Step 1: Understanding the Question:
Four transition elements from the 3d series are listed, and we need to determine which one displays the greatest range of oxidation states, specifically extending continuously from $+2$ through $+7$.
Step 2: Key Formula or Approach:
The variety of oxidation states a transition element can exhibit depends on the availability of electrons in both the outermost $\mathrm{4s}$ subshell and the unpaired electrons residing in the inner $\mathrm{3d}$ subshell. The highest oxidation state achievable by a $3\text{d}$ element is typically expressed as:
$$\text{Maximum Oxidation State} = (\text{Number of electrons in }\mathrm{4s}) + (\text{Number of unpaired electrons in }\mathrm{3d})$$
Step 3: Detailed Explanation:
Let's analyze the valence electronic configuration of Manganese ($\mathrm{Mn}$, $Z = 25$):
$$\mathrm{Mn} = [\mathrm{Ar}]\, \mathrm{3d^5\, 4s^2}$$ Manganese possesses $2$ electrons in its $\mathrm{4s}$ orbital and $5$ unpaired electrons in its half-filled $\mathrm{3d}$ orbital. Because all $7$ valence electrons can actively engage in bonding, manganese can demonstrate the largest number of oxidation states across the first transition series:
$+2$: e.g., $\mathrm{Mn^{2+}}$ in $\mathrm{MnSO_4}$
$+3$: e.g., $\mathrm{Mn_2O_3}$
$+4$: e.g., $\mathrm{MnO_2}$
$+6$: e.g., $\mathrm{MnO_4^{2-}}$ (Manganate ion)
$+7$: e.g., $\mathrm{MnO_4^-}$ (Permanganate ion)
The other choices like $\mathrm{Cr}$ typically reach a maximum of $+6$, $\mathrm{V}$ reaches $+5$, and $\mathrm{Ni}$ reaches $+4$.
Step 4: Final Answer:
The element showing oxidation states from $+2$ to $+7$ is Manganese ($\mathrm{Mn}$), corresponding to option (A).