# if there were ten periods in the periodic table then how many elements would this period can maximum comprise of ?give answer with explanation

16 Points
12 years ago
all elements hav orbits and orbits can hav 4 types of orbitals..

s   p   d   f

each orbital has its own capacity for electrons..

s=2

p=6

d=10

f=14

max no of elements in a period = max no of electrons in its outer most shell

since period no increases, the no of orbitals in the outermost orbit increase.

so max electons inc and so capacity of each period increases.

Eg:  1st orbit has only s orbital so only 2  ELEMENTS in the 1st period

2nd orbit has s and p orbitals so 8 electrons.. so 8 elements in 2nd period..

so on...

since an  orbit can hav only s  p  d  f orbitals 10th period will hav 2+6+10+14 elements
abhishek agarwal
16 Points
11 years ago

the correct ans is 72... but even i dn''t kw the explanation......

Salil
19 Points
8 years ago
#Method 1
n*s n-4*h n-3*g n-2*f n-1*d n*p
Put value of n. If orbital exists then count 2e- per orbital (2h or 3g do not exist). Sum up the electrons. Gives the no. of elements in a period.
#Method 2
Period no even. 2*(n+2/2)^2
Period no odd. 2*9n+1/2)^2
Here n = 10
Putting directly in formula.
Elements = 72
Benja
11 Points
6 years ago
Trick to find no. of elements in a period.For odd periods: (1st, 3rd, 5th....periods)(n+1)^2/2 For even periods (2nd, 4th, 6th....periods)So coming to the question, no. Of elements present in 10th period will be-(n+2)^2/2{USING THIS FORMULA SINCE 10th PERIOD IS EVEN)= (10+2)^2/2=(12)^2/2=144/2=72 elements will be present in the 10th period.
Tejus
35 Points
6 years ago
sir is not with s p d f also there will be g and others which would lead to the correct answer

I would have done it but don't known to me no. of electrons in g and h orbital plz notify whether I am on correct way to answer or not.
Harsh
11 Points
5 years ago
Following afbau's principle
Energy shells are filled according to their energy order
i.e.
In the diagonal relation manner like the following
5s 5p 5d 5f 5g
4s 4p 4d 4f
3s 3p 3d
2s 2p
1s
Where electrons start from 1s (first shell) then 2s 2p (second shell), 3s 3p (3rd shell) 4s 3d 4p (4th shell) and so on..
So for the 10th shell..it ll be
10s 6h 5g 4f 9d 10p
And so adding the no. Of electeons
Total electrons =total no. Of elements in that period
= 2+22+18+14+10+6=72

ankit singh