Question
Match List-I with List-II.
| List-I Electronic configuration of neutral atom (where $\mathbf{n = 2}$ ) |
List-II $\mathbf{1}^{\text{st}\text{~}}$ Ionization |
|---|---|
| A. ${ns}^{2}$ | I. 2080 |
| B. $ns^{2}np^{1}$ | II. 899 |
| C. $ns^{2}np^{3}$ | III. 800 |
| D. $ns^{2}np^{6}$ | IV. 1402 |
List-II
$\mathbf{1}^{\text{st}\text{~}}$ Ionization
Energy ( ${\mathbf{kJ}\text{ }\mathbf{mol}}^{\mathbf{- 1}}$ )
Choose the correct answer from the options given below:
Options
Solution
Sol. [Key Concept: Match ${ns}^{2}$ (Be, $899$), ${ns}^{2}{np}^{1}$ (B, $800$), ${ns}^{2}{np}^{3}$ (N, $1402$) and ${ns}^{2}{np}^{6}$ (Ne, $2080$): the IE order $B < Be < N < Ne$ gives A-II, B-III, C-IV, D-I.]
(A) $2{\text{ }s}^{2} \Rightarrow Be$
(B) $2{\text{ }s}^{2}2p^{1} \Rightarrow \text{ }B$
(C) $2s^{2}2p^{3} \Rightarrow \text{ }N$
(D) $2{\text{ }s}^{2}2p^{6} \Rightarrow Ne$
IE order : $B < Be < N < Ne$
A-II, B-III, C-IV, D-I
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