A) 16 and 4
B) 12 and 5
C) 12 and 4
D) 16 and 5
Correct Answer: B
Solution :
\[_{24}Cr\to 1{{s}^{2}},\,2{{s}^{2}},\,\underset{l=1}{\mathop{2{{p}^{6}}}}\,,\,3{{s}^{2}},\,\underset{l=1}{\mathop{3{{p}^{6}}}}\,,\underset{l=2}{\mathop{\,3{{d}^{5}}}}\,,\,4{{s}^{1}}\] (We know that for p the value of \[l=1\] and for \[d,\] \[l=2)\] For \[l=1\] total number of electron = 12 For \[l=2\] total number of electron = 5.You need to login to perform this action.
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