A) \[n=2,l=0,\text{ }m=0\]
B) \[n=2,l=1,m=0\]
C) \[n=3,l=1,m=-1\]
D) \[n=3,l=2,\text{ }m=+2\]
E) \[n=4,\text{ }l=0,\text{ }m=0\]
Correct Answer: D
Solution :
\[K(Z=19):\text{ 1}{{s}^{2}},\text{ }2{{s}^{2}}2{{p}^{6}},3{{s}^{2}}3{{p}^{6}},4{{s}^{1}}\] In the ground state the value of I can be either zero or one. Hence, the set (d) of quantum numbers i.e.,\[(n=3,\text{ }Z=2,m=+2)\]cannot possible in the ground state.
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