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1. the number of sublevels in a principal energy level (increases or de…

Question

  1. the number of sublevels in a principal energy level (increases or decreases) as n increases
  2. according to the pauli exclusion principle, a given orbital can contain only ______ electrons.
  3. to what element does each of the following electron configurations correspond?

a. 1s² 2s² 2p⁶ 3s² 3p²
b. 1s² 2s²
c. 1s² 2s² 2p⁶
d. 1s² 2s² 2p⁶ 3s² 3p⁵
e. 1s² 2s² 2p⁶ 3s² 3p²4s¹ 3d¹⁰

  1. write the full electron configuration for each of the following elements.

a. phosphorus, z = 15
b. calcium, z = 20
c. potassium, z = 19
d. boron, z = 5
e. manganese, z = 25

  1. to what element does each of the following electron configurations correspond?

a. 1s² 2s² 2p⁶ 3s² 3p⁶4s² 3d¹⁰ 4p⁴
b. 1s² 2s² 2p⁶ 3s² 3p⁶4s² 3d¹
c. 1s² 2s² 2p⁶ 3s² 3p⁴
d. 1s² 2s² 2p⁶ 3s² 3p⁶4s² 3d¹⁰ 4p⁶5s² 4d¹⁰ 5p⁵
e. 1s² 2s² 2p⁶ 3s² 3p⁶4s² 3d¹⁰ 4p⁶5s² 4d⁵

  1. write the complete orbital box diagrams for each of the following elements, using orbital boxes to represent orbitals and arrows to represent electrons.

a. aluminum, z = 13
b. phosphorus, z = 15
c. bromine, z = 35
d. argon, z = 18

Explanation:

Step1: Recall energy - level sublevel relationship

The number of sublevels in a principal energy level is equal to n. So as n increases, the number of sublevels increases.

Step2: Apply Pauli - exclusion principle

The Pauli exclusion principle states that a given orbital can contain only 2 electrons with opposite spins.

Step3: Determine elements from electron configurations (a)

Count the total number of electrons in \(1s^{2}2s^{2}2p^{6}3s^{2}3p^{2}\). \(2 + 2+6 + 2+2=14\). The element with 14 electrons is silicon (Si).

Step4: Determine elements from electron configurations (b)

For \(1s^{2}2s^{2}\), \(2 + 2 = 4\). The element with 4 electrons is beryllium (Be).

Step5: Determine elements from electron configurations (c)

In \(1s^{2}2s^{2}2p^{6}\), \(2+2 + 6=10\). The element with 10 electrons is neon (Ne).

Step6: Determine elements from electron configurations (d)

For \(1s^{2}2s^{2}2p^{6}3s^{2}3p^{5}\), \(2+2 + 6+2 + 5=17\). The element with 17 electrons is chlorine (Cl).

Step7: Determine elements from electron configurations (e)

In \(1s^{2}2s^{2}2p^{6}3s^{2}3p^{2}4s^{1}3d^{10}\), \(2+2 + 6+2+2 + 1+10 = 25\). The element with 25 electrons is manganese (Mn).

Step8: Write electron - configurations for elements (a)

For phosphorus (\(Z = 15\)), the electron - configuration is \(1s^{2}2s^{2}2p^{6}3s^{2}3p^{3}\).

Step9: Write electron - configurations for elements (b)

For calcium (\(Z = 20\)), the electron - configuration is \(1s^{2}2s^{2}2p^{6}3s^{2}3p^{6}4s^{2}\).

Step10: Write electron - configurations for elements (c)

For potassium (\(Z = 19\)), the electron - configuration is \(1s^{2}2s^{2}2p^{6}3s^{2}3p^{6}4s^{1}\).

Step11: Write electron - configurations for elements (d)

For boron (\(Z = 5\)), the electron - configuration is \(1s^{2}2s^{2}2p^{1}\).

Step12: Write electron - configurations for elements (e)

For manganese (\(Z = 25\)), the electron - configuration is \(1s^{2}2s^{2}2p^{6}3s^{2}3p^{6}4s^{2}3d^{5}\).

Step13: Determine elements from electron configurations (5a)

In \(1s^{2}2s^{2}2p^{6}3s^{2}3p^{6}4s^{2}3d^{10}4p^{4}\), \(2+2 + 6+2+6 + 2+10 + 4=34\). The element with 34 electrons is selenium (Se).

Step14: Determine elements from electron configurations (5b)

For \(1s^{2}2s^{2}2p^{6}3s^{2}3p^{6}4s^{2}3d^{1}\), \(2+2 + 6+2+6 + 2+1=21\). The element with 21 electrons is scandium (Sc).

Step15: Determine elements from electron configurations (5c)

In \(1s^{2}2s^{2}2p^{6}3s^{2}3p^{4}\), \(2+2 + 6+2 + 4=16\). The element with 16 electrons is sulfur (S).

Step16: Determine elements from electron configurations (5d)

For \(1s^{2}2s^{2}2p^{6}3s^{2}3p^{6}4s^{2}3d^{10}4p^{6}5s^{2}4d^{10}5p^{5}\), \(2+2 + 6+2+6 + 2+10 + 6+2+10 + 5=53\). The element with 53 electrons is iodine (I).

Step17: Determine elements from electron configurations (5e)

In \(1s^{2}2s^{2}2p^{6}3s^{2}3p^{6}4s^{2}3d^{10}4p^{6}5s^{2}4d^{5}\), \(2+2 + 6+2+6 + 2+10 + 6+2+5=43\). The element with 43 electrons is technetium (Tc).

Answer:

  1. Increases
  2. 2
  3. a. Si; b. Be; c. Ne; d. Cl; e. Mn
  4. a. \(1s^{2}2s^{2}2p^{6}3s^{2}3p^{3}\); b. \(1s^{2}2s^{2}2p^{6}3s^{2}3p^{6}4s^{2}\); c. \(1s^{2}2s^{2}2p^{6}3s^{2}3p^{6}4s^{1}\); d. \(1s^{2}2s^{2}2p^{1}\); e. \(1s^{2}2s^{2}2p^{6}3s^{2}3p^{6}4s^{2}3d^{5}\)
  5. a. Se; b. Sc; c. S; d. I; e. Tc