Question

Suppose you take a trip to a distant universe and find that the periodic table there is derived from an arrangement of quantum numbers different from the one on Earth. The rules in that universe are:

1. principal quantum number n = 1, 2, . . . (as on Earth);
2. angular momentum quantum number l = 0, 1, 2,. . . , n – 1 (as on Earth);
3. magnetic quantum number ml = 0, 1, 2, . . . , l (only positive integers up to and including l are allowed);
4. spin quantum number ms = –1, 0, 1 (that is, three allowed values of spin).

(a) Assuming that the Pauli exclusion principle remains valid, what is the maximum number of electrons that can populate a given orbital?
(b) Write the electronic configuration of the element with atomic number 8 in the periodic table.
(c) What is the atomic number of the second noble gas?

Please explain everything explicitly!

Thanks!

Answer 1

Concepts and reason

The problem is based on the concept of quantum numbers of an electron and the Pauli’s exclusion principle. An electron in an orbital occupies four set of quantum numbers that are n, l, ${m_l}$ and ${m_s}$

Fundamentals

The four set of quantum numbers are n, l, ${m_l}$ and${m_s}$. Here, n is principal quantum number, l is the angular momentum quantum number, ${m_l}$ is magnetic quantum number and ${m_s}$ is spin quantum number.

According to Pauli’s exclusion principle, no two electrons in an atom can simultaneously have the same set of four quantum numbers.

Part a

Since on Earth, the Pauli’s exclusion principle is followed, hence the spin quantum number has two values $+ \frac{1}{2}$ and $- \frac{1}{2}$ and only 2 electrons of opposite spin can occupy one orbital. If the number spin quantum number has 3 values: -1, 0 and +1, then it will have 3 electrons of each spin in each orbital.

Part b

Each orbital can occupy 3 electrons. Then s orbital also occupies 3 electrons then electronic configuration of the element with atomic number 8 becomes $1{s^3}2{s^3}2{p^2}$

Part c

For an element to behave as a noble gas it should have a completely filled orbital. Therefore, the first noble gas will have the configuration$1{s^3}$.

The second noble gas will have a completely filled 1s, 2s and 2p orbital and p orbital will have two subshells because for $l = 1$, ${m_l}$ will be 0 and 1. So, its configuration is $1{s^3}2{s^3}2{p^6}$and its atomic number is 12.

Ans: Part a

The maximum number of electrons that can populate a given orbital are 3.

Part b

Electronic configuration of the element with atomic number 8 is $1{s^3}2{s^3}2{p^2}$.

Part c

The atomic number of the second noble gas is 12.