Single Choice

The correct set of quantum numbers for valence electron of $$Rb$$ (atomic number 37) is:

A$$5,0,0,-\cfrac{1}{2}$$
Correct Answer
B$$5,1,0,\cfrac{1}{2}$$
C$$6,0,1,\cfrac{1}{2}$$
D$$5,1,1,\cfrac{1}{2}$$

Solution

Electronic configuration of $$Rb=[Kr] 5s^1$$

$$\displaystyle Rb$$ is alkali metal with valence electron in 5s subshell. Hence, $$\displaystyle n=5$$

For s electron, $$\displaystyle l=0$$ and $$\displaystyle m=0$$.

$$s$$ can have value $$\displaystyle \pm \dfrac {1}{2}$$.

Thus, the correct set of quantum numbers for Rb (atomic number 37) is $$\displaystyle 5,0,0,-\cfrac{1}{2}$$.

SIMILAR QUESTIONS

Atomic Structure

What is the maximum number of orbitals that can be identified with the following quantum numbers? $$n=3,l=+1, m_l=0$$

Atomic Structure

The number of orbitals associated with quantum numbers $$n=5,m_s=+\dfrac{1}{2}$$ is :

Atomic Structure

The quantum number of four electrons are given below- $$I. n=4,l=2, m_l=-2, m_s=-1/2$$ $$II. n=3,l=2, m_l=1, m_s=+1/2$$ $$III. n=4,l=1, m_l=0, m_s=+1/2$$ $$IV. n=3,l=1, m_l=1, m_s=-1/2$$ The correct order of their increasing energy will be:

Atomic Structure

The total number of orbitals associated with the principal quantum number 5 is :

Atomic Structure

What is the maximum number of orbitals that can be identified with the following quantum number? $$n = 3, l = 1, m = 0$$

Atomic Structure

The maximum number of electrons that can have principal quantum number, $$n =3$$ , and spin quantum number, $$m_{s}=-1/2$$, is:

Atomic Structure

The electrons identified by quantum numbers $$n$$ and $$l$$ can be placed in the order of increasing energy as: 1. $$n=4, l=1$$ 2. $$n=4, l=0$$ 3. $$n=3, l=2$$ 4. $$n=3, l=1$$

Atomic Structure

What is the lowest value of n that allows g orbitals to exist?

Atomic Structure

An electron is in one of the 3d orbitals. Give the possible values of n, l and $$\displaystyle { m }_{ l }$$ for this electron.

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