Classification of Elements Part 1

(A) Option A is true. Here in $\mathrm{PbO}_2$ oxidation number of $\mathrm{Pb}$ is $+4$, due to inert pair effect the oxidation number $+4$ is not stable for $\mathrm{Pb}$, So, to become stable oxide $+2\,\mathrm{Pb}$ will oxidize other and reduced to $\mathrm{Pb}^{+2.}$. So, $\mathrm{PbO}_2$ has the highest oxidizing power. (B) Option B is true. Bond length from top to bottom in periodic table increases, So the bond length order is$H > \mathrm{HBr} > \mathrm{HCl} > \mathrm{HF}$ The compound which can give $\mathrm{H}^{+}$ion easily that compound has higher acidic strength. As the bond length between $\mathrm{H}$ and $\mathrm{I}$ in $\mathrm{H}\ \mathrm{I}$ is longest then it will be carrier to break the bond between $\mathrm{HI}$ and $\mathrm{H}$. So $\mathrm{I}$ will have highest acidic strength.$\mathrm{HI}$ Option $(\mathrm{C})$ is false.The structure of those 4 molecules are $\mathrm{C}$ $\overset{\;\;\;..}{N}\overset{}{H}\;\;\;\overset{\;\;\;\;..}{P}\overset{}{H_3}$ Due to the lone pair electron all of them are basic nature. The molecule which can easily donate the lone pair electron will have more basic strength.Here, Nitrogen ( $\overset{..}{As}\overset{}{H_3}\;\;\;\overset{..}{Sb}\overset{}{H_3}$ ) is a element of 2 nd period so in $\mathrm{N}$ new electron is added in the second period and as size of $\mathrm{N}$ is small so electron density is more on the outer shell, so electron become unstable due to repulsion with each other. That is why electron pair is easily removable from $\mathrm{N}$.On the other hand $\mathrm{NH}_3$ is a 3 rd block, As is a 4 th block and $p$ is a 5 th block elements. So, their size is bigger. So, electron density on the outer shell is less and their repulsion also decreases. So, outer shell electrons are more stable and $\mathrm{S}\ \mathrm{b}$ and $\mathrm{PH}_3, \mathrm{AsH}_3$ will have less ability to donate the electron.So the correct order is $\mathrm{SbH}_3$(D) Option (D) is true Electronic configuration of$\mathrm{SbH}_3 < \mathrm{AsH}_3 < \mathrm{PH}_3 < \mathrm{NH}_3$$\;\mathrm{B}(5)=1s^2\ 2s^2\ 2p^1$$\;\mathrm{C}(6)=1s^2\ 2s^2\ 2p^2$$\;\mathrm{N}(7)=1s^2\ 2s^2\ 2p^3$ In general in periodic table from left to right effective nuclear change increase and it becomes more difficult to remove electron from an atom. That is why ionization enthalpy increases.But for those atoms which has half filled or full filled outer most shell, those atoms will be more stable and more energy is needed to remove electron from those atoms. Here $\;\mathrm{O}(8)=1s^2\ 2s^2\ 2p^4$ has stable half filled $\mathrm{N}\ (1s^2\ 2s^2\ 2p^3)$ subshell so it is more stable than $2p$.So, correct order is $\mathrm{O}$