
When $KMn{O_4}$ reacts with $KBr$ in alkaline medium gives bromated ions. Then oxidation state of $Mn$ changes from $ + 7$ to:
A.$ + 6$
B. $ + 4$
C. $ + 3$
D. $ + 2$
Answer
550.5k+ views
Hint: The oxidation number is known as oxidation state, in which the atoms either gains electrons or loses its electrons to form a chemical bond with another atom. When permanganate ions react with halogen in an alkaline medium the oxidation state of manganese changes due to the formation of bond pairs between the outermost electrons of the metal ion.
Complete step-by-step answer:
We are provided with $2.76g$ of silver carbonate which is strongly ignited. Now, when Silver Carbonate is strongly ignited, it leaves a residue of Silver while gases like Carbon Dioxide and Oxygen get released into air.
\[A{g_2}C{O_3}(s) \to 2Ag(s) + C{O_2}(g) + \dfrac{1}{2}{O_2}(g)\]
We know that,
Molecular weight of Silver Carbonate (\[A{g_2}C{O_3}\]) = \[
2 \times {\text{Molecular Weight of Ag + 1}} \times {\text{Molecular Weight of Carbon + 3}} \times {\text{Molecular Weight of Oxygen }} \\
{\text{ = 2}} \times {\text{108 + 1}} \times {\text{12 + 3}} \times {\text{16 = 276g}} \\
\]
From $276g$ of Silver Carbonate (\[A{g_2}C{O_3}\]) we get $ 216g$ of Silver (\[Ag\])
Therefore, from$2.76g$ of \[A{g_2}C{O_3}\] on strong ignition, we will get \[\dfrac{{216}}{{276}} \times 2.76g = 2.16g{\text{ Ag}}\] as residue.
Thus, $2.76g$ of Silver Carbonate \[A{g_2}C{O_3}\] will give $2.16g$ of \[Ag\]
Hence, the correct option for the given question is $b)2.16g$
Additional information:
Unlike other metal carbonates that usually decompose into metal oxides liberating carbon dioxide, silver carbonate on strong ignition decomposes into elemental silver liberating mixture of carbon dioxide and oxygen gas which mixes with air, leaving behind a residue of silver.
Note: Before solving the sum, one should be properly acquainted with the concept of decomposition of metal oxides. This sum can be easily solved, if the student understands the question properly and realizes which the reaction formed during the decomposition of silver carbonate. Thus, before solving such questions students need to keep all this in mind.
Complete step-by-step answer:
We are provided with $2.76g$ of silver carbonate which is strongly ignited. Now, when Silver Carbonate is strongly ignited, it leaves a residue of Silver while gases like Carbon Dioxide and Oxygen get released into air.
\[A{g_2}C{O_3}(s) \to 2Ag(s) + C{O_2}(g) + \dfrac{1}{2}{O_2}(g)\]
We know that,
Molecular weight of Silver Carbonate (\[A{g_2}C{O_3}\]) = \[
2 \times {\text{Molecular Weight of Ag + 1}} \times {\text{Molecular Weight of Carbon + 3}} \times {\text{Molecular Weight of Oxygen }} \\
{\text{ = 2}} \times {\text{108 + 1}} \times {\text{12 + 3}} \times {\text{16 = 276g}} \\
\]
From $276g$ of Silver Carbonate (\[A{g_2}C{O_3}\]) we get $ 216g$ of Silver (\[Ag\])
Therefore, from$2.76g$ of \[A{g_2}C{O_3}\] on strong ignition, we will get \[\dfrac{{216}}{{276}} \times 2.76g = 2.16g{\text{ Ag}}\] as residue.
Thus, $2.76g$ of Silver Carbonate \[A{g_2}C{O_3}\] will give $2.16g$ of \[Ag\]
Hence, the correct option for the given question is $b)2.16g$
Additional information:
Unlike other metal carbonates that usually decompose into metal oxides liberating carbon dioxide, silver carbonate on strong ignition decomposes into elemental silver liberating mixture of carbon dioxide and oxygen gas which mixes with air, leaving behind a residue of silver.
Note: Before solving the sum, one should be properly acquainted with the concept of decomposition of metal oxides. This sum can be easily solved, if the student understands the question properly and realizes which the reaction formed during the decomposition of silver carbonate. Thus, before solving such questions students need to keep all this in mind.
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