Li Be BC N O F Nel In a period from left to right, ionization potential increases. But in case of nitrogen
and oxygen, nitrogen has more ionization potential than oxygen atom due to stable electronic configuration of nitrogen. $7 N=1 s^{2}, 2 s^{2} 2 p^{3} \mid$
$\uparrow\|\| \|$
(Stable configuration) $80=1 s^{2}, 2 s^{2} 2 p^{4}$ (Unstable configuration) Nitrogen has stable configuration as
the p-orbitals are half-filled completely. So, more energy is required to remove an electron from the outermost orbit of nitrogen atom. So, $\mathrm{N}_{2}$, has 14.5 value of ionisation potential.
Generate a complete, print-ready paper with questions like this in minutes — across 16+ boards, with answer keys.
$\mathrm{A}+\mathrm{B} \rightarrow \mathrm{C}$ (Reaction 1$)$
$\mathrm{P} \rightarrow \mathrm{Q}$ (Reaction $2$)
The ratio of the half life of Reaction $1$ : Reaction $2$ is $5: 2$. If $t_1$ and $t_2$ represent the time taken to complete $2 / 3^{\text {dd }}$ and $4 / 5^{\text {dd }}$ of Reaction $1$ and
Reaction $2$, respectively, then the value of the ratio $\mathrm{t}_1: \mathrm{t}_2$ is . . . .$\times 10^{-1}$ (nearest integer).
[Given: $\log _{10}(3)=0.477$ and $\log _{10}(5)=0.699$ ]
$Image$
$1.$ Which one of the following reagents is used in the above reaction?
$(A)$ aq. $\mathrm{NaOH}+\mathrm{CH}_3 \mathrm{Cl}$
$(B)$ aq. $\mathrm{NaOH}+\mathrm{CH}_2 \mathrm{Cl}_2$
$(C)$ aq. $\mathrm{NaOH}+\mathrm{CHCl}_3$
$(D)$ aq. $\mathrm{NaOH}+\mathrm{CCl}_4$
$2.$ The electrophile in this reaction is
$(A)$ : $\mathrm{CHCl}$ $(B)$ ${ }^{+} \mathrm{CHCl}_2$
$(C)$ $\mathrm{CCl}_2$ $(D)$. $\mathrm{CCl}_3$
$3.$ The structure of the intermediate $I$ is
$Image$
Give the answer question $1, 2$ and $3.$