Turnbull-s blue and Prussian -s blue- respectively- are FeII-FeIICN6-2- FeIII-FeIIICN6- I II FeII-FeIIICN6-3- FeIII-FeIICN6-4- III IV -

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Byju's Answer

Standard XII

Physics

Multiplication with Vectors

Turnbull's bl...

Question

Turnbull's blue and Prussian 's blue, respectively, are
$\begin{matrix} F e^{I I} {[F e^{I I} {(C N)}_{6}]}^{2 -} & F e^{I I I} [F e^{I I I} {(C N)}_{6}] I & I I F e^{I I} {[F e^{I I I} {(C N)}_{6}]}^{3 -} & F e^{I I I} {[F e^{I I} {(C N)}_{6}]}^{4 -} I I I & I V \end{matrix}$ :

I, II

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I, III

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III, IV

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IV, III

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Solution

The correct option is B III, IV
Option (C) is correct.
Turnbull's blue is $F e^{I I} {[F e^{I I I} {(C N)}_{6}]}^{3 -}$
Prussian 's blue is $F e^{I I I} {[F e^{I I} {(C N)}_{6}]}^{4 -}$ .

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Similar questions

Q. Turnbull's blue and Prussian's blue respectively are :
(I).

{F e}^{I I} [{F e}^{I I} ({C N)}_{6}]^{2 -}

(II).

{F e}^{I I I} [{F e}^{I I I} ({C N)}_{6}]

(III).

{F e}^{I I} [{F e}^{I I I} ({C N)}_{6}]^{-}

(IV).

{F e}^{I I I} [{F e}^{I I} ({C N)}_{6}]^{-}

Q. Select correct statement(s):
I) When excess

F e C l_{3}

solution is added to

K_{4} [F e (C N)_{5}]

solution, in addition to

F e^{I I I} [F e^{I I} (C N)_{6}]^{-}, F e^{I I} [F e^{I I I} (C N)_{6}]^{-}

is also formed due to side redox reaction
(II) When

F e C l_{2}

is added to

K_{3} [F e (C N)_{6}]

solution, in addition to

F e^{I I} [F e^{I I I} (C N)_{6}]^{-}, F e^{I I I} [F e^{I I} (C N)_{6}]^{-}

is also formed due to side redox reaction
(III)

F e^{I I I} [F e^{I I} (C N)_{6}]^{-}

is paramagnetic while

F e^{I I} [F e^{I I I} (C N)_{6}]^{-}

is diamagnetic
(IV)

F e^{I I I} [F e^{I I} (C N)_{6}]^{-}

is diamagnetic while

F e^{I I} [F e^{I I I} (C N)_{6}]^{-},

is paramagnetic

Q. If equal volumes of

11

K M n O_{4}

and

1

K_{2} C r_{2} O_{7}

solutions are allowed to oxidise

F e (I I)

F e (I I I)

, then

F e (I I)

oxidised will be:

Q. A sample of wustite

{F e}_{x} O

contains one

F e (I I I)

for every three

F e (I I)

. The value of

x

is:

Q. Aqueous

F e (I I I)

ion develops intense red colour with

S C N^{-}

while

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does not. It is due to:

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Turnbull's blue and Prussian 's blue, respectively, areFeII[FeII(CN)6]2−FeIII[FeIII(CN)6]IIIFeII[FeIII(CN)6]3−FeIII[FeII(CN)6]4−IIIIV :

The correct option is B III, IVOption (C) is correct.Turnbull's blue is FeII[FeIII(CN)6]3− Prussian 's blue is FeIII[FeII(CN)6]4−.

Turnbull's blue and Prussian 's blue, respectively, are
$\begin{matrix} F e^{I I} {[F e^{I I} {(C N)}_{6}]}^{2 -} & F e^{I I I} [F e^{I I I} {(C N)}_{6}] I & I I F e^{I I} {[F e^{I I I} {(C N)}_{6}]}^{3 -} & F e^{I I I} {[F e^{I I} {(C N)}_{6}]}^{4 -} I I I & I V \end{matrix}$ :

The correct option is B III, IV
Option (C) is correct.
Turnbull's blue is $F e^{I I} {[F e^{I I I} {(C N)}_{6}]}^{3 -}$
Prussian 's blue is $F e^{I I I} {[F e^{I I} {(C N)}_{6}]}^{4 -}$ .