Why first ionisation enthalpy of $\mathrm{Cr}$ is lower than that of $\mathrm{Zn}$ ?

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${ }_{24} \mathrm{Cr}:[\mathrm{Ar}] 3 d^{5} 4 s^{1} \quad{ }_{30} \mathrm{Zn}:[\mathrm{Ar}] 3 d^{10} 4 s^{2}$

After loss of one element, $\mathrm{Cr}$ attains stable $\left(d^{5}\right)$ configuration while in zinc, the electron has to be removed from $4 s$ orbital. Hence, first ionization enthalpy of $\mathrm{Cr}$ is lower than $\mathrm{Zn}$.

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Which of the following statements are correct about $\mathrm{Zn}, \mathrm{Cd}$ and $\mathrm{Hg}$ ?

$A$. They exhibit high enthalpy of atomization as the d-subshell is full.

$B$. $\mathrm{Zn}$ and $\mathrm{Cd}$ do not show variable oxidation state while $\mathrm{Hg}$ shows $+\mathrm{I}$ and + $II.$

$C$. Compounds of $\mathrm{Zn}, \mathrm{Cd}$ and $\mathrm{Hg}$ are paramagnetic in nature.

$D$. $\mathrm{Zn}, \mathrm{Cd}$ and $\mathrm{Hg}$ are called soft metals.

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