Thank you for registering.

One of our academic counsellors will contact you within 1 working day.

Please check your email for login details.
MY CART (5)

Use Coupon: CART20 and get 20% off on all online Study Material

ITEM
DETAILS
MRP
DISCOUNT
FINAL PRICE
Total Price: Rs.

There are no items in this cart.
Continue Shopping

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

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


give answer with explanation

Grade:11

7 Answers

Emaad Mulla
16 Points
9 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
8 years ago

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

Salil
19 Points
5 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
3 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
3 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
3 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
askIITians Faculty 614 Points
10 months ago
The maximum number of elements in a period is equal to the maximum number of electrons in its outer most shell.  As we move from one period to the other the number of electrons it can accommodate in its orbital increases (in the order of s, p, d and f).

But the maximum number of electrons a period can accommodate is = 2 + 6 + 10 + 14 = 32

So, we would have 32 elements in the 10th period.

Think You Can Provide A Better Answer ?

Provide a better Answer & Earn Cool Goodies See our forum point policy

ASK QUESTION

Get your questions answered by the expert for free