Explain why cesium is more reactive than Rubidium or Barium?

Answer 1

if you look in the periodic table of elements cesium is in the period 1 group and the lower the element in that group the more reactive it is. so francium would be more reactive than cesium. another area where it's reactive is the last coloumn on the top left.

Answer 2

maybe the nuclear force of cesium in their electons is smaller than the nuclear force of rubidium and barium in their electrons....so the electrons of cesium is more easy to be lost and to participate of the reactions

Answer 3

Reactivity Of Barium

Answer 4

Barium is a silvery white metal. The surface of barium metal is covered with a thin layer of oxide that helps protect the metal from attack by air, but to a lesser extent than the corresponding layer in magnesium. Once ignited, barium metal burns in air to give a mixture of white barium oxide, BaO, and barium nitride, Ba3N2. Barium oxide is more normally made by heating barium carbonate. The superoxide BaO2 apparently is also formed in this reaction. Barium, three places below magnesium in the periodic table is more reactive with air than magnesium.

2Ba(s) + O2(g) → 2BaO(s)

Ba(s) + O2(g) → BaO2(s)

3Ba(s) + N2(g) → Ba3N2(s)

Reaction of barium with water
Barium reacts readily with water to form barium hydroxide, Ba(OH)2 and hydrogen gas (H2). The reaction is quicker than that of strontium (immediately above barium in the periodic table) but probably slower than that of radium (immediately below barium in the periodic table).

Ba(s) + 2H2O(g) → Ba(OH)2(aq) + H2(g)

Reaction of barium with the halogens
I expect barium is very reactive towards the halogens, but I've not yet found explicit references to that effect. So, it is epxected that chlorine, Cl2 bromine, Br2, or iodine, I2, would burn to form the dihalides barium(II) chloride, BaCl2, barium(II) bromide, BaBr2, and barium(II) iodide, BaI2 respectively. The reactions with bromine and with iodine wolud probably require heat.

Ba(s) + Cl2(g) → BaCl2(s)

Ba(s) + Br2(g) → BaBr2(s)

Ba(s) + I2(g) → BaI2(s)

Answer 5

The simplest answer is that it has the lowest first ionisation energy, so it loses its outer electron the most readily.