Addition of ${\rm{FeC}}{{\rm{l}}_3}$ to ${{\rm{K}}_4}\left[ {{\rm{Fe}}{{\left( {{\rm{CN}}} \right)}_6}} \right]$ in dilute and cold solution gives :

${\rm{F}}{{\rm{e}}_4}{\left[ {{\rm{Fe}}{{\left( {{\rm{CN}}} \right)}_6}} \right]_3}$ sol

NEET UG CHEMISTRY MCQ: SURFACE CHEMISTRY

1 Easy

1 of 1

Understanding +4 | -1

Addition of

\mathrm{FeC}{\mathrm{l}_3}

{\mathrm{K}_4}\left[ {\mathrm{Fe}{{\left( {\mathrm{CN}} \right)}_6}} \right]

in dilute and cold solution gives :

View answer and explanation

Correct option: D

All of these

Option D (All of these) is correct because the addition of

\mathrm{FeC}{\mathrm{l}_3}

{\mathrm{K}_4} {\mathrm{Fe}{{\left( {\mathrm{CN}} \right)}_6}}

can lead to the formation of Prussian blue sol,

\mathrm{F}{\mathrm{e}_4} {\mathrm{Fe}{{\left( {\mathrm{CN}} \right)}_6}} _3}

sol, and a positive sol.

Practice All Chapter MCQs →

10^{-3} \mathrm{M}

1 \mathrm{~mm}^3

2 \times 10^{13}

\mathrm{N}_{\mathrm{A}}=6 \times 10^{23}

\log \frac{{x}}{{m}}

\log {P}

\theta=45^{\circ}

0.5

\log {k}= 0.699,

\mathrm{g}

\log \;x/m\mathrm{ } \;vs.

\log \;P

19\;\mathrm{g/c}{\mathrm{m}^3}

{1.9 \times {{10}^{ - 4}}\;\mathrm{g}}

10\;\mathrm{ }nm

m{m^3}