Correct set of four quantum numbers for ...

Correct set of four quantum numbers for the valence (outermost) electron of rubidium `(Z = 37)` is

`n=5, l=0, m=0, s=+1//2`

`n=5, l=1, m=0, s=+1//2`

`n=5, l=1, m=1, s=+1//2`

`n=6, l=0, m=0, s=+1//2`

Text Solution

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The correct Answer is:

I : for `n=5, l_(min)=0, :.` Orbital angular momentum `=sqrt(l(+1)) ħ=0`. (False)
II : Outermost electronic configuration `=3s^(1)` or `3s^(2)`. :. possible atomic number `=11` or `12` (False).
III : `Mn_(25)=[Ar] 3d^(5) 4s^(2) :. 5` unpaired electrons. :. Total spin `=+- 5/2` (False).
IV : Inert gases have no unpaired electrons. :. Spin magnetic moment `=0` (True).

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The correct set of four quantum number for the valence (outermost) electron of radiation (Z = 37) is

`5,0,0,+1//2`

`5,1,0,+1//2`

`5,1,1,+1//2`

`6,0,0,+1//2`

The correct set of four quantum numbers for the valence electron of rubidium (Z = 37) is

5, 1, 1, `+1//2`

6, 0, 0, `+1//2`

5, 0, 0, `+1//2`

5, 1, 0, `+1//2`

The correct set of four quantum numbers for the valence electron of rubidium (Z=37) is

`n=5, l=0, m=0, m_(s)=+1/2`

`n=5, l=1, m=0, m_(s)=+1/2`

`n=5, l=1,m=1,m_(2)=+1/2`

`n=6,l=0,m=0,m_(s)=+1/2`