Question

The four quantum numbers of the valence electron of potassium are :

• A. $4,0,1,\frac{1}{2}$
• B. $4,1,0,\frac{1}{2}$
• C. $4,0,0,\frac{1}{2}$
• D. $4,1,1,\frac{1}{2}$

Answer 1

Answer: (C)

$4,0,0,\frac{1}{2}$

Answer 2

The electronic configuration of potassium (At. no. 19) is: $1{{s}^{2}},2{{s}^{2}}2{{p}^{6}},3{{s}^{2}}3{{p}^{6}},4{{s}^{1}}$ The valence electron is present in $4{{s}^{1}}.$ Hence, quantum numbers for $4{{s}^{1}}$ are : n (Principal quantum no.) = 4 $l$ (Azimuthal quantum no.) = 0 Note: for s sub shell $l=0$ for p sub shell $l=1$ for d sub shell $l=2$ for $f$ sub shell $l=3$ m (magnetic quantum no.) = 0 The value of m depends upon $l.$ s (spin quantum no.) $=+\frac{1}{2}.$