Predict the position of die element in the periodic table satisfying the electronic configuration (n -1) ${{d}^{1}}$n${{s}^{2}}$ for n = 4

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#1

(n -1) ${{d}^{1}}$n${{s}^{2}}$ for n = 4 becomes 3${{d}^{1}}$4${{s}^{2}}$. It lies in the 4th period and in the 3rd group.