Electron Configuration worksheet

1. Electron configuration: 1s² 2s² 2p⁶ 3s² 3p⁶ 4s² 3d¹⁰ 4p⁶ 5s² 4d¹⁰ 5p⁶ 4f¹⁴ 5d⁴

1. The charge tells you how many electrons the ion lost or gained, so subtract it from the atomic number for W²⁺

   $74-2$$=72$

2. Fill the subshells in the standard (Aufbau) order, lowest energy first. Each s holds up to 2 electrons, p 6, d 10 and f 14

   $\text{1s → 2s → 2p → 3s → 3p → 4s → 3d → 4p → 5s → 4d → 5p → 6s → 4f → 5d}$

3. A positive ion loses electrons from the outermost shell first (p before s), because those are held most loosely. Then it loses inner d and f electrons

   $\text{remove 2 from 6s}$

4. Write out the full lowest-energy (ground-state) configuration of W²⁺

   $\text{1s² 2s² 2p⁶ 3s² 3p⁶ 4s² 3d¹⁰ 4p⁶ 5s² 4d¹⁰ 5p⁶ 4f¹⁴ 5d⁴}$

5. For the short version, swap the inner electrons that match Xe (54 electrons) for [Xe]

   $\text{[Xe] 4f¹⁴ 5d⁴}$

6. Count the electrons in the outermost shell (shell 5), since these do most of the chemistry

   $\text{5s² + 5p⁶ + 5d⁴ →}12\text{\hspace{0.17em} electrons}$

2. Electron configuration: 1s² 2s² 2p⁶ 3s² 3p⁶ 4s² 3d¹⁰ 4p⁶ 5s² 4d¹⁰ 5p⁶ 6s² 4f¹⁴ 5d¹⁰ 6p⁶ 7s² 5f¹⁴ 6d¹⁰ 7p⁴

1. A neutral atom has as many electrons as protons: the atomic number of livermorium is 116

   $\text{electrons}$$=116$

2. Fill the subshells in the standard (Aufbau) order, lowest energy first. Each s holds up to 2 electrons, p 6, d 10 and f 14

   $\text{1s → 2s → 2p → 3s → 3p → 4s → 3d → 4p → 5s → 4d → 5p → 6s → 4f → 5d → 6p → 7s → 5f → 6d → 7p}$

3. Write out the full lowest-energy (ground-state) configuration of Lv

   $\text{1s² 2s² 2p⁶ 3s² 3p⁶ 4s² 3d¹⁰ 4p⁶ 5s² 4d¹⁰ 5p⁶ 6s² 4f¹⁴ 5d¹⁰ 6p⁶ 7s² 5f¹⁴ 6d¹⁰ 7p⁴}$

4. For the short version, swap the inner electrons that match Rn (86 electrons) for [Rn]

   $\text{[Rn] 7s² 5f¹⁴ 6d¹⁰ 7p⁴}$

5. Count the electrons in the outermost shell (shell 7), since these do most of the chemistry

   $\text{7s² + 7p⁴ →}6\text{\hspace{0.17em} electrons}$

3. Electron configuration: 1s² 2s² 2p⁶ 3s² 3p⁶ 4s² 3d¹⁰ 4p⁴

1. A neutral atom has as many electrons as protons: the atomic number of selenium is 34

   $\text{electrons}$$=34$

2. Fill the subshells in the standard (Aufbau) order, lowest energy first. Each s holds up to 2 electrons, p 6, d 10 and f 14

   $\text{1s → 2s → 2p → 3s → 3p → 4s → 3d → 4p}$

3. Write out the full lowest-energy (ground-state) configuration of Se

   $\text{1s² 2s² 2p⁶ 3s² 3p⁶ 4s² 3d¹⁰ 4p⁴}$

4. For the short version, swap the inner electrons that match Ar (18 electrons) for [Ar]

   $\text{[Ar] 4s² 3d¹⁰ 4p⁴}$

5. Count the electrons in the outermost shell (shell 4), since these do most of the chemistry

   $\text{4s² + 4p⁴ →}6\text{\hspace{0.17em} electrons}$

4. Electron configuration: 1s² 2s² 2p⁶ 3s² 3p⁶ 4s² 3d¹⁰ 4p⁶ 5s² 4d¹⁰ 5p⁶ 6s² 4f¹⁴ 5d¹⁰ 6p⁶

1. A neutral atom has as many electrons as protons: the atomic number of radon is 86

   $\text{electrons}$$=86$

2. Fill the subshells in the standard (Aufbau) order, lowest energy first. Each s holds up to 2 electrons, p 6, d 10 and f 14

