Chemistry - PERIODIC TABLE Question with Solution | TestHub

The ionization process is given by:

$M(g)\stackrel{I{P}_1}{\to }{M}^{+}(g)\stackrel{I{P}_2}{\to }{M}^{+2}(g)\stackrel{X}{\to }{M}^{+3}(g)$

The total energy change for $M(g)\to M^{+3}(g)$ is the sum of the individual ionization energies:

$\Delta H=I{P}_1+I{P}_2+X$

We are given $\Delta H=5X$.

So, $I{P}_1+I{P}_2+X=5X$

This simplifies to $I{P}_1+I{P}_2=4X$ (Equation 1)

We are also given the ratio:

$\frac{I{P}_1}{I{P}_2}=\frac{X}{2}$

From this, $I{P}_1=\frac{X}{2}I{P}_2$ (Equation 2)

Substitute Equation 2 into Equation 1:

$\frac{X}{2}I{P}_2+I{P}_2=4X$

$I{P}_2\left(\frac{X}{2}+1\right)=4X$

$I{P}_2\left(\frac{X+2}{2}\right)=4X$

$I{P}_2=\frac{4X\times 2}{X+2}$

$I{P}_2=\frac{8X}{X+2}$

Now, substitute the value of $I{P}_2$ back into Equation 2 to find $I{P}_1$:

$I{P}_1=\frac{X}{2}\times \frac{8X}{X+2}$

$I{P}_1=\frac{8X^2}{2(X+2)}$

$I{P}_1=\frac{4X^2}{X+2}$

Therefore, $I{P}_1=\frac{4X^2}{2+X}$ and $I{P}_2=\frac{8X}{2+X}$.

The final answer is $\boxed{C}$.