Due to lanthanoid contraction, the atomic radii of the second and third row transition elements is almost same. So, they resemble each other much more as compared to first row elements and show similar character.

$E^{\mathrm{s}}$ value of Cu is positive because of the fact that sum of sublimation enthalpy and ionisation enthalpy to convert $\mathrm{Cu}(\mathrm{s})$ to $\mathrm{Cu}^{2+}(\mathrm{aq})$ is so high that it is not compensate by its hydration enthalpy. $E^{\circ}$ value for Zn is negative because of the fact that after removal of electrons from 4 s orbital, stable $3 d^{10}$ configuration is obtained.

As the oxidation state increases, size of the ion of transition element decreases. As per Fajan's rule, as the size of metal ion decreases, covalent character of the bond formed increases.

Therefore, the halide of transition elements become more covalent with increasing oxidation state of the metal.

During filling up of electrons follow $(n+l)$ rule. Here $4 s$ has lower energy than $3 d$ orbital. After the orbitals are filled 4 s goes beyond $3 d$, i.e., $4 s$ is farther from nucleus than $3 d$. So, electron from $4 s$ is removed earlier than from $3 d$.

Reactivity of transition elements depends mostly upon their ionisation enthalpies. As we move from left to right in the periodic table ( Se to Cu ), ionisation enthalpies increase almost regularly. Hence, their reactivity decreases almost regularly from Se to Cu .