   $\text{1s → 2s → 2p → 3s → 3p → 4s → 3d → 4p → 5s → 4d → 5p → 6s → 4f → 5d → 6p}$

3. Write out the full lowest-energy (ground-state) configuration of Rn

   $\text{1s² 2s² 2p⁶ 3s² 3p⁶ 4s² 3d¹⁰ 4p⁶ 5s² 4d¹⁰ 5p⁶ 6s² 4f¹⁴ 5d¹⁰ 6p⁶}$

4. For the short version, swap the inner electrons that match Xe (54 electrons) for [Xe]

   $\text{[Xe] 6s² 4f¹⁴ 5d¹⁰ 6p⁶}$

5. Count the electrons in the outermost shell (shell 6), since these do most of the chemistry

   $\text{6s² + 6p⁶ →}8\text{\hspace{0.17em} electrons}$

5. Electron configuration: 1s² 2s² 2p⁶ 3s² 3p⁶ 4s² 3d¹⁰ 4p⁶ 5s² 4d¹⁰ 5p⁶ 6s² 4f¹⁴ 5d¹⁰ 6p⁶ 7s² 5f¹⁴ 6d¹⁰ 7p⁶

1. A neutral atom has as many electrons as protons: the atomic number of oganesson is 118

   $\text{electrons}$$=118$

2. Fill the subshells in the standard (Aufbau) order, lowest energy first. Each s holds up to 2 electrons, p 6, d 10 and f 14

   $\text{1s → 2s → 2p → 3s → 3p → 4s → 3d → 4p → 5s → 4d → 5p → 6s → 4f → 5d → 6p → 7s → 5f → 6d → 7p}$

3. Write out the full lowest-energy (ground-state) configuration of Og

   $\text{1s² 2s² 2p⁶ 3s² 3p⁶ 4s² 3d¹⁰ 4p⁶ 5s² 4d¹⁰ 5p⁶ 6s² 4f¹⁴ 5d¹⁰ 6p⁶ 7s² 5f¹⁴ 6d¹⁰ 7p⁶}$

4. For the short version, swap the inner electrons that match Rn (86 electrons) for [Rn]

   $\text{[Rn] 7s² 5f¹⁴ 6d¹⁰ 7p⁶}$

5. Count the electrons in the outermost shell (shell 7), since these do most of the chemistry

   $\text{7s² + 7p⁶ →}8\text{\hspace{0.17em} electrons}$

6. Electron configuration: 1s² 2s² 2p⁶ 3s² 3p⁶ 4s² 3d¹⁰ 4p⁶ 5s² 4d¹⁰ 5p⁶ 4f¹ 5d¹

1. The charge tells you how many electrons the ion lost or gained, so subtract it from the atomic number for Ce²⁺

   $58-2$$=56$

2. Fill the subshells in the standard (Aufbau) order, lowest energy first. Each s holds up to 2 electrons, p 6, d 10 and f 14

   $\text{1s → 2s → 2p → 3s → 3p → 4s → 3d → 4p → 5s → 4d → 5p → 6s → 4f}$

3. Cerium is a known exception. Measurements show its lowest-energy state has 4f¹, 5d¹ instead of the simple filling order

   $\text{Ce: 1s² 2s² 2p⁶ 3s² 3p⁶ 4s² 3d¹⁰ 4p⁶ 5s² 4d¹⁰ 5p⁶ 6s² 4f¹ 5d¹}$

4. A positive ion loses electrons from the outermost shell first (p before s), because those are held most loosely. Then it loses inner d and f electrons

   $\text{remove 2 from 6s}$

5. Write out the full lowest-energy (ground-state) configuration of Ce²⁺

   $\text{1s² 2s² 2p⁶ 3s² 3p⁶ 4s² 3d¹⁰ 4p⁶ 5s² 4d¹⁰ 5p⁶ 4f¹ 5d¹}$

6. For the short version, swap the inner electrons that match Xe (54 electrons) for [Xe]

   $\text{[Xe] 4f¹ 5d¹}$

7. Count the electrons in the outermost shell (shell 5), since these do most of the chemistry

   $\text{5s² + 5p⁶ + 5d¹ →}9\text{\hspace{0.17em} electrons}$

7. Electron configuration: 1s² 2s² 2p⁶ 3s² 3p⁶ 3d¹⁰

1. The charge tells you how many electrons the ion lost or gained, so subtract it from the atomic number for Zn²⁺

   $30-2$$=28$

2. Fill the subshells in the standard (Aufbau) order, lowest energy first. Each s holds up to 2 electrons, p 6, d 10 and f 14

