# Alkali metal acts as strong reducing agents explain.

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Hint: Reducing agent is an element or compound which loses an electron to an electron recipient in a redox chemical reaction. A reducing agent thus gets oxidized. The reducing agents "reduce" the oxidizing agents. For an element to be a strong reducing agent it should be able to get oxidised easily which means it should lose an electron very easily.

Alkali metals belong to group 1 of the periodic table.
Reducing agents are those compounds or elements which can reduce others and get oxidized themselves.
Alkali Metals act as strong reducing agents, this is because alkali metals have only one valence electron in their valence shell which they can easily lose and attain nearest noble gas configuration and become more stable than they were earlier. Thus they lose an electron and oxidize themselves while reducing others, hence they are called reducing agents.
This makes them a strong reducing agent.Their general outermost electronic configuration is $n{{s}^{1}}$. This means that they have only one electron in the outermost shell.So they can very easily lose this electron to other electron-accepting species.In this way, they themselves get oxidised then and hence act as a reducing agent. Consider the reaction:
$2Na+C{{l}_{2}}\to 2NaCl$
Here, the $Na$ metal is getting oxidised and$Cl$ is getting reduced. $Na$ metal loses one electron and forms$N{{a}^{+}}$ and $Cl$accepts one electron forming .$C{{l}^{-}}$ Thus, here $Na$ metal itself gets oxidised and $Cl$reduces the atom. Hence, $Na$ metal acts as a strong reducing agent.

Note: Group 1 metals in the Modern / long-form periodic tables are called Alkali Metals.
All the alkali metals are mentioned below:
Lithium $\left[ Li \right]$
Sodium $\left[ Na \right]$
Potassium $\left[ k \right]$
Rubidium $\left[ Rb \right]$
Caesium $\left[ Cs \right]$
Francium $\left[ Fr \right]$