The lowest first ionization energy would be associated with which of the following electronic configurations

1s² 2s² 2p⁶ 3s¹

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1s² 2s² 2p⁵

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1s² 2s² 2p⁶

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1s² 2s² 2p⁶ 3s² 3p²

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Solution

The correct option is A
1s² 2s² 2p⁶ 3s¹

Among elements of same period (a & d) ; alkali

metal (a) have least ionization energy due to larger atomic size. b and c belong to II period element. II period

elements have more I.e than III period elements.

Order of I.E. a<d<b<c

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Similar questions

{1 s}^{2} {2 s}^{2} {2 p}^{6} {3 s}^{2} {3 p}^{2}

{1 s}^{2} {2 s}^{2} {2 p}^{6} {3 s}^{2} {3 p}^{1}

{1 s}^{2} {2 s}^{2} {2 p}^{6} {3 s}^{2} {3 p}^{5}

{1 s}^{2} {2 s}^{2} {2 p}^{6} {3 s}^{2} {3 p}^{6} {4 s}^{1}

{(I E)}_{1}

{(I E)}_{4}

The electronic configuration of Ti is 1 $s^{2}$ 2 $s^{2}$ 2 $p^{6}$ 3 $s^{2}$ 3 $p^{6}$ 3 $d^{2}$ 4 $s^{2}$ and it represents

\begin{matrix} List-I & List - II P. & 1 s^{2} & 1. & Element shows highest negative oxidation state Q. & 1 s^{2} 2 s^{2} 2 p^{5} & 2. & Element shows highest first ionisation enthalpy R. & 1 s^{2} 2 s^{2} 2 p^{6} 3 s^{2} 3 p^{5} & 3. & Element shows highest electronegativity on Pauling scale S. & 1 s^{2} 2 s^{2} 2 p^{2} & 4. & Element shows maximum electron gain enthalpy (most exothermic) \end{matrix}

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