rubidium (Rb).

$[Kr]5s^{2}4d^{10}5p^{3}$

Kr has 36 electrons. Adding 2 + 10+3 = 15 more electrons gives 36+15 = 51. The element is antimony (Sb).

$[Xe]6s^{2}4f^{14}5d^{8}$

Xe has 54 electrons. Adding 2+14 + 8=24 more electrons gives 54+24 = 78. The element is platinum (Pt).

$[Rn]7s^{2}5f^{11}$

Rn has 86 electrons. Adding 2+11 = 13 more electrons gives 86+13 = 99. The element is einsteinium (Es).

Step6: Check validity of electron configurations

$1s^{2}2s^{2}2p^{6}3s^{2}3p^{6}4s^{2}4d^{10}4p^{5}$

This is valid.

$1s^{2}2s^{2}2p^{6}3s^{2}3d^{5}$

This is not valid. After filling 3s, the next sub - shell to fill is 3p, not 3d.

$[Ra]7s^{2}5f^{8}$

Ra is not a noble gas, so this is not a valid abbreviated form.

$[Kr]5s^{2}4d^{10}5p^{6}$

This is valid.

$[Xe]$

This is valid as an abbreviated form for elements starting from Cs.

Answer:

1. Sodium: $1s^{2}2s^{2}2p^{6}3s^{1}$
2. Iron: $1s^{2}2s^{2}2p^{6}3s^{2}3p^{6}4s^{2}3d^{6}$
3. Bromine: $1s^{2}2s^{2}2p^{6}3s^{2}3p^{6}4s^{2}3d^{10}4p^{5}$
4. Barium: $1s^{2}2s^{2}2p^{6}3s^{2}3p^{6}4s^{2}3d^{10}4p^{6}5s^{2}4d^{10}5p^{6}6s^{2}$
5. Neptunium: $1s^{2}2s^{2}2p^{6}3s^{2}3p^{6}4s^{2}3d^{10}4p^{6}5s^{2}4d^{10}5p^{6}6s^{2}4f^{14}5d^{10}6p^{6}7s^{2}5f^{4}6d^{1}$
6. Cobalt: $[Ar]4s^{2}3d^{7}$
7. Silver: $[Kr]5s^{1}4d^{10}$
8. Tellurium: $[Kr]5s^{2}4d^{10}5p^{4}$
9. Radium: $[Rn]7s^{2}$
10. Lawrencium: $[Rn]7s^{2}5f^{14}6d^{1}$
11. Sulfur (S)
12. Rubidium (Rb)
13. Antimony (Sb)
14. Platinum (Pt)
15. Einsteinium (Es)
16. Valid
17. Not valid
18. Not valid
19. Valid
20. Valid