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electronic configuration practice- use periodic table with electron con…

Question

electronic configuration practice- use periodic table with electron configuration table
in the space below, write the unabbreviated electron configurations of the following elements:

  1. sodium
  2. magnesium
  3. iron
  4. potassium
  5. selenium

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

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

determine what elements are denoted by the following electron configurations:
11.1s² 2s² 2p⁶ 3s² 3p⁴
12.1s² 2s² 2p⁶ 3s² 3p⁶ 4s² 3d¹⁰4p⁶ 5s¹
13.kr 5s² 4d¹⁰5p¹
explain what is wrong with the following electron configurations:

Explanation:

Step1: Recall electron - filling rules

Use the Aufbau principle, Pauli - exclusion principle, and Hund's rule.

Step2: Determine number of electrons

For neutral atoms, the number of electrons is equal to the atomic number.

Step3: Write 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)

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

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)

$[Rn]7s^{2}$

Iodine (I, atomic number = 53)

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

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 $2+2 + 6+2+6+2+10+6 + 1=39$. The element is yttrium (Y).

$[Kr]5s^{2}4d^{10}5p^{1}$

The number of electrons from $[Kr]$ is 36, and adding $2 + 10+1=13$ more electrons gives a total of 49. The element is indium (In).

Answer:

  1. $1s^{2}2s^{2}2p^{6}3s^{1}$
  2. $1s^{2}2s^{2}2p^{6}3s^{2}$
  3. $1s^{2}2s^{2}2p^{6}3s^{2}3p^{6}4s^{2}3d^{6}$
  4. $1s^{2}2s^{2}2p^{6}3s^{2}3p^{6}4s^{1}$
  5. $1s^{2}2s^{2}2p^{6}3s^{2}3p^{6}4s^{2}3d^{10}4p^{4}$
  6. $[Ar]4s^{2}3d^{7}$
  7. $[Kr]5s^{1}4d^{10}$
  8. $[Kr]5s^{2}4d^{10}5p^{4}$
  9. $[Rn]7s^{2}$
  10. $[Kr]5s^{2}4d^{10}5p^{5}$
  11. Sulfur (S)
  12. Yttrium (Y)
  13. Indium (In)