Question

The correct set of four quantum number for the valence electrons for rubidium atom (z = 37) is

Answer 1

$\Large\color{darkblue}\bullet\underline{\underline{\sf Rubidium-}}$

Rubidium Electronic Configuration is $\color{red}{\sf [Kr]5s^1}$

1. Principal Quantum Number (n)

As we can see last electron enters in 5th orbital therefore its principal quantum number will be $\color{red}{\sf 5}$.

2. Azimuthal Quantum Number ${\sf (\ell)}$

Last electron enters in s-block therefore its azimuthal Quantum Number will be $\color{red}{\sf 0}$.

3. Magnetic Quantum Number (m)

It is +ve to -ve of $\ell$ . And $\ell$ is 0 therefore Magnetic Quantum Number will also be $\color{red}{\sf 0}$.

4. Spin Quantum Number (s)

It will be $\color{red}{\sf +\dfrac{1}{2}}$

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$\Large\color{darkblue}\bullet\underline{\underline{\sf Extra \: Info.-}}$

Azimuthal Quantum Number for

$</p><p>\begin{tabular}{c | 1}</p><p></p><p> Orbital & Azimuthal\: Quantum\: Number \\</p><p></p><p>\cline {1-2}</p><p></p><p>s & 0 \\</p><p>p & 1 \\</p><p>d & 2 \\</p><p>f & 3 \\</p><p></p><p>\end{tabular}</p><p>$

Answer 2

Answer:

E.C of Rubidium is (Z = 37) = 1s22s22p63s23p64s23d10 4p65s1 The valence electron is in 5s orbital.Its 4 quantum numbers are - n = 5, l = 0, m = 0 , ...

E.C of Rubidium is (Z = 37) = 1s22s22p63s23p64s23d10 4p65s1 The valence electron is in 5s orbital.Its 4 quantum numbers are - n = 5, l = 0, m = 0 , ... More