In the following balanced reaction, what volume of a 3.0 M H2SO4 is required to completely neutralize 200.0 ml of a 2.5 M NaOH?
2 answers:
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Answer:
1. Orbital diagram
2p⁴ ║ ↑↓ ║ "↑" ║ ↑
2s² ║ ↑↓ ║
1s² ║ ↑↓ ║
2. Quantum numbers
- <em>n </em>= 2,
- <em>l</em> = 1,
= 0,
= +1/2
Explanation:
The fill in rule is:
- Follow shell number: from the inner most shell to the outer most shell, our case from shell 1 to 2
- Follow the The Aufbau principle, 1s<2s<2p<3s<3p<4s<3d<4p<5s<4d<5p<6s<4f<5d<6p<7s<5f<6d<7p
- Hunds' rule: Every orbital in a sublevel is singly occupied before any orbital is doubly occupied. All of the electrons in singly occupied orbitals have the same spin (to maximize total spin).
So, the orbital diagram of given element is as below and the sixth electron is marked between " "
2p⁴ ║ ↑↓ ║ "↑" ║ ↑
2s² ║ ↑↓ ║
1s² ║ ↑↓ ║
The quantum number of an electron consists of four number:
- <em>n </em>(shell number, - 1, 2, 3...)
- <em>l</em> (subshell number or orbital number, 0 - orbital <em>s</em>, 1 - orbital <em>p</em>, 2 - orbital <em>d...</em>)
In our case, the electron marked with " " has quantum number
- <em>n </em>= 2, shell number 2,
- <em>l</em> = 1, subshell or orbital <em>p,</em>