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Consider the hypothetical situation where the azimuthal quantum number, l, takes values 0,1,2,.....n+1, where n is the principal quantum number. Then, the element with atomic number :
Option: 1 9 is the first alkali metal
Option: 2 13 has a half-filled valence subshell
Option: 3 8 is the first noble gas
Option: 4 6 has a 2p-valence subshell

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GIven l= 0 to (n + 1)

Now for n = 1,
l = 0, 1, 2

Orbitals :

1s  , l =0 

1p , l = 1

1d , l = 2

 

And for

n = 2

l = 0, 1, 2, 3

\mathrm{Orbitals} : 2s,2p,2d,2f

 

And for

n = 3

l = 0, 1, 2, 3,4

\mathrm{Orbitals} : 3s,3p,3d,3f,3g

Now, in order to wire electronic configuration, we need to apply (n + l) rule.

Energy order : 1s < 1p < 2s < 1 d < 2p < 3s < 2d ...

1) 9 : 1s2 1p6 2s1 is the first alkali metal because after losing one electron, it will achieve the first noble gas configuration

2) 13 : 1s2 1p6 2s2 1d3 is not half filled. (Showing incorrect statement)

3) 8 : 1s2 1p6 is the first noble gas.

4) 6 : 1s2 1p4 has 1p valence subshell.

 

Therefore, Option 2 is correct.

Posted by

Kuldeep Maurya

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