Consider the following transformation - Co H 2 O x -3- Pink complex Excess HCl - CoCl y - n - Blue complex- Find the value of expression x - y - n-

The following reaction takes place with excess HCl. (pink)[Co(H_2O)_6]^3+excess HCl⟶(blue)[CoCl_4]^2 So we can clearly see that x = 6 y = 4 n = 2 Hence x+y/n = ...

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Byju's Answer

Standard XII

Chemistry

Colours in Coordination Compounds

Consider the ...

Question

Consider the following transformation ${[C o (H_{2} O)_{x}]}_{2}^{3 +}$ (Pink complex) $Excess HCl - ----- \to {[C o C l_{y}]}_{y}^{n -}$ (Blue complex). Find the value of expression $\frac{x + y}{n}$ .

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Solution

The following reaction takes place with excess $H C l$ .
$[C o (H_{2} O)_{6}]^{3 +} (p i n k) e x c e s s H C l ⟶ [C o C l_{4}]^{2 -} (b l u e)$
So we can clearly see that
x = 6
y = 4
n = 2
Hence $\frac{x + y}{n} = \frac{6 + 4}{2} = 5$

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Similar questions

Q. Consider the following transformation

{[C o (H_{2} O)_{x}]}_{2}^{3 +}

(Pink complex)

Excess HCl - ----- \to {[C o C l_{y}]}_{y}^{n -}

(Blue complex). Find the value of expression

\frac{x + y}{n}

Q. Consider the following complexes.
(i)

[C r (E D T A)]^{-}

(ii)

[R u (e n)_{3}]^{2 +}

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[P t (g l y)_{2} C l_{2}]^{- 2}

(iv)

C i s - [C r (C l)_{2} (o x)_{2}]^{3 -}

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[P t (e n) (N H_{3})_{2} (p y)_{2}]^{4 +}

Then find out the value of expression

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q = total number of optically inactive complexes

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Q. A (Light pink colour complex)

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Consider the following transformation [Co(H2O)x]3+ (Pink complex) Excess HCl−−−−−−→[CoCly]n− (Blue complex). Find the value of expression x+yn.

The following reaction takes place with excess HCl. [Co(H2O)6]3+(pink)excessHCl⟶[CoCl4]2−(blue) So we can clearly see that x = 6 y = 4 n = 2 Hence x+yn=6+42=5

Consider the following transformation ${[C o (H_{2} O)_{x}]}_{2}^{3 +}$ (Pink complex) $Excess HCl - ----- \to {[C o C l_{y}]}_{y}^{n -}$ (Blue complex). Find the value of expression $\frac{x + y}{n}$ .

The following reaction takes place with excess $H C l$ .
$[C o (H_{2} O)_{6}]^{3 +} (p i n k) e x c e s s H C l ⟶ [C o C l_{4}]^{2 -} (b l u e)$
So we can clearly see that
x = 6
y = 4
n = 2
Hence $\frac{x + y}{n} = \frac{6 + 4}{2} = 5$