Question

The electronic configuration of an element is $1s^{2}2s^{2}2p^{6}3s^{2}3p^3$

What is the atomic number of the element which is just below the above element in the periodic table?

• A. 33
• B. 34
• C. 31
• D. 39

Answer 1

The correct option is A

33

$1s^{2}2s^{2}2p^{6}3s^{2}3p^3$

You can notice that the above element has 15 electrons in its structure. Now for the element below this.

Remember the magic numbers - 2, 8, 8, 18, 32...?

This is the increase in electrons when we go down the group in the corresponding period. For example, Li would have 2 more electrons as compared to H and Na would have 8 more electrons as compared to Li.

Applying this logic to our problem, we know that this element is in the 3rd period. So, adding 18 would give us the next element's electrons.

So, 15 + 18 = 33