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Byju's Answer

Standard XII

Chemistry

Reactivity with Halogens (Group 1)

I2 dissolve i...

Question

$I_{2}$ dissolve in KI solution due to formation of:

K I_{2}

and

I^{-}

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K^{+}; I^{-}

and

I_{2}

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I_{3}^{-}

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None of the above

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Solution

The correct option is D $I_{3}^{-}$
$I_{2}$ form complex ion $I_{3}^{-}$ in $K I$ solution due to which it dissolves in it.

$K I + I_{2} ⟶ K I_{3}$

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Similar questions

Q. A saturated solution of

I_{2}

in water contains

0.33 g l i t r e^{- 1}

I_{2}

more than this can be dissolved in a

K I

solution because of the following equilibrium:

I_{2} + I^{-} ⇌ I_{3}^{-}

. A

0.1 M K I

solution

(0.1 M I^{-})

actually dissolves

12.5 g / l i t r e I_{2}

, most of which is converted to Assuming that the concentration of

I_{2}

in all saturated solution is the same, calculate the equilibrium constant

(K_{c})

for the above reaction. What is the effect of adding water to clear saturated solution of

I_{2}

K I

solution.

Q. In KI solution,

I_{2}

readily dissolves and forms :

Q. A saturated solution of iodine in water is

1.25 \times 10^{- 3} (M)

. In any saturated solution of

I_{2}

, concentration of

I_{2}

1.25 \times 10^{- 3} (M)

. In

1 L

0.1 (M)

solution of

I^{-}

, it is seen maximum

51.25 \times 10^{- 3} mole

I_{2}

can be dissolved. In the aqueous solution,

I^{-} (a q), I_{2} (a q)

undergo complex formation,

What is the value of

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Q. A saturated solution of iodine in water contains

0.33

g iodine/L. More than this can dissolve in a

K I

solution because of the following equilibrium:

I_{2} (a q) + I^{-} \to I_{3} - (a q)

. A

0.1 M

K I

solution

(0.1 M

I^{-})

actually dissolves

12.5

g of Iodine/L, most of which is converted to

I_{3}^{-} .

Assuming that the concentration of

I_{2}

in all saturated solutions is the same, calculate the equilibrium constant for the above reaction: (Mol. wt. of

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Q. Consider the following statements for the reaction:

K I O_{3} + 5 K I + 6 H C l \to 3 I_{2} + 6 K C l + 3 H_{2} O

1.

I^{-}

is reduced to

I_{2}

2.

{I O_{3}}^{-}

is oxidized to

I_{2}

3.

{I O_{3}}^{-}

is reduced to

I_{2}

4.

Oxidation number of

I

increases from

- 1

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)

The correct statements are:

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I2 dissolve in KI solution due to formation of:

The correct option is D I−3I2 form complex ion I−3 in KI solution due to which it dissolves in it.KI+I2⟶KI3

$I_{2}$ dissolve in KI solution due to formation of:

The correct option is D $I_{3}^{-}$
$I_{2}$ form complex ion $I_{3}^{-}$ in $K I$ solution due to which it dissolves in it.

$K I + I_{2} ⟶ K I_{3}$