Question

Common oxidation state of scandium, a transition element is / are
a. \[ + 3\]
b. $ + 1$
c. $ + 2$ and $ + 3$
d.$ + 4$ and $ + 1$

Answer 1

Hint: Scandium has atomic number $21$. According to its electronic configuration, we have to determine which oxidation state will be stable for scandium. It is a silvery white metallic element. It changes its color to slightly pink or yellow when exposed to air.

Complete answer:
Scandium has atomic number $21$. Valence shell configuration is $3{d^1}4{s^2}$. When we calculate the energy for vacant orbitals, we see that $4s$ orbital has less energy. That’s why $4s$ orbital gets filled at first. Then $3d$ orbital gets filled. But in case of fulfilled orbitals, $4s$ orbital has greater energy than $3d$ orbital. That’s why electrons are removed from the $4s$ orbital first. By calculating z effectively, we can easily determine that. If only one electron is removed making the $ + 1$oxidation state, it will not be stable as there will be $3{d^1}4{s^1}$ configuration. If two electrons are removed, only one electron will be remaining in $3d$ orbital. That is not stable. If three electrons are removed, there will be zero electrons in $3d$ and $4s$ orbitals. There will be fulfilled orbitals upto $3p$ orbital. This will be stable. If four electrons are removed, that will not be stable because there will be four electrons in $3p$ orbital. Fulfilled and half-filled configurations are only stable. From this, we can easily tell that $ + 3$ oxidation state is only stable for scandium. In other states, Scandium does not have full filled or half-filled configurations.

Hence, option (A) is the correct answer.

Note:
Thus, it can be concluded that $ + 3$ oxidation state is stable for scandium. The fact that full filled and half-filled valence shell configurations are stable, is applicable for all the elements. Scandium is used to make high intensity lighting and also mercury vapor lamps. The light produced by scandium is similar to sunlight.