An elements is in `M^(3+)` form. Its electronic configuration is `[Ar]3d^(1)` the ion is

Options

A. `Ti^(3+)`
B. `Ti^(4+)`
C. `Ca^(2+)`
D. `Sc^(+)`

Explanation

To determine which element corresponds to the given electronic configuration of \( [Ar] 3d^1 \) in the \( M^{3+} \) form, we can follow these steps:

### Step 1: Identify the Element's Atomic Number
The electronic configuration provided is \( [Ar] 3d^1 \). The \( [Ar] \) indicates that the element has the same electron configuration as Argon, which has an atomic number of 18. The \( 3d^1 \) indicates that there is one electron in the 3d subshell.

### Step 2: Determine the Total Number of Electrons
The total number of electrons for the element can be calculated as follows:
- Argon contributes 18 electrons.
- The \( 3d^1 \) indicates 1 additional electron.

Thus, the total number of electrons = \( 18 + 1 = 19 \).

### Step 3: Identify the Element
The atomic number of the element is 19, which corresponds to Potassium (K). However, since we are looking for the \( M^{3+} \) ion, we need to consider the transition metals.

### Step 4: Consider Transition Metals
The transition metals have their electrons filled in the d-subshell. The element that has a \( 3d^1 \) configuration after losing 3 electrons must be Scandium (Sc), which has an atomic number of 21.

### Step 5: Determine the Ion Form
Scandium has the following electron configuration:
- Neutral Scandium: \( [Ar] 4s^2 3d^1 \)

When Scandium loses 3 electrons to form \( Sc^{3+} \):
- The two \( 4s \) electrons are removed first, followed by the \( 3d \) electron.
- The resulting electronic configuration is \( [Ar] 3d^0 \), but we are interested in the \( 3d^1 \) configuration, which indicates that the element in question is indeed Scandium before ionization.

### Conclusion
Thus, the ion with the electronic configuration \( [Ar] 3d^1 \) in the \( M^{3+} \) form is **Scandium (Sc)**.