- ✓$S > O > N$
- B$O > S > N$
- C$N > O > S$
- D$S > N > O$
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, $S \,>\, O\, >\, N$
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$SO_3(g) \rightleftharpoons SO_2(g)+ \frac{1}{2} O_2(g)$
is $K_c= 4.9 \times 10^{-2}.$ The value of $K_c$ for the reaction
$2SO_2(g) + O_2(g) \rightleftharpoons 2SO_3(g)$
will be

Given : $K _{ a }\left( CH _{3} CH _{2} COOH \right)=1.3 \times 10^{-5}$
Assertion $A$ :Magnesium can reduce $Al _{2} O _{3}$ at a temperature below $1350^{\circ} C$, while above $1350^{\circ} C$ aluminium can reduce $MgO$.
Reason $R$ : The melting and boiling points of magnesium are lower than those of aluminium.
In light of the above statements. choose most appropriate answer from the options given below