$[F e (C N)_{6}]^{3 -}$ is less stable than $[F e (C N)_{6}]^{4 -}$
If true 1 else 0.
0

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Solution

The correct option is A 0
From their stability complex constant, it can be seen that $[F e (C N)_{6}]^{3 -}$ is more stable than $[F e (C N)_{6}]^{4 -}$

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

is more stable than

[F e (C N)_{6}]^{4 -}

If true 1 else 0.

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

is more stable than

[F e (C N)_{6}]^{4 -}

Which complex is more stable between

[F e (C N)_{6}]^{4 -}

and

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

Which of the following statements is/are true for the complexes, $[F e (H_{2} O)_{6}]^{2 +}, [F e (C N)_{6}]^{3 -}, [F e (C_{2} O_{4})_{3}]^{3 -}$ and $[F e (C O)_{5}]$ ?

{F e}^{2 +}

gives blue colour, called Turnbull's blue with

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

While

{F e}^{3 +}

gives blue colour, called Prussian blue with

K_{4} [{F e (C N)}_{6}]

If the given statement is true enter 1 else enter 0.

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[Fe(CN)6]3− is less stable than [Fe(CN)6]4−If true 1 else 0.0

The correct option is A 0From their stability complex constant, it can be seen that [Fe(CN)6]3− is more stable than [Fe(CN)6]4−

$[F e (C N)_{6}]^{3 -}$ is less stable than $[F e (C N)_{6}]^{4 -}$
If true 1 else 0.
0

The correct option is A 0
From their stability complex constant, it can be seen that $[F e (C N)_{6}]^{3 -}$ is more stable than $[F e (C N)_{6}]^{4 -}$