Question

Which one of the atomic orbital has higher ENERGY? Why?
I) n=3,l=2, m=+1 or
II) n=4,l=0,m=0

Answer 1

1)3p
2)4s
as n represent number of orbital 123etc
and l represent S, p, d, f for l=o=s
l=1=p
l=2=d
l=3=f so one with n+lmore is having high energy so 3p have not energy
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Answer 2

Final answer:

    I) n=3,l=2, m=+1 has highest energy because of high (n + l)  value.

Given that: We are given atomic orbitals I) n=3,l=2,m=+1  & II) n=4,l=0,m=0

To find: We have to find which atomic orbital possess highest energy and its reason.

Explanation:

• The energy of a given subshell = n + l

n = principle quantum number, l = azimuthal quantum number

• Higher the value of (n + l), higher is its energy.

• Here    I) n=3,l=2, m=+1   corresponding orbital is 3d

Energy = n + l

            = 3+2

            = 5

II)  n=4,l=0,m=0 corresponding orbital is 4s

Energy = n + l

      = 4 + 0

     = 4

• Hence, the option I) of quantum numbers (n=3,l=2,m= +1)  represents the highest energy orbital due to high n+l value.

• The decreasing order of penetrating effect of orbitals is  s > p > d > f

• There by 4s orbital penetrates the inner electrons more effectively than 3d orbital. In accordance with aufbau principle, 4s orbital will be filled first.

To know more about the concept please go through the links

https://brainly.in/question/14073354

https://brainly.in/question/861473

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