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electron configuration practice chemistry name: due date: i. determine …

Question

electron configuration practice
chemistry
name:
due date:
i. determine which of the following electron configurations are not valid. state which rule has been violated.

  1. 1s²2s²2p⁶3s²3p⁶4s¹4d¹⁰4p⁶
  2. 1s²2s²2p⁶3s²3d⁸
  3. rn 7s¹5f⁸
  4. kr 5s²4d¹⁰5p⁵
  5. xe

ii. write the full electron configuration, short - hand electron configuration, and fill in the orbital diagrams, for the following elements.

  1. nitrogen
  2. chlorine
  3. sodium
  4. neon

Explanation:

Step1: Recall electron - filling rules

Electrons fill orbitals in order of increasing energy (Aufbau principle), with a maximum of 2 electrons per orbital (Pauli - exclusion principle) and electrons in degenerate orbitals occupy them singly first with parallel spins (Hund's rule).

Step2: Analyze Nitrogen (N, atomic number = 7)

Full electron configuration: $1s^{2}2s^{2}2p^{3}$. Short - hand: $[He]2s^{2}2p^{3}$. Orbital diagram: In the 1s box, 2 electrons with opposite spins; in the 2s box, 2 electrons with opposite spins; in the three 2p boxes, 1 electron in each with parallel spins.

Step3: Analyze Chlorine (Cl, atomic number = 17)

Full electron configuration: $1s^{2}2s^{2}2p^{6}3s^{2}3p^{5}$. Short - hand: $[Ne]3s^{2}3p^{5}$. Orbital diagram: 1s and 2s boxes have 2 electrons with opposite spins each, 2p boxes are full (6 electrons), 3s box has 2 electrons with opposite spins, and in the three 3p boxes, 2 boxes have 2 electrons and 1 box has 1 electron.

Step4: Analyze Sodium (Na, atomic number = 11)

Full electron configuration: $1s^{2}2s^{2}2p^{6}3s^{1}$. Short - hand: $[Ne]3s^{1}$. Orbital diagram: 1s and 2s boxes have 2 electrons with opposite spins each, 2p boxes are full (6 electrons), and 3s box has 1 electron.

Step5: Analyze Neon (Ne, atomic number = 10)

Full electron configuration: $1s^{2}2s^{2}2p^{6}$. Short - hand: $[He]2s^{2}2p^{6}$. Orbital diagram: 1s and 2s boxes have 2 electrons with opposite spins each, and 2p boxes are full (6 electrons).

Step6: Analyze electron configurations in part H

  1. $1s^{2}2s^{2}2p^{6}3s^{2}3p^{6}4s^{2}4d^{10}4p^{6}$ is invalid. The 4d sub - shell cannot be filled before the 5s sub - shell according to the Aufbau principle.
  2. $1s^{2}2s^{2}2p^{6}3s^{2}3d^{9}$ is invalid. The 3d sub - shell cannot be filled before the 4s sub - shell according to the Aufbau principle.
  3. $[Rn]7s^{2}5f^{4}$ is valid.
  4. $[Kr]5s^{2}4d^{10}5p^{5}$ is valid.
  5. $[Xe]$ is just a noble gas core and not a complete electron configuration for an element.

Answer:

  1. Nitrogen:
  • Full electron configuration: $1s^{2}2s^{2}2p^{3}$
  • Short - hand: $[He]2s^{2}2p^{3}$
  • Orbital diagram: 1s: ↑↓, 2s: ↑↓, 2p: ↑ ↑ ↑
  1. Chlorine:
  • Full electron configuration: $1s^{2}2s^{2}2p^{6}3s^{2}3p^{5}$
  • Short - hand: $[Ne]3s^{2}3p^{5}$
  • Orbital diagram: 1s: ↑↓, 2s: ↑↓, 2p: ↑↓ ↑↓ ↑↓, 3s: ↑↓, 3p: ↑↓ ↑↓ ↑
  1. Sodium:
  • Full electron configuration: $1s^{2}2s^{2}2p^{6}3s^{1}$
  • Short - hand: $[Ne]3s^{1}$
  • Orbital diagram: 1s: ↑↓, 2s: ↑↓, 2p: ↑↓ ↑↓ ↑↓, 3s: ↑
  1. Neon:
  • Full electron configuration: $1s^{2}2s^{2}2p^{6}$
  • Short - hand: $[He]2s^{2}2p^{6}$
  • Orbital diagram: 1s: ↑↓, 2s: ↑↓, 2p: ↑↓ ↑↓ ↑↓
  1. Invalid, Aufbau principle violated.
  2. Invalid, Aufbau principle violated.
  3. Valid.
  4. Valid.
  5. Incomplete electron configuration.