D & F Block Elements MCQs for NEET — Chemistry Questions with Answers

(B)

     $$\begin{gathered} 2{\mathrm{CuSO}}_4+4\mathrm{Kl}\stackrel{ }{\to }2{\mathrm{K}}_2{\mathrm{SO}}_4+{\mathrm{Cu}}_2{\mathrm{I}}_2+{\mathrm{I}}_2 \\ 2{\mathrm{Na}}_2{\mathrm{S}}_2{\mathrm{O}}_3+{\mathrm{I}}_2\stackrel{ }{\to }{\mathrm{Na}}_2{\mathrm{S}}_4{\mathrm{O}}_6+2\mathrm{NaI}; {\mathrm{CuI}}_2 \mathrm{is} \mathrm{not} \mathrm{formed} \mathrm{according} \mathrm{to} \mathrm{following} \mathrm{reaction}. \\ 2{\mathrm{CuI}}_2\left(\mathrm{unstable}\right)\stackrel{ }{\to }{\mathrm{Cu}}_2{\mathrm{I}}_2+{\mathrm{I}}_2. \end{gathered}$$

(D)

     (A) ${\mathrm{AgNO}}_3+{\mathrm{PH}}_3\stackrel{ }{\to }\mathrm{Ag}\downarrow +{\mathrm{HNO}}_3+{\mathrm{H}}_3{\mathrm{PO}}_3$

     (B) ${\mathrm{AgNO}}_3\stackrel{\mathbf{Organic}\mathbf{ }\mathbf{tissue}\mathbf{ }\left(\mathbf{proteins}\right)}{\to }\mathrm{Ag}\downarrow \left(\mathrm{black}\right)$

     (C) $\mathrm{Ag}+2\mathrm{KCN}\stackrel{ }{\to }{\left[\mathrm{Ag}{\left(\mathrm{CN}\right)}_2\right]}^{+}$

(C)  ${\mathrm{CO}}_2+{\mathrm{H}}_2\mathrm{O}\stackrel{ }{\to }{\mathrm{H}}_2{\mathrm{CO}}_3\rightleftharpoons 2{\mathrm{H}}^{+}+{\mathrm{CO}}_4^{2-}; {\mathrm{CrO}}_4^{2-}+2{\mathrm{H}}^{+}\rightleftharpoons {\mathrm{Cr}}_2{\mathrm{O}}_7^{2-}\left(\mathrm{orange} \mathrm{red}\right)+{\mathrm{H}}_2\mathrm{O}$

(D)

$4{\mathrm{FeCr}}_2{\mathrm{O}}_4+8{\mathrm{Na}}_2{\mathrm{CO}}_3+7{\mathrm{O}}_2\stackrel{\mathbf{Fusion}}{\to }{\mathrm{Na}}_2\stackrel{\mathbf{+}\mathbf{VI}}{\mathrm{Cr}}{\mathrm{O}}_4\underset{{\mathbf{H}}_{\mathbf{2}}\mathbf{O}}{\overset{{\mathbf{H}}^{\mathbf{+}}}{\to }}{\mathrm{Na}}_2\stackrel{\mathbf{+}\mathbf{VI}}{{\mathrm{Cr}}_2}{\mathrm{O}}_7\stackrel{ {\mathbf{H}}^{\mathbf{+}}\mathbf{/}{\mathbf{H}}_{\mathbf{2}}{\mathbf{O}}_{\mathbf{2}}}{\to }\stackrel{\mathbf{+}\mathbf{VI}}{\mathrm{Cr}}\mathrm{O}{\left({\mathrm{O}}_2\right)}_2\left(\mathrm{deep} \mathrm{blue} \mathrm{violet}\right)\stackrel{{\mathbf{H}}_{\mathbf{2}}\mathbf{O}}{\to }{\mathrm{O}}_2+{\mathrm{H}}_2\mathrm{O}+{\mathrm{Cr}}^{3+}.$

${\mathrm{Na}}_2{\mathrm{Cr}}_2{\mathrm{O}}_7+{\mathrm{H}}_2{\mathrm{O}}_4\stackrel{ }{\to }2{\mathrm{CrO}}_3\left(\mathrm{chronic} \mathrm{anhydride}\right) + {\mathrm{Na}}_2{\mathrm{SO}}_4+{\mathrm{H}}_2\mathrm{O}.$

(A) 

     $2{\mathrm{KMnO}}_4+3{\mathrm{H}}_2{\mathrm{SO}}_4\stackrel{ }{\to }2{\mathrm{KHSO}}_4+{\left({\mathrm{MnO}}_3\right)}_2 {\mathrm{SO}}_4+2{\mathrm{H}}_2\mathrm{O}$

     ${\left({\mathrm{MnO}}_3\right)}_2{\mathrm{SO}}_4+{\mathrm{H}}_2\mathrm{O}\stackrel{ }{\to }{\mathrm{Mn}}_2{\mathrm{O}}_7+{\mathrm{H}}_2{\mathrm{SO}}_4$

      ${\mathrm{Mn}}_2{\mathrm{O}}_7\stackrel{ ∆ }{\to }2{\mathrm{MnO}}_2+3{\mathrm{O}}_2$

(B) ${\mathrm{Mn}}^{3+}={\left[\mathrm{Ar}\right]}^{18} 3{\mathrm{d}}^4, {\mathrm{Cr}}^{3+}={\left[\mathrm{Ar}\right]}^{18} 3{\mathrm{d}}^3, {\mathrm{V}}^{3+}={\left[\mathrm{Ar}\right]}^{18} 3{\mathrm{d}}^2, {\mathrm{Ti}}^{3+}={\left[\mathrm{Ar}\right]}^{18} 3{\mathrm{d}}^1 \mathrm{and} {\mathrm{Cr}}^{3+}$ is most

      stable due to greater CFSE of ${\mathrm{t}}_{2\mathrm{g}}^3$ configuration.

(C) Cupric ion produces blue colour in solution due to the presence of unpaired electron (d-d transition)

      while cuprous ion does not have unpaired electron.

$$\begin{gathered} \left(\mathrm{C}\right) 2\stackrel{+6}{\mathrm{Cr}}{\mathrm{O}}_4^{2-}+2{\mathrm{H}}^{+}\stackrel{+6}{{\mathrm{Cr}}_2}{\mathrm{O}}_7^{2-}+{\mathrm{H}}_2\mathrm{O} \\ \left(\mathrm{yellow}\right) \left(\mathrm{orange}\right) \end{gathered}$$

           There is no change in oxidation state. So the explanation is wrong.

Always transition metals combines with more than one carbonyl group.

By comparing the value of ionization energies the thermodynamic stability of transition metal compound can be compared. Ionization value shows that Ni(II) compound are more stable than Pt(II) compund while Pt(IV) compound are relatively more stable than Ni(IV).