A) V > Mn > Cr > Ti
B) Mn > Cr > Ti > V
C) Ti > V > Cr > Mn
D) Cr > Mn > V > Ti
Correct Answer: D
Solution :
| \[{{\,}_{23}}Ti:3{{s}^{2}},4{{s}^{2}}\xrightarrow{I{{E}_{1}}}3{{d}^{2}},4{{s}^{1}}\] |
| \[{{\,}_{23}}V:3{{d}^{3}},4{{s}^{2}}\xrightarrow{I{{E}_{1}}}3{{d}^{3}},4{{s}^{1}}\] |
| \[{{\,}_{24}}Cr:3{{d}^{5}},4{{s}^{1}}\xrightarrow{I{{E}_{1}}}3{{d}^{5}}\xrightarrow[half-filled]{I{{E}_{2}}\,from}\text{maximum}\] |
| \[{{\,}_{25}}Mn:3{{d}^{5}},4{{s}^{2}}\xrightarrow{I{{E}_{1}}}3{{d}^{5}},4{{s}^{1}}\] |
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