Two different methods to express the equilibrium constant give different results $\pu{2.0M}$ of $\ce{I2}$ and $\pu{2.0M}$ of $\ce{Br2}$ are initially present in a reaction mixture. Calculate the equilibrium concentration of $[\ce{I2}]$ , $[\ce{Br2}]$ and $[\ce{IBr}]$ at $\pu{425K}$. $K_c=100$
Equation: $\ce{I2 + Br2 -&gt; 2IBr }$



1st method

Equation:$\ce{I2 \ \ \ + \ \ \ Br2 -&gt;\ \ \ \ 2IBr}$
Initial: $\ \ \pu{2.0M}\ \ \ \ \ \ \pu{2.0M} \ \ \ \ \ \ \ \ \ \ 0$
Change:   $\ \ x \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ x \ \ \ \ \ \ \ \ \ \ \ \ \ 2x$
Equilib:     $\ \ 2-x \ \ \ \ \ \ \ 2-x \ \ \ \ \ \ \ \ \ \ 2x$
$K_\mathrm c =\frac{(2x)^2}{(2-x)^2}=100$


2nd method what if I separate $2\ce{IBr}$

Equation: $\ce{I2 \ \ \ + \ \ \ Br2 -&gt;\ \ \ IBr \ \ + \ \ \ IBr }$
Initial:     $\pu{2.0M}\ \ \ \ \ \ \ \pu{2.0M}\ \ \ \ \ \ \ \ \ \ \ \ 0 \ \ \ \ \ \ \ \ \ \ \ \ 0$
Change:    $\ \ \ x \ \ \ \ \ \ \ \ \ \ \ \ \ x \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ x \ \ \ \ \ \ \ \ \ \ \ \ \ x$
Equilib:       $2-x \ \ \ \ \ \ 2-x \ \ \ \ \ \ \ \ \ \ \ \ \ \ x \ \ \ \ \ \ \ \ \ \ \ \ x $
$K_\mathrm c =\frac{(x)(x)}{(2-x)^2}=100$
So which one is actually right and which one is wrong? Why?
The idea to use the second method came to my mind because of the the definition:

The equilibrium constant is a ratio of the concentration of the products to the concentration of the reactants

If $\ce{aA +bB \ —&gt; cC + dD}$ (1)
$K_\mathrm c = \frac{[\ce{C}]^c.[\ce{D}]^d}{[\ce{A}]^a.[\ce{B}]^b}$
If I separate : $a\ce{A} = \ce{A}+\ce{A}+...+\ce{A}$ ($\ce{A} a$ times )
Same thing with $b\ce{B}, c\ce{C}$ and $d\ce{D}$, the equation (1) becomes:
$(\ce{A}+\cdots+\ce{A})+(\ce{B}+\cdots+\ce{B}) \ce{—&gt;}(\ce{C}+\cdots+\ce{C})+(\ce{D}+\cdots+\ce{D})$
$K_\mathrm c = \frac{([\ce{C}]\cdots[\ce{C}])([\ce{D}]\cdots[\ce{D}])}{([\ce{A}]\cdots[\ce{A}])([\ce{B}]\cdots[\ce{B}])}$=$ \frac{[\ce{C}]^c.[\ce{D}]^d}{[\ce{A}]^a.[\ce{B}]^b}$ (Proved)
So, by using the second method, how and why is it not right ?
So which one is actually right and which one is wrong? Why?


The second method is wrong.  In your second method, you've written: $\ce{I2 \+ Br2 -&gt;IBr  + IBr }$ and then assigned the concentration $x$ and $x$ to each of the two (identical!) $\ce{IBr}$'s. Till now, it's fine. 

The $K_\mathrm c$ has been used by you as:

$$K_\mathrm c=\frac{[\ce{IBr}][\ce{IBr}]}{[\ce{I2}][\ce{Br2}]}$$

which is okay but a better way to write it would have been 

$$K_\mathrm c=\frac{[\ce{IBr}]^2}{[\ce{I2}][\ce{Br2}]}$$

as you've taken the single concentration term for $\ce{IBr}$ (=$2x$) from the correct formula, split it up into two parts ($x$ and $x$), and then used those two parts separately in $K_\mathrm c$, as if there were not one but two distinct $\ce{IBr}$'s being formed! It is just confusing.

Anyways, the first formula will still work. The main problem here is that $[\ce{IBr}]=\color{blue}{\text{total }}\text{concentration of}~\ce{IBr}$. When you put any reactant/product inside the square brackets, you're supposed to write their total concentration, even if they have been split up in parts. So, when in the derivation of $K_\mathrm c$, you write:

$$K_\mathrm c = \frac{([\ce{C}]\cdots[\ce{C}])([\ce{D}]\cdots[\ce{D}])}{([\ce{A}]\cdots[\ce{A}])([\ce{B}]\cdots[\ce{B}])}= \frac{[\ce{C}]^c.[\ce{D}]^d}{[\ce{A}]^a.[\ce{B}]^b}$$

for the reaction: $$\ce{aA + bB —&gt; cC + dD}$$

each of those individual $[\ce{C}]$'s will refer to the total concentration of $\ce{C}$. That means, as $\ce{C}$ was separately written $c$ times by you in the reaction, the final $[\ce{C}]$ that you use will be the sum of all those $c$ $[\ce{C}]$s. 

So, in the end of your second method, while applying the correct formula for $K_\mathrm c$, you must add those $x$ and $x$ together, and then write:

$$K_\mathrm c=\frac{(x+x)\times(x+x)}{(2-x)\times(2-x)}$$

into the formula you were using.