Please help with the following equations. Here is my question, please answer in

Question

Please help with the following equations. Here is my question, please answer in a way that is really simple, or detailed, because I am just learning and cannot understand this. 1. I understand that H and I are diatomic. So, with that being said, why does the H in the first equation stay Diatomic, but the I in the second equation does not? I thought the subscripts do not carry over on the product side after the reaction? 2. Please note, the equations are not balanced, I do know how to balance them but of course since I am not getting subscript correct, that will throw the answer off. Completc and lbalance the fallwing chimical ua Actual anu tHO KOH2 not balanced o. 0 balanced) _ actual answer. Na t 1-2-NT

Explanation / Answer

1. Ans:-

It's true that hydrogen exists as H2 and iodine as I2 , but it's valid in their Standard State.

In the first reaction, reactants are K i.e. Potassium (a S-block metal) and water(H2O). Due to the domination of hydration energy over the ionisation energy of potassium, the reaction is violent with evolution of hydrogen gas. In this reaction, first nascent hydrogen ([H]) is formed along with KOH , but as nascent hydrogen is very reactive, it reacts with other nascent hydrogen to form hydrogen gas.

K + H2O = KOH + H

K + H2O = KOH + H. ( H + H -----> H2)

___________________

2K + 2H2O = 2KOH + H2 (balanced chemical reaction)

2. Ans :-

In the second reaction, the reactants are Na (Sodium, which is a S-BLOCK metal) and I2(Iodine, which is a non metal).

The reaction between a metal and a non-metal is ionic in nature.In ionic reaction, the balance of charge is very important.Ionic reaction proceeds through the formation of cation and anion followed by combination of cation and with anion with charge balanced. In this given reaction, 'Na' looses one electron and iodine gains the electron. As Sodium has only one electron in its valence shell, it looses only one electron to attain the nearest stable inert gas configuration.

So, Na ----> Na+ + e-. Also, another Sodium undergoes the same. So, again

. Na -----> Na+ + e-

So, 2Na ------> 2 Na+ + 2e-

Then, iodine being a non-metal, accepts the electrons. So, I2 + 2e- -----> 2I-

( The reason is that I- fulfils the octate conconfiguration. )

Then, the cation Na+  and anion I- , reacts with each other to form a neutral ionic cocompound 'NaI'.

Thus, the overall balanced chemical reaction becomes

2Na + I2 ------> 2NaI

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