   $\text{1s → 2s → 2p → 3s → 3p → 4s → 3d}$

3. A positive ion loses electrons from the outermost shell first (p before s), because those are held most loosely. Then it loses inner d and f electrons

   $\text{remove 2 from 4s}$

4. Write out the full lowest-energy (ground-state) configuration of Zn²⁺

   $\text{1s² 2s² 2p⁶ 3s² 3p⁶ 3d¹⁰}$

5. For the short version, swap the inner electrons that match Ar (18 electrons) for [Ar]

   $\text{[Ar] 3d¹⁰}$

6. Count the electrons in the outermost shell (shell 3), since these do most of the chemistry

   $\text{3s² + 3p⁶ + 3d¹⁰ →}18\text{\hspace{0.17em} electrons}$

8. Electron configuration: 1s² 2s² 2p⁴

1. A neutral atom has as many electrons as protons: the atomic number of oxygen is 8

   $\text{electrons}$$=8$

2. Fill the subshells in the standard (Aufbau) order, lowest energy first. Each s holds up to 2 electrons, p 6, d 10 and f 14

   $\text{1s → 2s → 2p}$

3. Write out the full lowest-energy (ground-state) configuration of O

   $\text{1s² 2s² 2p⁴}$

4. For the short version, swap the inner electrons that match He (2 electrons) for [He]

   $\text{[He] 2s² 2p⁴}$

5. Count the electrons in the outermost shell (shell 2), since these do most of the chemistry

   $\text{2s² + 2p⁴ →}6\text{\hspace{0.17em} electrons}$

9. Electron configuration: 1s² 2s² 2p⁶ 3s² 3p⁶ 4s² 3d¹⁰ 4p⁶ 5s² 4d¹⁰ 5p⁶ 6s² 4f¹⁴ 5d¹⁰ 6p⁶ 7s² 5f² 6d¹

1. A neutral atom has as many electrons as protons: the atomic number of protactinium is 91

   $\text{electrons}$$=91$

2. Fill the subshells in the standard (Aufbau) order, lowest energy first. Each s holds up to 2 electrons, p 6, d 10 and f 14

   $\text{1s → 2s → 2p → 3s → 3p → 4s → 3d → 4p → 5s → 4d → 5p → 6s → 4f → 5d → 6p → 7s → 5f}$

3. Protactinium is a known exception. Measurements show its lowest-energy state has 5f², 6d¹ instead of the simple filling order

   $\text{Pa: 1s² 2s² 2p⁶ 3s² 3p⁶ 4s² 3d¹⁰ 4p⁶ 5s² 4d¹⁰ 5p⁶ 6s² 4f¹⁴ 5d¹⁰ 6p⁶ 7s² 5f² 6d¹}$

4. Write out the full lowest-energy (ground-state) configuration of Pa

   $\text{1s² 2s² 2p⁶ 3s² 3p⁶ 4s² 3d¹⁰ 4p⁶ 5s² 4d¹⁰ 5p⁶ 6s² 4f¹⁴ 5d¹⁰ 6p⁶ 7s² 5f² 6d¹}$

5. For the short version, swap the inner electrons that match Rn (86 electrons) for [Rn]

   $\text{[Rn] 7s² 5f² 6d¹}$

6. Count the electrons in the outermost shell (shell 7), since these do most of the chemistry

   $\text{7s² →}2\text{\hspace{0.17em} electrons}$

10. Electron configuration: 1s² 2s² 2p⁶ 3s² 3p⁶ 3d³

1. The charge tells you how many electrons the ion lost or gained, so subtract it from the atomic number for V²⁺

   $23-2$$=21$

2. Fill the subshells in the standard (Aufbau) order, lowest energy first. Each s holds up to 2 electrons, p 6, d 10 and f 14

   $\text{1s → 2s → 2p → 3s → 3p → 4s → 3d}$

3. A positive ion loses electrons from the outermost shell first (p before s), because those are held most loosely. Then it loses inner d and f electrons

   $\text{remove 2 from 4s}$

4. Write out the full lowest-energy (ground-state) configuration of V²⁺

   $\text{1s² 2s² 2p⁶ 3s² 3p⁶ 3d³}$

5. For the short version, swap the inner electrons that match Ar (18 electrons) for [Ar]

   $\text{[Ar] 3d³}$

6. Count the electrons in the outermost shell (shell 3), since these do most of the chemistry

   $\text{3s² + 3p⁶ + 3d³ →}11\text{\hspace{0.17em} electrons}$