NEET Practice Questions (MCQs) with Answers & Solutions

(b) 

      $$\begin{gathered} {\mathrm{AgNO}}_3 \stackrel{}{\to } \mathrm{Ag}+\frac{1}{0}{\mathrm{O}}_2+{\mathrm{NO}}_2 \\ 2{\mathrm{NaNO}}_3\stackrel{500°\mathrm{C}}{\to }2{\mathrm{NaNO}}_2+{\mathrm{O}}_2 \\ 4{\mathrm{NaNO}}_3\stackrel{800°\mathrm{C}}{\to }2{\mathrm{Na}}_2\mathrm{O}+5{\mathrm{O}}_2+2{\mathrm{N}}_2 \end{gathered}$$

(b)

     $$\begin{gathered} \underset{\left(\mathrm{A}\right)}{{\mathrm{PCl}}_5}+{\mathrm{H}}_2\mathrm{O}\underset{}{\to }\underset{\left(\mathrm{B}\right)}{{\mathrm{POCl}}_3}+2\mathrm{HCl} \\ \underset{\left(\mathrm{B}\right)}{{\mathrm{POCl}}_3}+3{\mathrm{H}}_2\mathrm{O}\stackrel{}{\to }\underset{\left(\mathrm{C}\right)}{{\mathrm{H}}_3{\mathrm{PO}}_4}+3\mathrm{HCl} \end{gathered}$$

(a)     (F); As the size of halogen atom increases crowding on Si atom will

          increase, hence, tendency of attack of Lewis base decreases.

(b)     (T); M.P. of ${\mathrm{NH}}_3$ is higgest due to intermolecular H-bonding in it.

          Next lower M.P. will be of ${\mathrm{SbH}}_3$ followed by ${\mathrm{AsH}}_3$ due to high mol.wt.

          of ${\mathrm{SbH}}_3$.

(c)      (F); M.P. and M.P. of increase from ${\mathrm{PH}}_3$ to ${\mathrm{SbH}}_3$ via ${\mathrm{AsH}}_3$ due to 

          increase  in mol. wt. ${\mathrm{NH}}_3$ does not follow this trend due to inter 

          molecular H-bonding.

(d)      (T); Value of bond moment decreases.

(a)   ${\mathrm{NF}}_3+{\mathrm{H}}_2\mathrm{O}\underset{}{\to }\mathrm{S} {\mathrm{NCl}}_3+{\mathrm{H}}_2\mathrm{O}\underset{}{\to }{\mathrm{S}}_{{\mathrm{N}}^2}$

(b)    ${\mathrm{P}}_4{\mathrm{O}}_{10}\underset{}{\to }{\mathrm{S}}_{{\mathrm{N}}^{\mathrm{AE}}} {\mathrm{SiCl}}_4\underset{}{\to }{\mathrm{S}}_{{\mathrm{N}}^2}$

(c)    ${\mathrm{SF}}_4, {\mathrm{TeF}}_6\underset{}{\to }\mathrm{Both} {\mathrm{S}}_{{\mathrm{N}}^2}$ 

(d)    ${\mathrm{SiCl}}_4, {\mathrm{SiF}}_4\underset{}{\to }\mathrm{Both} {\mathrm{S}}_{{\mathrm{N}}^2}$

(d)  ${\mathrm{PH}}_3$ (Lewis base) can react with ${\mathrm{B}}_2{\mathrm{H}}_6$ (Lewis acid).

(d)    In ${\mathrm{NCl}}_3$ molecule, ${\mathrm{H}}_2\mathrm{O}$ molecule attacks on less electronegative Cl-atom 

         rather than central N-atom. ${\mathrm{NCl}}_3+3{\mathrm{H}}_2\mathrm{O}\underset{}{\to }{\mathrm{NH}}_3+3\mathrm{HOCl}$ 

(c)  Reducing character $\propto \frac{1}{\text{'}\mathrm{E}—\mathrm{H}\text{'} \mathrm{bond} \mathrm{dissociation} \mathrm{enthalpy}}$

       Hence, correct reducing character is ${\mathrm{NH}}_3<{\mathrm{PH}}_3<{\mathrm{AsH}}_3<{\mathrm{SbH}}_3<{\mathrm{BiH}}_3$.

(c)    M exhibits two oxidation states +3 and +5 but covalency can not be 5;

        hence M can not expand its valence shell. Therefore, M will be nitrogen 

        having atomicity two.  

(a)     a : $2\mathrm{OF}\stackrel{\mathrm{Dimerisation}}{\to }$$\mathrm{F}—\stackrel{-\mathrm{I}}{\mathrm{O}}—\stackrel{-\mathrm{I}}{\mathrm{O}}—\mathrm{F}$

         b : ${\mathrm{H}}_4{\mathrm{P}}_2{\mathrm{O}}_8\stackrel{+{\mathrm{H}}_2\mathrm{O}}{\to }$$2{\mathrm{H}}_3{\mathrm{PO}}_4+\mathrm{H}—\stackrel{-\mathrm{I}}{\mathrm{O}}—\stackrel{-\mathrm{I}}{\mathrm{O}}—\mathrm{H}$

         c : $2\mathrm{Na}\underset{∆}{\overset{\mathrm{excess} {\mathrm{O}}_2}{\to }}$$\mathrm{Na} {}_{+}\mathrm{O}_{-}—\mathrm{O} {}_{-}\mathrm{Na}_{+}$

(c)   ${\mathrm{SO}}_2$ soluble in water. So can not be collected over water.

         ${\mathrm{SO}}_2+{\mathrm{H}}_2\mathrm{O}\stackrel{}{\to }{\mathrm{H}}_2{\mathrm{SO}}_4$