Element 'X' , whose atoms have an outer - shell electronic configuration `ns^(2) np^(4)`, is most likely to react chemically to form ions, which have a charge of ________.

Options

A. `2+`
B. `1+`
C. `1-`
D. `2-`

Answer

To determine the charge of the ion formed by element 'X' with the outer-shell electronic configuration `ns^(2) np^(4)`, we can follow these steps:

### Step 1: Identify the Electronic Configuration
The outer-shell electronic configuration given is `ns^(2) np^(4)`. This means that the element has:
- 2 electrons in the s subshell (ns)
- 4 electrons in the p subshell (np)

### Step 2: Determine the Total Number of Valence Electrons
The total number of valence electrons can be calculated by adding the electrons in the s and p subshells:
- Total valence electrons = 2 (from ns) + 4 (from np) = 6 valence electrons.

### Step 3: Assess Stability and Ion Formation
Elements tend to react in order to achieve a stable electron configuration, often resembling the nearest noble gas configuration. For an element with 6 valence electrons, it is 2 electrons short of achieving a full p subshell (which can hold a maximum of 6 electrons).

### Step 4: Determine the Charge of the Ion
To achieve stability, element 'X' will need to gain 2 electrons to fill its p subshell. When an atom gains electrons, it becomes negatively charged. Therefore, if element 'X' gains 2 electrons, the charge will be:
- Charge = -2 (since it gains 2 electrons).

### Conclusion
Thus, the charge of the ion formed by element 'X' is **-2**.

---