Question

Alkali metal (M, group 1) react with water to form aqueous alkali metal hydroxide $\left( {MOH} \right)$ and hydrogen gas as shown by the equation below: $2M(s) + 2{H_2}O \to 2MOH(aq) + {H_2}(g)$
Which element would react most rapidly with water?
A.Lithium (Li)
B.Sodium (Na)
C.Potassium (K)
D.Rubidium (Rb)

Answer 1

Hint: Alkali metals lose one electron when they are reacted with water. Then the element which can lose electrons easily will react most rapidly with water. And the energy required to lose an electron is ionization energy and the element with the least energy reacts most rapidly.

Complete step by step answer:
The reaction of alkali metal with water is shown below:
$2M(s) + 2{H_2}O \to 2MOH(aq) + {H_2}(g)$
Here we can see that alkali metal loses one electron during the above reaction. And the energy required to lose an electron is ionization energy which is dependent on size of atom and screen effect. Ionization energy is inversely proportional to size of atom and screening effect.
And we know that size of atom and screening effect both increase down the group and ionization energy decreases down the group.
The element which can lose electrons easily and make cation react most rapidly with water.
Here ionization energy of Rubidium is minimum and it can lose electrons more easily than all other elements then Rubidium is an element which can react with water rapidly.
Hence the correct answer is option A.

Note:Alkali metals are very reactive and stored in oil to prevent their reaction with water of the water vapor present in air. Alkali metal produces hydrogen gas when reacted with water and hydrogen gas is a very inflammable gas and burns with a blast. So, always alkali metal should be reacted in small amounts.