# CORRECT SET OF QUANTUM NUMBERS - ELECTRONIC CONFIGURATION IIT JEE - NEET

## Correct set of four quantum numbers for the valence (outermost) electron of Rubidium (Z=37) is :

(IIT JEE 1986)

1) 5, 0, 0, +½

2) 5, 1, 0, +½

3) 5, 1, 1, +½

4) 6, 0, 0, +½

Logic:

We may write the electronic configuration of Rubidium, Rb and then find out the quantum numbers for the last electron (Rb valence electron).

Or

Remember that, Rb belongs to alkali metals (1st group) and 5th period of modern periodic table. For alkali metals, the outer electronic configuration is ns1 and for the 5th period n = 5.

 Period Elements of 1st group 2 3Li 3 11Na 4 19K 5 37Rb 6 55Cs 7 87Fr

Solution:

Electronic configuration of Rb = 1s2 2s2 2p6 3s2 3p6 4s2 3d10 4p6 5s1

Check that there are 37 electrons.

Therefore,

Principal quantum number, n = 5

Azimuthal quantum number, l = 0 (since for s orbital l = 0)

Magnetic quantum number, m = 0 (since for l = 0, there is only one m value i.e. 0)

Spin quantum number, s = +½ or -½ (there is only one electron and it can have clockwise or anticlockwise spins)

Conclusion:

The correct set of quantum numbers will be: 5, 0, 0, +½ or -½. Therefore correct option is 1.

## SET OF QUANTUM NUMBERS Followup  questions

### Q-1: The correct set of four quantum numbers for unpaired electron chlorine is :

1) 2, 0, +1, +½

2) 3, 0, -1, +½

3) 3, 1, -1, +½

4) 2, 0, 0, +½

Logic:

Write the outer electronic configuration of Chlorine atom.

It belongs to halogens i.e. 17th group and 3rd period of periodic table

Solution:

The outer electronic configuration of Chlorine atom in the ground state is: 3s2 3p5

The valence electron resides in one of the 3p orbital.

Hence the correct set of quantum numbers is: n=3, l=1, m=+1 or 0 or -1, s=+½ or -½.

Conclusion:

The correct set of quantum numbers among the given options is: 3, 1, -1, +½  i.e correct option is 3.

### Q-2: The values of four quantum numbers of valence electron of an element are n=4, l=0, m=0 and s=+1/2. The element is :

(Eamcet - 2004-M)

a) K

b) Ti

c) Na

d) Sc

Logic:

n = 4 and l = 0 represent 4s orbital. Since it is valence electron, the element belongs to s-block and 4th period of periodic table.

Solution:

There are two elements that belong to s-block i.e. K and Na. On close inspection of periodic table, the quantum number data for the valence electron  fits for potassium, K, since it belongs to 4th period.

Remember that the principal quantum number of valence electron represents the period number also that is especially true for s and p block elements.

Note: Na belongs to 3rd period and hence n should be 3 for the valence electron of this element. Ti and Sc belong to d-block.

### Homework

1) Write all possible values of each quantum number for the outermost electron in an Rb atom.

2) What is the set of quantum numbers for unpaired electron of Fluorine atom?

3) Write the set of quantum numbers for the electron present in the first excited state of hydrogen atom.

4) How many sublevels are possible for n = 4 quantum level?

5) What is the principal quantum level of the highest energy electron of rubidium atom in the ground state?

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