Question

If each orbital contain 3 electrons then how many elements present in 5th period

Answer 1

First we need to calculate :

i) No of subshells

ii) No of orbitals

iii) No of elements

Formula to calculate the number of subshell if number of period is known :

If number of period is even : (n+2)/2

If number of period is odd : (n+1)/2

Here it's the second case i.e,

the number of period is even=2

Therefore (n+2)/2 = (2+2)/2 = 2

Which means it consists 2 types of subshell

Thus no of orbitals in various subshell are :

s= 1 orbital

p= 3 orbitals

Not related to this question in bold & italics

( just general information)

( d= 5 orbitals ; f= 7 orbitals ; g= 9 orbitals h=11 orbitals & so on— if total no. of subshell , by the formula would have been greater)

Therefore total no of orbital are 4

Thus maximum no. of elements = 4×3 =12

Answer 2

Answer:

27

Explanation:

no of element= electron * [( period+1)/2]^2