Main 2014 | JEE 2014 | Discussion Page

Loading [MathJax]/jax/output/HTML-CSS/jax.js

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

A) 5,0,0,

$+\frac{1}{2}$

B) 5,1,0,

$+\frac{1}{2}$

C) 5,1,1 ,

$+\frac{1}{2}$

D) 5,0,1,

$+\frac{1}{2}$

Option A

Given, atomic number of Rb, Z=37

Thus, its elecronic configuration is [Kr]5s¹. Since the last electron or valence electron enter in 5s subshell.

So, the quantum numbers are n=5, l=0, (for s orbital) m=0

$(\because m=+1 to -1),s=+\frac{1}{2}or -\frac{1}{2}$

Discussion Forum