A) V
B) Cr
C) Mn
D) Fe
Correct Answer: B
Solution :
\[V=1{{s}^{2}}2{{s}^{2}}2{{p}^{6}}3{{s}^{2}}3{{p}^{6}}3{{d}^{3}}4{{s}^{2}}\] \[Cr=1{{s}^{2}}2{{s}^{2}}2{{p}^{6}}3{{s}^{2}}3{{p}^{6}}3{{d}^{5}}4{{s}^{1}}\] \[Mn=1{{s}^{2}}2{{s}^{2}}2{{p}^{6}}3{{s}^{2}}3{{p}^{6}}3{{d}^{5}}4{{s}^{2}}\] \[Fe=1{{s}^{2}}2{{s}^{2}}2{{p}^{6}}3{{s}^{2}}3{{p}^{6}}3{{d}^{6}}4{{s}^{2}}\] In second ionization enthalpy \[C{{r}^{+}}\] has exact half filled d-sub shell.
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