The correct of four quantum numbers the valence electrons of rubidium atom left-Z-37right- is-5- 1- 0- -dfrac-1-2-5- 0- 1- -dfrac-1-2-5- 1- 1- -dfrac-1-2-5- 0- 0- -dfrac-1-2-

Question

Toppr Admin · Accepted Answer

E-C of Rubidium is -Z-37- - 1s22s22p63s23p64s23d104p65s1-xA0-The valence electron is in 5s orbital-Its 4 quantum numbers are-xA0-n-5-l-0-m-0-s-12Therefore- the correct option is C

The correct set of four quantum numbers for the valence electrons of rubidium atom $(Z = 37)$ is:
$5, 1, 0, + \frac{1}{2}$
$5, 0, 1, + \frac{1}{2}$
$5, 1, 1, + \frac{1}{2}$
$5, 0, 0, + \frac{1}{2}$

E.C of Rubidium is (Z=37) = $1 s^{2} 2 s^{2} 2 p^{6} 3 s^{2} 3 p^{6} 4 s^{2} 3 d^{10} 4 p^{6} 5 s^{1}$
The valence electron is in 5s orbital.

Its 4 quantum numbers are- $n = 5, l = 0, m = 0, s = + \frac{1}{2}$

Therefore, the correct option is $C$

The correct set of four quantum numbers for the valence electrons of rubidium atom (Z=37) is:5,1,0,+125,0,1,+125,1,1,+125,0,0,+12

E.C of Rubidium is (Z=37) = 1s22s22p63s23p64s23d104p65s1The valence electron is in 5s orbital.Its 4 quantum numbers are- n=5,l=0,m=0,s=+12Therefore, the correct option is C

The correct set of four quantum numbers for the valence electrons of rubidium atom $(Z = 37)$ is:
$5, 1, 0, + \frac{1}{2}$
$5, 0, 1, + \frac{1}{2}$
$5, 1, 1, + \frac{1}{2}$
$5, 0, 0, + \frac{1}{2}$

E.C of Rubidium is (Z=37) = $1 s^{2} 2 s^{2} 2 p^{6} 3 s^{2} 3 p^{6} 4 s^{2} 3 d^{10} 4 p^{6} 5 s^{1}$
The valence electron is in 5s orbital.

Its 4 quantum numbers are- $n = 5, l = 0, m = 0, s = + \frac{1}{2}$

Therefore, the correct option is $C$