ACl2+BCl2⟶ACl4+B↓(Excess)BOHeatB+21O2(A and B are metals.)Their ore of B is :

ACl2+BCl2⟶ACl4+B↓ (Excess)In this reaction, we can see that the oxidation number of A changes from +2 to +4. So, A should be Sn. BOHeat B+21O2The oxidation number of B changes from +2 to 0. B should be less reactive metal. So it is Hg. The ore of Hg is cinnabar. Option (A) is correct.

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ACl_2 +BCl_2 ACl_4 +B (Excess) BO [ ]{ Heat } B+12{ O }_{

Question

$A C l_{2} + BC l_{2} ⟶ A C l_{4} + B ↓$
(Excess)

$BO He a t B + \frac{1}{2} O_{2}$
( $A$ and $B$ are metals.)

Their ore of $B$ is :

cinnabar

azurite

galena

siderite

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A C l_{2} + BC l_{2} ⟶ A C l_{4} + B ↓

(Excess)

BO He a t B + \frac{1}{2} O_{2}

(

A

and

B

are metals.)

Their ore of

B

is :

8 mins ago

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Text solutionVerified

A C l_{2} + BC l_{2} ⟶ A C l_{4} + B ↓

(Excess)

In this reaction, we can see that the oxidation number of A changes from +2 to +4. So, A should be Sn.

BO He a t B + \frac{1}{2} O_{2}

The oxidation number of B changes from +2 to 0. B should be less reactive metal. So it is Hg. The ore of Hg is cinnabar.

Option (A) is correct.

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Question Text	$A C l_{2} + BC l_{2} ⟶ A C l_{4} + B ↓$ (Excess) $BO He a t B + \frac{1}{2} O_{2}$ ( $A$ and $B$ are metals.) Their ore of $B$ is :
Answer Type	Text solution:1
Upvotes	150

ACl2​+BCl2​⟶ACl4​+B↓(Excess)BOHeat​B+21​O2​(A and B are metals.)Their ore of B is :

Text solutionVerified

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$A C l_{2} + BC l_{2} ⟶ A C l_{4} + B ↓$
(Excess)

$BO He a t B + \frac{1}{2} O_{2}$
( $A$ and $B$ are metals.)

Their ore of $B$ is :