The electronic configuration of metal M is `1s^(2)2s^(2)2p^(6)3s^(1).` The formula of its oxide will be :

Options

A. MO
B. `M_(2)o`
C. `M_(2)O_(3)`
D. `MO_(3)`

Explanation

To determine the formula of the oxide of the metal M with the electronic configuration `1s^(2)2s^(2)2p^(6)3s^(1)`, we can follow these steps:

### Step-by-Step Solution:

1. **Identify the Metal:**
The given electronic configuration is `1s^(2)2s^(2)2p^(6)3s^(1)`. The outermost shell (n=3) has one electron in the 3s subshell. This indicates that the metal M is in Group 1 of the periodic table, which consists of alkali metals. The metal with this configuration is Sodium (Na).

2. **Determine the Valency of the Metal:**
Sodium (Na) has one electron in its outermost shell (3s^1). To achieve a stable electronic configuration (octet), sodium will lose this one electron, resulting in a +1 oxidation state (Na^+).

3. **Identify the Oxide Ion:**
The oxide ion is derived from oxygen. Oxygen has the electronic configuration `1s^(2)2s^(2)2p^(4)` and needs two additional electrons to complete its octet. Therefore, the oxide ion has a charge of -2 (O^2-).

4. **Combine the Metal and Oxide Ion:**
To form a neutral compound, we need to balance the charges from the sodium cation and the oxide anion. Since Na^+ has a +1 charge and O^2- has a -2 charge, we will need two sodium ions to balance one oxide ion.

5. **Write the Formula:**
The formula for the oxide formed from sodium and oxygen is Na2O. This indicates that two sodium ions combine with one oxide ion to form the compound.

### Final Answer:
The formula of the oxide of metal M is **Na2O**.

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