Question

An atom of an element has 28 innermost electrons and 7 outermost electrons. In which period of the Periodic Table is this element located?
1. 5

2. 2

3. 3

4. 4

Answer 1

The answer is four because if you multiply seven times four you will get twenty eight

Answer 2

Answer: The given element belongs to Period 4.

Explanation:

We are given:

Number of innermost electrons = 28

Number of outermost electrons = 7

Total number of electrons = 28 + 7 = 35

Electronic configuration is defined as the representation of electrons around the nucleus of an atom.

Number of electrons in an atom is determined by the atomic number of that atom.

The electronic configuration of the element having atomic number '35' is:  $1s^22s^22p^63s^23p^64s^23d^{10}4p^5$

To identify the period, we look at the highest occupied principle quantum number which is 'n'. Here, the value of highest occupied shell or 'n' is 4.

Hence, the given element belongs to Period 4.