The element that usually does not show variable oxidation states is
(a) Sc
(b) Cu
(c) Ti
(d) V

Q: Why doesn't Scandium show a $+2$ oxidation state by just losing its $4s$ electrons?

While it's theoretically possible to lose just the two $4s$ electrons, the remaining single $3d$ electron is very easily lost because removing it results in a highly stable noble gas configuration ($[\text{Ar}]$). The stability gained far outweighs the energy required to remove that third electron.

Q: Is Zinc considered a transition metal if it doesn't show variable oxidation states?

According to IUPAC, a transition metal must have an incompletely filled d-subshell in its neutral state or in any of its common oxidation states. Zinc ($[\text{Ar}] 3d^{10} 4s^2$) has a completely filled d-subshell in both its neutral state and its $+2$ oxidation state, so it is technically not considered a true transition metal.

Q: Why does Copper show a $+2$ state if losing one electron gives it a stable $3d^{10}$ configuration?

Although $\text{Cu}^+$ has a stable $3d^{10}$ configuration, the $\text{Cu}^{2+}$ ion is actually more stable in aqueous solutions. This is because the hydration enthalpy of $\text{Cu}^{2+}$ is much higher (more negative) than that of $\text{Cu}^+$, which more than compensates for the second ionization energy required to remove a $3d$ electron.

Q: Do all elements in group 3 behave like Scandium?

Yes, Yttrium (Y) and Lanthanum (La), which are below Scandium in group 3, also predominantly show a $+3$ oxidation state for the exact same reason: losing three electrons gives them a highly stable noble gas core.

Q: What is the highest oxidation state shown by any 3d transition metal?

Manganese (Mn, $Z=25$) shows the highest oxidation state in the 3d series, which is $+7$. It achieves this by losing all two of its $4s$ electrons and all five of its $3d$ electrons, as seen in compounds like potassium permanganate ($\text{KMnO}_4$).

Question

The element that usually does not show variable oxidation states is

Options:

(A) Sc

(B) Cu

(C) Ti

(D) V

Sigma Percentile · Accepted Answer

The transition metals of the d-block are famous for their ability to exhibit a wide variety of oxidation states. This unique property arises because the energy difference between the $(n-1)d$ and $ns$ orbitals is very small, allowing electrons from both subshells to participate in chemical bonding. However, there are a few notable exceptions to this rule, and understanding them is crucial for mastering d-block chemistry.

### Analyzing the Setup

Let's break down the electronic configurations of the elements given in the options to see why most of them show variable oxidation states.

Titanium (Ti, $Z=22$) has the configuration $[	ext{Ar}] 3d^2 4s^2$. It can lose its two $4s$ electrons to form a $+2$ state, and it can also lose one or both of its $3d$ electrons to form $+3$ and $+4$ states.

Vanadium (V, $Z=23$) has the configuration $[	ext{Ar}] 3d^3 4s^2$. With five valence electrons available, it can exhibit a wide range of oxidation states from $+2$ all the way up to $+5$.

Copper (Cu, $Z=29$) has an anomalous configuration of $[	ext{Ar}] 3d^{10} 4s^1$. It commonly loses its single $4s$ electron to form a $+1$ state, and it can also lose one $3d$ electron to form a more stable $+2$ state in aqueous solutions.

### The Master Concept

Now, let's focus on Scandium (Sc, $Z=21$), the very first element of the 3d transition series. Its electronic configuration is $[	ext{Ar}] 3d^1 4s^2$.

When Scandium forms compounds, it has a very strong tendency to lose all three of its valence electrons (two from the $4s$ orbital and one from the $3d$ orbital) simultaneously.

Why does it do this? Because losing these three electrons leaves Scandium with the highly stable, fully filled noble gas core of Argon ($[	ext{Ar}]$). The energy required to remove the third electron is more than compensated by the immense stability gained from achieving this noble gas configuration.

### Final Conclusion

Because the $	ext{Sc}^{3+}$ ion is so exceptionally stable, Scandium almost exclusively exhibits a **$+3$ oxidation state**. It does not show $+2$ or $+1$ states under normal conditions. Therefore, unlike the vast majority of transition metals, Scandium does not show variable oxidation states.

**Bonus Tip:** Zinc (Zn, $Z=30$) at the end of the 3d series is another classic exception. With a configuration of $[	ext{Ar}] 3d^{10} 4s^2$, it only loses its two $4s$ electrons to form a $+2$ state, as its $3d^{10}$ subshell is completely filled and highly stable. Both Scandium and Zinc are frequent targets in competitive exams for this exact reason!

Animated Solution for Chemistry - d and f-Block Elements: The element that usually does not show variable oxidation states is

Select Answer:

Visualized Solution

Variable Oxidation States

Energy Difference

Titanium ( $Z = 22$ )

Vanadium ( $Z = 23$ )

Copper ( $Z = 29$ )

Scandium ( $Z = 21$ )

Stability of $Sc^{3 +}$

Conclusion & Exceptions

The Sigma Insight: d-block Elements

Analyzing the Setup

The Master Concept

Final Conclusion

Similar Questions

Iron exhibits +2 and +3 oxidation states. Which of the following statements about iron is incorrect?

The common positive oxidation states for an element with atomic number 24, are

In context with the transition elements, which of the following statements is incorrect?

The transition element having least enthalpy of atomisation is

The incorrect statement(s) among (1)-(3) is (are) 1. W(VI) is more stable than Cr(VI). 2. in the presence of HCl, permanganate titrations provide satisfactory results. 3. some lanthanoid oxides can be used as phosphors.

Of the following outer electronic configurations of atoms, the highest oxidation state is achieved by which one of them ?

Which of the following groups of transition metals is called coinage metals?

In which of the following pairs, the outer most electronic configuration will be the same?

The correct order of following $3 d$ -metal oxides, according to their oxidation numbers is (A) $CrO_{3}$ (B) $Fe_{2} O_{3}$ (C) $MnO_{2}$ (D) $V_{2} O_{5}$ (E) $Cu_{2} O$

The set that contains atomic numbers of only transition elements, is