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