An element M having mass number 27 has 14 neutrons in its nucleus. The formula for oxide of this element will be

Options

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

Explanation

To solve the problem of determining the formula for the oxide of element M with a mass number of 27 and 14 neutrons, we can follow these steps:

### Step 1: Determine the Atomic Number
The atomic number (Z) can be calculated using the formula:
\[ \text{Atomic Number} (Z) = \text{Mass Number} - \text{Number of Neutrons} \]

Given:
- Mass Number (A) = 27
- Number of Neutrons (N) = 14

Substituting the values:
\[ Z = 27 - 14 = 13 \]

### Step 2: Identify the Element
The atomic number 13 corresponds to the element Aluminum (Al).

### Step 3: Determine the Valency of the Element
Aluminum has 3 electrons in its outermost shell (valence shell). Therefore, the valency of Aluminum is +3.

### Step 4: Determine the Valency of Oxygen
Oxygen typically has a valency of -2.

### Step 5: Write the Formula for the Oxide
To combine Aluminum (Al) and Oxygen (O) to form a neutral compound, we need to balance the total positive and negative charges.

Let’s denote the formula for the oxide as \( \text{Al}_x\text{O}_y \):
- The total positive charge from Aluminum: \( +3x \)
- The total negative charge from Oxygen: \( -2y \)

Setting the charges equal for neutrality:
\[ 3x = 2y \]

To find the simplest whole number ratio, we can set \( x = 2 \) and \( y = 3 \):
- This gives \( 3(2) = 2(3) \), which balances.

Thus, the formula for the oxide of Aluminum is:
\[ \text{Al}_2\text{O}_3 \]

### Final Answer
The formula for the oxide of element M is \( \text{Al}_2\text{O}_3 \).

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