Metal at the top of reactivity series and metal at the bottom of the reactivity series are :

potassium and gold respectively

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gold and potassium respectively

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sodium and silver respectively

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silver and sodium respectively

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Solution

The correct option is B potassium and gold respectively
The reactivity series follows the order:
$K > N a > C a > M g > A l > Z n > F e > P b > H > C u > H g > A g > A u$
Thus, potassium is the most reactive metal so placed at the top in the reactivity series. Gold is least reactive so, placed at the bottom of reactivity series.

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Similar questions

The metal, gold (Au) is present at the bottom of the reactivity series and potassium (K) is present at the top of the reactivity series.

Statement 1: It is comparatively easier to extract the metals that are present at top of the reactivity series than the metals present at the bottom of the reactivity series.
Statement 2: Metals at the top of the reactivity series are highly reactive in comparison with the metals at the bottom.

Q. The metals sodium, calcium, and magnesium at the top of the reactivity series can be obtained by:

Q. Conceptual Questions
Name the metal which has been placed:
(a) at the top of the reactivity series.
(b) at the bottom of the reactivity series.

The figure represents the reactivity of metals then which of the following are correct?

(i) Sodium is more reactive than magnesium.

(ii) Magnesium ad aluminium metals displace lead.

(iii) Magnesium and aluminium metals displace calcium.

(iv) Iron and tin metals can displace magnesium.

(v) Most reactive metal is sodium.

(vi) Gold is least reactive.

(vii) Calcium and zinc metals displace iron.

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Metal at the top of reactivity series and metal at the bottom of the reactivity series are :

The correct option is B potassium and gold respectivelyThe reactivity series follows the order:K>Na>Ca>Mg>Al>Zn>Fe>Pb>H>Cu>Hg>Ag>AuThus, potassium is the most reactive metal so placed at the top in the reactivity series. Gold is least reactive so, placed at the bottom of reactivity series.