Question

if the value of n + l is not greater than 3 then the maximum number of electron in the orbital would be​

Answer 1

Answer:

Let n=1 then l=0

n=2 then l=0,1

n+l=2+1=3

p subject has 3 orbitals hence max no of electrons=3*2=6

Answer 2

Answer:

Required answer is 3s.

Explanation:

Given,

$n + 1 < 3$

So,

$n + 1 = 1 \: or \: n + 1 = 2 \: and \: n + 1 = 3$

$1 + 0 = 1s \: and \: 2 + 0 = 2s \: \: and \: 3 + 0 = 3s$

And

$2 + 1 = 2p \: and \: 3 + 0 = 3s$

According to your question, (n+l) is not more than 3, i.e it can be 1,2,

When (n+l) is 1, we have a 1s orbital.

When (n+l) is 2, we have a 2s orbital.

When (n+l) is 3, we have 2p and 3s orbitals.