
Which of the following ions does not liberate hydrogen gas on reaction with dilute acids?
A. \[C{r^{2 + }}\]
B. \[M{n^{2 + }}\]
C. \[T{i^{2 + }}\]
D. \[{V^{2 + }}\]
Answer
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Hint: The comparison of electronegativities of the first-row transition metals will provide us the correct answer as we go from left to write the electronegativity increases. Thus, Sc is a rather active metal, on the other hand, Cu is much less reactive.
Complete step-by-step answer:
Exceptions to the overall trends are rather dependent on the stability associated with filled and half-filled subshells. The electron configuration of zinc is \[4{s^2}3{d^{10}}\] results in its strong tendency to form the stable \[Z{n^{2 + }}\;\] ion, with the \[3{d^{10}}\;\] electron configuration, whereas \[C{u^ + }\], which also has the \[3{d^{10}}\;\] electron configuration, in first-row transition metal only copper will from such positive ion. Similarly, with a half-filled subshell, \[M{n^{2 + }}\;(3{d^5})\] is much more difficult to oxidize than\[F{e^{2 + }}\;(3{d^6})\]. The chemistry of manganese is therefore primarily that of the \[M{n^{2 + }}\;(3{d^5})\] ion, whereas both the \[F{e^{2 + }}\;and{\rm{ }}F{e^{3 + }}\;\]ions are important in the chemistry of iron. The reactivity of d-block metals increases from \[M{n^{2 + }}\;(3{d^5})\]to\[T{i^{ + 2}}\;\left( {3{d^2}} \right)to\;{V^{ + 2}}\;\left( {3{d^3}} \right)to\;C{r^{ + 2}}\left( {3{d^4}} \right)\]. Therefore, the reactivity of \[M{n^{2 + }}\]is very less. Therefore, \[M{n^{2 + }}\] does not liberate hydrogen gas on reaction with dilute acids.
Hence, the option is (B) \[M{n^{2 + }}\], it does not liberate hydrogen gas on reaction with dilute acids.
Note: Metal reactivity series is the series based on the reactivity of the metals. The more reactive metals are kept at the top of the series followed by the ones with the lower reactivity. The more reactive a metal will be, the more will be its tendency to lose electrons. The metals in the metal reactivity series are arranged in the following manner; Group1 metals > group 2 metals > transition metals > least reactive metals.
Complete step-by-step answer:
Exceptions to the overall trends are rather dependent on the stability associated with filled and half-filled subshells. The electron configuration of zinc is \[4{s^2}3{d^{10}}\] results in its strong tendency to form the stable \[Z{n^{2 + }}\;\] ion, with the \[3{d^{10}}\;\] electron configuration, whereas \[C{u^ + }\], which also has the \[3{d^{10}}\;\] electron configuration, in first-row transition metal only copper will from such positive ion. Similarly, with a half-filled subshell, \[M{n^{2 + }}\;(3{d^5})\] is much more difficult to oxidize than\[F{e^{2 + }}\;(3{d^6})\]. The chemistry of manganese is therefore primarily that of the \[M{n^{2 + }}\;(3{d^5})\] ion, whereas both the \[F{e^{2 + }}\;and{\rm{ }}F{e^{3 + }}\;\]ions are important in the chemistry of iron. The reactivity of d-block metals increases from \[M{n^{2 + }}\;(3{d^5})\]to\[T{i^{ + 2}}\;\left( {3{d^2}} \right)to\;{V^{ + 2}}\;\left( {3{d^3}} \right)to\;C{r^{ + 2}}\left( {3{d^4}} \right)\]. Therefore, the reactivity of \[M{n^{2 + }}\]is very less. Therefore, \[M{n^{2 + }}\] does not liberate hydrogen gas on reaction with dilute acids.
Hence, the option is (B) \[M{n^{2 + }}\], it does not liberate hydrogen gas on reaction with dilute acids.
Note: Metal reactivity series is the series based on the reactivity of the metals. The more reactive metals are kept at the top of the series followed by the ones with the lower reactivity. The more reactive a metal will be, the more will be its tendency to lose electrons. The metals in the metal reactivity series are arranged in the following manner; Group1 metals > group 2 metals > transition metals > least reactive metals.
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