Question
Arrange $N, O$ and $S$ in order of decreasing electron affinity

Answer

a
Electron affinity increases as you add more valence electron. That puts oxygen $(O)$ as having more electron affinity than $( N )$. So, $O \,> \,N$

Electron affinity would typically decrease as you move down the periodic table. But there is a factor in the second period of elements due to the close distance or the orbital from the nucleus so that repulsion of an electron from each other reduce electron affinity.

So, $S \,>\, O\, >\, N$

Need a full question paper?

Generate a complete, print-ready paper with questions like this in minutes — across 16+ boards, with answer keys.

Start Generating Free

Similar questions

Ozonolysis of which one of the following will give two molecules of acetaldehyde
The bond order of $He_2^ + $ molecule ion is
Which among the following reactions is favoured in forward direction by increase of temperature?
Which $IUPAC$ name is incorrectly matched
An organic compound  $X $ on treatment with acidified ${K_2}C{r_2}{O_7}$ gives a compound  $Y $ which reacts with ${I_2}$ and sodium carbonate to form tri-iodomethane. The compound $X $ is
$200\, \mathrm{~mL}$ of $0.2\, \mathrm{M} \mathrm{HCl}$ is mixed with $300\, \mathrm{~mL}$ of $0.1\, \mathrm{M} \mathrm{NaOH}$. The molar heat of neutralization of this reaction is $-57.1 \,\mathrm{~kJ}$. The increase in temperature in ${ }^{\circ} \mathrm{C}$ of the system on mixing is $\mathrm{x} \times 10^{-2}$. The value of $\mathrm{x}$ is ....... . (Nearest integer)

[Given : Specific heat of water $=4.18\, \mathrm{~J} \,\mathrm{~g}^{-1}\, \mathrm{~K}^{-1}$

Density of water $=1.00\, \mathrm{~g}\, \mathrm{~cm}^{-3}$ ]

(Assume no volume change on mixing)

Ozone is prepared by passing silent electric discharge through oxygen. In this reaction
The set of quantum number for the $19^{th}$ electrons in chromium is
 ${H_3}C - \mathop {\mathop {C{\mkern 1mu} {\mkern 1mu}  - {\mkern 1mu} }\limits_{|{\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} \,\,\,{\kern 1pt} {\kern 1pt} {\kern 1pt} }^{|{\kern 1pt} {\kern 1pt} {\kern 1pt} \,\,\,{\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} } }\limits_{C{H_{3\,\,\,}}}^{C{H_3}\,\,\,} Br + KOH(Aq.) \to {H_3}C - \mathop {\mathop {C{\mkern 1mu} {\mkern 1mu}  - {\mkern 1mu} }\limits_{|{\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} \,\,\,{\kern 1pt} {\kern 1pt} {\kern 1pt} }^{|{\kern 1pt} {\kern 1pt} \,\,\,{\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} } }\limits_{C{H_{3\,\,\,}}}^{C{H_{3\,\,\,}}} OH + KBr$   above reaction is
In the reaction  $A_{(g)} + 2B_{(g)}$ $\rightleftharpoons$ $C_{(g)} + Q\,kJ$, greater product will be obtained or the forward reaction is favoured by