The element that usually does NOT show variable oxidation states is:

We recall that in the transition series the possibility of showing many oxidation states comes from the participation of both the outer $$ns$$ electrons and the inner $$(n-1)d$$ electrons in bonding. Whenever more electrons of similar energy are available, an atom can lose different numbers of them and thus display variable oxidation states.

However, the very first element of a transition series possesses only one $$d$$ electron (electronic configuration $$[Ar]\;3d^{1}4s^{2}$$ for scandium, $$[Kr]\;4d^{1}5s^{2}$$ for yttrium, etc.). After losing all three of these outer-shell electrons, the ion attains the noble-gas core and becomes exceptionally stable. Because no other $$d$$ or $$s$$ electrons of comparable energy remain, the atom cannot normally lose a different number of electrons. Therefore such first-row elements generally show a single common oxidation state of $$+3$$.

Let us examine each option one by one.

For copper, the ground-state configuration is $$[Ar]\;3d^{10}4s^{1}$$. Removing only the $$4s$$ electron gives $$\mathrm{Cu^{+}}$$ with oxidation state $$+1$$, while removing one more $$3d$$ electron gives $$\mathrm{Cu^{2+}}$$ with oxidation state $$+2$$. So copper definitely shows variable oxidation states.

For titanium, the configuration is $$[Ar]\;3d^{2}4s^{2}$$. It can lose $$2,\;3,$$ or $$4$$ electrons, leading respectively to $$\mathrm{Ti^{2+}},\;Ti^{3+},\;Ti^{4+}$$, so titanium also exhibits several oxidation states.

For scandium, the configuration is $$[Ar]\;3d^{1}4s^{2}$$. Losing all three outer electrons gives $$\mathrm{Sc^{3+}}$$ with oxidation state $$+3$$, and this ion has the stable noble-gas configuration $$[Ar]$$. Because removal of any fewer or any additional electrons is highly unfavorable, scandium is found almost exclusively in the $$+3$$ state and thus does not show variable oxidation states.

For vanadium, the configuration is $$[Ar]\;3d^{3}4s^{2}$$. It can form $$\mathrm{V^{2+}},\;V^{3+},\;V^{4+},\;V^{5+}$$, corresponding to oxidation states $$+2,\;+3,\;+4,\;+5$$, clearly demonstrating variability.

Among the four given elements, only scandium lacks such variability.

Hence, the correct answer is Option C.