Classification of Elements and Periodicity in Properties MCQ Questions & Answers in Inorganic Chemistry | Chemistry

Phosphorus has vacant $$'d'$$ - orbitals due to which it has higher electron affinity than nitrogen. As also has vacant $$d$$ - orbitals but its atomic size is very big thus its electron affinity will be lesser than $$N.$$

On passing from left to right in a period, acidic character of the normal oxides of the elements increases with increase in electronegativity.

For isoelectronic species, size of anion increases as negative charge increases. Thus the correct order is
$$\eqalign{ & \,\,\,\,{N^{3 - }} > {O^{2 - }} > {F^ - } \cr & \left( {1.71} \right)\,\left( {1.40} \right)\,\,\,\left( {1.36} \right) \cr} $$

It is difficult to remove an electron from a highly electronegative element.

$${O^{ - - }}\,{\text{and}}\,{F^ - }$$   are isoelectronic. Hence have same number of shells, therefore greater the nuclear charge smaller will be the size i.e. $$\,{O^{ - - }}\, > \,{F^ - }$$
further $$L{i^ + }\,{\text{and}}\,{B^{3 + }}$$   are isoelectronic. therefore $$L{i^ + }\, > {B^{3 + }}$$
Hence the correct order of atomic size is $${O^{ - - }}\, > \,{F^ - }\, > \,L{i^ + }\, > {B^{3 + }}$$

Non-metals easily gain electrons and hence, they form negative ions, so they are electronegative in nature.

$$N,O,F$$   are more electronegative element, so they accept electrons more easily and form negative ions ( anions ).

The valency of $$Mg$$  is + 2, and $$N$$ is - 3.
The formula becomes $$ \to M{g_3}{N_2}.$$
The valency of $$Si$$  is + 4, and $$O$$ is - 2
The formula becomes $$ \to Si{O_2}.$$

No explanation is given for this question. Let's discuss the answer together.

$$\eqalign{ & A \Rightarrow H\left( {1{s^1}} \right) \cr & B \Rightarrow He\left( {1{s^2}} \right) \cr & C \Rightarrow Li\left( {1{s^2}2{s^1}} \right) \cr & {A_1} = I{E_1}\left( A \right)\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,{B_2} = I{E_2}\left( B \right) \cr & {B_1} = I{E_1}\left( B \right)\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,{C_2} = I{E_2}\left( C \right) \cr & {C_1} = I{E_1}\left( C \right)\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,{C_3} = I{E_3}\left( C \right) \cr & {B_1} > {A_1} > {C_1}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,{C_3} > {B_2} > {A_1} \cr & He > H > Li\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,L{i^{2 + }}H{e^ + }H \cr & 1{s^2}\,\,1{s^1}\,\,2{s^1}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,1{s^1}\,\,1{s^1}\,\,1{s^1} \cr & {C_3} > {C_2} > {B_2}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,L{i^{2 + }}\,L{i^ + }\,\,H{e^ + } \cr & 1{s^2}\,\,1{s^2}\,\,1{s^1} \cr} $$