Question

Suppose an orbital may accommodate three electrons, then the number of elements in 4th period will be:

• A. 12
• B. 27
• C. 18
• D. 32

Answer 1

The correct option is B 27

In the fourth period of the periodic table, there is a filling of 4s,3d,4p orbitals. So the total number of orbitals to be filled $=1\left(4s\right)+5\left(3d\right)+3\left(4p\right)=9$

If each orbital can hold a maximum of 3 electrons the number of elements$=3\times 9=27$

Hence option B is the correct answer.