Match List -I with List - II

Question:

Match List -I with List - II

Choose the most appropriate answer from the options given below -

  1. (a) $\rightarrow$ (ii), (b) $\rightarrow$ (iii), (c) $\rightarrow$ (iv), (d) $\rightarrow$ (i)

  2. (a) $\rightarrow$ (i), (b) $\rightarrow$ (iv), (c) $\rightarrow$ (iii),(d) $\rightarrow$ (ii)

  3. (a) $\rightarrow$ (i), (b) $\rightarrow$ (iii), (c) $\rightarrow$ (iv),(d) $\rightarrow$ (ii)

  4. (a) $\rightarrow$ (iv), (b) $\rightarrow$ (i), (c) $\rightarrow$ (ii), (d) $\rightarrow$ (iii)

Correct Option: 1

Solution:

(a) $1 \mathrm{~s}^{2} 2 \mathrm{~s}^{2} \rightarrow \mathrm{Be}$

(b) $1 \mathrm{~s}^{2} 2 \mathrm{~s}^{2} 2 \mathrm{p}^{4} \rightarrow \mathrm{O}$

(c) $1 \mathrm{~s}^{2} 2 \mathrm{~s}^{2} 2 \mathrm{p}^{3} \rightarrow \mathrm{N}$

(d) $1 \mathrm{~s}^{2} 2 \mathrm{~s}^{2} 2 \mathrm{p}^{1} \rightarrow \mathrm{B}$

The ionization enthalpy order is

$\mathrm{B}<\mathrm{Be}<\mathrm{O}<\mathrm{N}$

Be has more IE compared to B due to extra stability \& $\mathrm{N}$ has more IE compared to oxygen due to extra stability

Hence, $\quad \mathrm{N} \rightarrow 1402 \mathrm{~kJ} / \mathrm{mol}$

$\mathrm{O} \rightarrow 1314 \mathrm{~kJ} / \mathrm{mol}$

$\mathrm{B} \rightarrow 801 \mathrm{~kJ} / \mathrm{mol}$

$\mathrm{Be} \rightarrow 899 \mathrm{~kJ} / \mathrm{mol}$

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