A) \[\text{ }\!\![\!\!\text{ }Ne\text{ }\!\!]\!\!\text{ }3{{s}^{2}}\text{ }3{{P}^{6}}\text{ }3{{d}^{1}}\text{ }4{{s}^{2}}\]
B) \[\text{ }\!\![\!\!\text{ }Ne\text{ }\!\!]\!\!\text{ }3{{s}^{2}}\text{ }3{{p}^{6}}\text{ }3{{d}^{2}}\text{ }4{{s}^{1}}\]
C) \[\text{ }\!\![\!\!\text{ }Ne\text{ }\!\!]\!\!\text{ }3{{s}^{2}}\text{ }3{{p}^{6}}\text{ }3{{d}^{5}}\text{ }4{{s}^{1}}\]
D) \[~[Ne]\text{ }3{{s}^{2}}\text{ }3{{p}^{6}}\text{ }4{{s}^{2}}\text{ }3{{d}^{4}}\]
Correct Answer: C
Solution :
Key Idea: Atomic number of \[Cr=24\]and its configuration is \[1{{s}^{2}},2{{s}^{2}},2{{p}^{6}},3{{s}^{2}},3{{p}^{6}},4{{s}^{1}},3{{d}^{5}}\] or \[\text{ }\!\![\!\!\text{ }Ne\text{ }\!\!]\!\!\text{ }3{{s}^{2}},\text{ }3{{p}^{6}},\text{ }3{{d}^{5}},\text{ }4{{s}^{1}}\]because exactly full filled orbitals are more stable than nearly fulfilled orbitals.
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