NEET Practice Questions (MCQs) with Answers & Solutions

(c) $\underset{\mathrm{X}}{2{\mathrm{MnO}}_2}+ 2{\mathrm{K}}_2{\mathrm{CO}}_3+\underset{\left(\mathrm{air}\right)}{{\mathrm{O}}_2}\stackrel{∆}{\to } \underset{\underset{\left(\mathrm{Y}\right)}{\mathrm{green}}}{2{\mathrm{K}}_2{\mathrm{MnO}}_4}+2{\mathrm{CO}}_2$

$\underset{\mathrm{Y}}{2{\mathrm{K}}_2{\mathrm{MnO}}_4}+{\mathrm{Cl}}_2\stackrel{}{\to } \underset{\left(\mathrm{Z}\right) \mathrm{Pink}}{2{\mathrm{KMnO}}_4}+2\mathrm{KCl}$

(c) $\mathrm{Cu}(\mathrm{CN}{)}_2$ and ${\mathrm{CuI}}_2$ being unstable spontaneously decomposes into $\mathrm{CuCN}\downarrow$ and $\mathrm{CuI}\downarrow$ respectively at room temperature.

(d) Practically ${\mathrm{F}}^{-}$ ion being very weak R.A. has no reducing property.

(b) $$\begin{gathered} \mathrm{NaCl}\left(\mathrm{s}\right)+{\mathrm{K}}_2{\mathrm{Cr}}_2{\mathrm{O}}_7\left(\mathrm{s}\right)+{\mathrm{H}}_2{\mathrm{SO}}_4(\mathrm{few} \mathrm{drops})\underset{\mathrm{reaction}}{\overset{\mathrm{Non} \mathrm{redox}}{\to }} \underset{\mathrm{orange} \mathrm{red} \mathrm{vapours}}{{\mathrm{CrO}}_2{\mathrm{Cl}}_2}\uparrow +\mathrm{HO} \\ {\mathrm{CrO}}_2{\mathrm{Cl}}_2+{\mathrm{OH}}^{-}\left(\mathrm{aq}\right)\to \underset{\mathrm{yellow} \mathrm{solution}}{{\mathrm{CrO}}_4^{2-}}+{\mathrm{Cl}}^{-}+2{\mathrm{H}}_2\mathrm{O} \end{gathered}$$

(D)

     (A) Higher the number of unpaired electrons higher will be the strength of metallic bonds. Hence

          transition elements have higher enthalpies of atomization.

     (B) In Cr the second electrons has to be removed from stable half filled electron configuration i.e. $3{\mathrm{d}}^5.$

           In Cu the second electron has to be removed from stable completely filled electron configuration i.e. $3{\mathrm{d}}^{10}.$

     (C) The sum of ${\mathrm{IE}}_1 \mathrm{and} {\mathrm{IE}}_2$ of Ni is less than that of Pt whereas the sum of ${\mathrm{IE}}_1, {\mathrm{IE}}_2, {\mathrm{IE}}_3 \mathrm{and} {\mathrm{IE}}_4$ are

           reversed.

     (D) Occurs in the middle, e.g., manganese has +7 oxidation state.

(C)

    ${\mathrm{Cr}}^{2+}$ is reducing as its configuration changes from ${\mathrm{d}}^4$ and ${\mathrm{d}}^3,$ the latter having a half-filled ${\mathrm{t}}_{2\mathrm{g}}$ level.

    On the other hand, the change from ${\mathrm{Mn}}^{3+}$ to ${\mathrm{Mn}}^{2+}$ results in the half-filled $\left({\mathrm{d}}^5\right)$ configuration which

    has extra stability.

(D)  ${\mathrm{VOSO}}_4={\mathrm{VO}}^{2+}+{\mathrm{SO}}_4^{2-}; \mathrm{x}+\left(-2\right)=+2=+4.$

(A)

     ${\left[{\mathrm{MnO}}_4\right]}^{2-}$; Mn is in +6 oxidation state. Electronic configuration is ${\left[\mathrm{Ar}\right]}^{18}3{\mathrm{d}}^{1}4{\mathrm{s}}^0.$ As it contains

      one unpaired electron it is paramagnetic and green coloured because of d-d transition in visible

      region. ${\left[{\mathrm{TiCl}}_6\right]}^{2-}, {\left[{\mathrm{VO}}_4\right]}^{3-} \mathrm{and} {\mathrm{CrO}}_2{\mathrm{Cl}}_2$ are diamagnetic as all electrons are paired.

(A)

    (I) ${\mathrm{VCl}}_3;$ ${}_{23}\mathrm{V}_{3+}; {\left[\mathrm{Ar}\right]}^{18}3{\mathrm{d}}^2=2$ unpaired electrons.

    (II) ${\mathrm{VO}}^{2+}; {\mathrm{V}}^{4+}; {\left[\mathrm{Ar}\right]}^{18}3{\mathrm{d}}^2=1$ unpaired electrons.

   (III) ${\left[{\mathrm{VO}}_4\right]}^{3-}; {\mathrm{V}}^{5+}; {\left[\mathrm{Ar}\right]}^{18}3{\mathrm{d}}^0=$ no unpaired electrons.

    (IV) ${\left[\mathrm{V}{\left({\mathrm{H}}_2\mathrm{O}\right)}_6\right]}^{2+}; {\mathrm{V}}^{2+}; {\left[\mathrm{Ar}\right]}^{18}3{\mathrm{d}}^3=3$ unpaired electrons.

(A)  This is due to the increasing stability of the lower species to which they are reduced.