Chemistry MCQs for NEET — Practice Questions with Answers

(d)       ${\mathrm{NH}}_3$ is a weak reducing agent than ${\mathrm{PH}}_3$, because $\mathrm{X}—\mathrm{H}$ bond strength

            decreases down the group. Due to absence of H-bonding, only weak van

            der waals force of attraction exists in ${\mathrm{PH}}_3$, it has lower critical

             temperature than ${\mathrm{NH}}_3$.

(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.  

(c)

     $$\begin{gathered} \mathrm{Acidic} \mathrm{character}\mathbf{ }\mathbf{:}\mathbf{ }\mathrm{HOCl}<{\mathrm{HClO}}_2<{\mathrm{HCIO}}_3<{\mathrm{HClO}}_4 \\ \mathrm{Oxidising} \mathrm{power} \mathbf{:}\mathbf{ }\mathrm{HOCl}>{\mathrm{HClO}}_2>{\mathrm{HCIO}}_3>{\mathrm{HClO}}_4 \\ \mathrm{Thermal} \mathrm{stability} \mathbf{:}\mathbf{ }\mathrm{HOCl}<{\mathrm{HClO}}_2<{\mathrm{HCIO}}_3<{\mathrm{HClO}}_4 \\ \text{'}\mathrm{Cl}—\mathrm{O}\text{'} \mathrm{bond} \mathrm{order} \mathbf{:} \mathrm{HOCl}<{\mathrm{HClO}}_2<{\mathrm{HCIO}}_3<{\mathrm{HClO}}_4 \end{gathered}$$

(c)

 $\Rightarrow \underset{\underset{\mathbf{Hyb}\mathbf{.}\mathbf{ }\mathbf{of}\mathbf{ }\mathbf{Cl}\mathbf{ }\mathbf{:}\mathbf{ }{\mathbf{sp}}^{\mathbf{3}}}{\mathrm{B}.\mathrm{O}.=1.0}}{\overset{+\mathrm{I}}{{\mathrm{ClO}}^{-}}} \underset{\underset{\mathbf{Hyb}\mathbf{.}\mathbf{ }\mathbf{of}\mathbf{ }\mathbf{Cl}\mathbf{ }\mathbf{:}\mathbf{ }{\mathbf{sp}}^{\mathbf{3}}}{\mathrm{B}.\mathrm{O}.=1.5}}{\overset{+\mathrm{III}}{{\mathrm{ClO}}_2^{-}}} \underset{\underset{\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{Hyb}\mathbf{.}\mathbf{ }\mathbf{of}\mathbf{ }\mathbf{Cl}\mathbf{ }\mathbf{:}\mathbf{ }{\mathbf{sp}}^{\mathbf{3}}}{\mathrm{B}.\mathrm{O}.=1.67}}{\overset{+\mathrm{V}}{{\mathrm{ClO}}_3^{-}}} \underset{\underset{\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{Hyb}\mathbf{.}\mathbf{ }\mathbf{of}\mathbf{ }\mathbf{Cl}\mathbf{ }\mathbf{:}\mathbf{ }{\mathbf{sp}}^{\mathbf{3}}}{\mathrm{B}.\mathrm{O}.=1.75}}{\overset{+\mathrm{VII}}{{\mathrm{ClO}}_4^{-}}}$

(d)    $6{\mathrm{CaOCl}}_2 \underset{\mathbf{oxidation}}{\overset{\mathbf{Auto}}{\to }} \mathrm{Ca}({\mathrm{CIO}}_3{)}_2+5{\mathrm{CaCl}}_2$