A) 4, 0, 0 \[\frac{1}{2}\]
B) 4, 1, -1, \[\frac{1}{2}\]
C) 3, 2, 2, \[\frac{1}{2}\]
D) 3, 2, -2, \[\frac{1}{2}\]
Correct Answer: A
Solution :
The ground state electronic configuration of Cr is as \[_{24}Cr=[Ar]\,\,3{{d}^{5}},4{{s}^{1}}\] \[\because \] The 19th electron enters in \[4{{s}^{1}}\] subshell. For \[4{{s}^{1}}\] \[n=4,\,\,l=0\] [For 5 orbital, \[l=0\]] \[m=0\] \[[\therefore m=-l\,to\,+l]\] \[s=+1/2\]
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