xCl₂ + yOH^- → zCl^- + pClO^- Balance the above reaction and find out values of x, y, z and p.

To find the correct values of the stoichiometric coefficients x, y, z, and p, we need to balance the given chemical equation:
$x{\mathrm{Cl}}_2+y{\mathrm{OH}}^{-}\to z{\mathrm{Cl}}^{-}+p{\mathrm{ClO}}^{-}+H_{2}O$

Note: In basic medium, the disproportionation of chlorine (${\mathrm{Cl}}_2$) produces chloride (${\mathrm{Cl}}^{-}$) and hypochlorite (${\mathrm{ClO}}^{-}$) ions, along with water ($H_{2}O$) to balance hydrogen and oxygen.

Let us balance the equation step-by-step using the half-reaction method or by balancing atoms and charges:

1. Balancing Chlorine (Cl) atoms:
The reactant side has $2x$ chlorine atoms.
The product side has $z+p$ chlorine atoms.
Therefore, we have the relation:
$2x=z+p$

2. Balancing Oxygen (O) and Hydrogen (H) atoms:
If we have y hydroxide ions (${\mathrm{OH}}^{-}$) on the reactant side, we have y hydrogen atoms and y oxygen atoms.
To balance the y hydrogen atoms, we must have $\frac{y}{2}$ water molecules ($H_{2}O$) on the product side.
Now, balancing the oxygen atoms:
Reactant side oxygen = y
Product side oxygen = p (from ${\mathrm{ClO}}^{-}$) + $\frac{y}{2}$ (from $H_{2}O$)
Setting them equal:
$y=p+\frac{y}{2}\Rightarrow \frac{y}{2}=p\Rightarrow y=2p$

3. Balancing the net charge:
Total charge on reactant side = $-y$
Total charge on product side = $-z-p$
Equating the charges:
$-y=-z-p\Rightarrow y=z+p$

4. Solving the equations:
From charge balance: $y=z+p$
From oxygen/hydrogen balance: $y=2p$
Equating these two expressions for y:
$2p=z+p\Rightarrow p=z$
Since $2x=z+p$, and we know $p=z$:
$2x=p+p=2p\Rightarrow x=p$

Thus, we have:
$x=p$
$z=p$
$y=2p$

Setting the simplest whole-number value $p=1$, we get:
$x=1$
$y=2$
$z=1$
$p=1$

Substituting these values back into the reaction gives a fully balanced equation:
${\mathrm{Cl}}_2+2{\mathrm{OH}}^{-}\to {\mathrm{Cl}}^{-}+{\mathrm{ClO}}^{-}+H_{2}O$

Therefore, the coefficients are x = 1, y = 2, z = 1, p = 1.