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In Chemistry / High School | 2025-07-04

What is the second quantum number of a [tex]$2 s^2$[/tex] electron in phosphorus, [tex]$1 s^2 2 s^2 2 p^6 3 s^2 3 p^3$[/tex]?
A. [tex]$I=3$[/tex]
B. [tex]$I=0$[/tex]
C. [tex]$I=1$[/tex]
D. [tex]$1=2$[/tex]

Asked by nyqbgyj659

Answer (2)

The second quantum number of a 2 s 2 electron in phosphorus is 0. This is because, for s orbitals, the azimuthal quantum number l is always 0. Therefore, the correct answer is B. I = 0 .
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Answered by Anonymous | 2025-07-04

The problem asks for the second quantum number ( l ) of a 2 s 2 electron in phosphorus.
The electron configuration of phosphorus is 1 s 2 2 s 2 2 p 6 3 s 2 3 p 3 .
For an s orbital, the second quantum number l is always 0.
Therefore, the second quantum number of a 2 s 2 electron is 0 ​ .

Explanation

Understanding the Problem The question asks for the second quantum number (also known as the azimuthal quantum number, or angular momentum quantum number) of a 2 s 2 electron in phosphorus. The electron configuration of phosphorus is given as 1 s 2 2 s 2 2 p 6 3 s 2 3 p 3 .

Understanding Quantum Numbers The second quantum number, denoted by l , describes the shape of an atomic orbital and has integer values from 0 to n − 1 , where n is the principal quantum number. For a given n , the possible values of l are 0, 1, 2, ..., n − 1 .

Identifying the Orbital The question specifically asks about a 2 s 2 electron. The '2' indicates the principal quantum number n = 2 , and 's' indicates the type of orbital.

Determining the Quantum Number For an s orbital, the second quantum number l is always 0. The s orbital is spherical in shape, which corresponds to l = 0 .

Final Answer Therefore, the second quantum number of a 2 s 2 electron in phosphorus is 0.


Examples
Understanding quantum numbers helps us predict the behavior of electrons in atoms, which is crucial in chemistry. For example, knowing the electron configuration and quantum numbers allows us to understand how atoms will interact to form chemical bonds. This knowledge is essential in designing new materials with specific properties, such as semiconductors for electronics or catalysts for chemical reactions.

Answered by GinnyAnswer | 2025-07-04