Question

1. which rule states that electrons must occupy the lowest energy orbital available?

a. aufbau principle
c. pauli exclusion principle
b. hund’s rule
d. uncertainity principle

1. how many electrons can occupy the s orbitals at each enery level?

a. two, if they have the same spin
c. no more than eight
b. two, if they have opposite spins
d. one

1. which of the following colors of visible light has the shortest wavelength?

a. red
c. orange
b. blue
d. violet

1. the major difference between a 1s orbital and a 2s orbital is that

a. the 1s orbital can have only one electron
c. the 2s orbital can hold more electrons
b. the 2s orbital is at a higher energy level
d. the 2s orbital has a slightly different shape

1. what is the electron confiuration for potassium?

a. 1s²2s²2p¹⁰3s²3p³
c. 1s²2s²2p³3s²3p²4s¹
b. 1s²2s²3s²3p⁶3d¹
d. 1s²2s²2p⁶3s²3p⁶4s¹

1. which rule states that a maximum of two electrons may occupy a single orbital and they have to have opposite spins?

a. aufbau principle
c. hund’s rule
b. pauli exclusion principle
d. uncertainty principle
short answer

1. write the full electron configuration for the following (4 points each):

flourine:
vanadium:
potassium:

Explanation:

1. 构造原理（Aufbau Principle）表明电子优先占据能量最低的轨道。
2. 每个s轨道最多可容纳2个自旋相反的电子。
3. 在可见光中，紫光（Violet）的波长最短。
4. 1s轨道和2s轨道的主要区别是2s轨道处于更高的能量级。
5. 钾（Potassium）的电子构型是$1s^{2}2s^{2}2p^{6}3s^{2}3p^{6}4s^{1}$。
6. 泡利不相容原理（Pauli Exclusion Principle）指出一个轨道最多容纳两个电子且自旋相反。
7. 氟（Fluorine）有9个电子，电子构型为$1s^{2}2s^{2}2p^{5}$；钒（Vanadium）有23个电子，电子构型为$1s^{2}2s^{2}2p^{6}3s^{2}3p^{6}3d^{3}4s^{2}$；钾（Potassium）有19个电子，电子构型为$1s^{2}2s^{2}2p^{6}3s^{2}3p^{6}4s^{1}$。

Answer:

1. a. Aufbau Principle
2. b. two, if they have opposite spins
3. d. Violet
4. b. the 2s orbital is at a higher energy level
5. d. $1s^{2}2s^{2}2p^{6}3s^{2}3p^{6}4s^{1}$
6. b. Pauli Exclusion Principle
7. Fluorine: $1s^{2}2s^{2}2p^{5}$

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