The geometry of `SiO_(4)^(4-)` ion is

Options

A. Triangular
B. Tetrahedral
C. Pentagonal bipyramidal
D. Linear

Answer

To determine the geometry of the `SiO4^(4-)` ion, we can follow these steps:

### Step 1: Identify the Ion
The ion in question is `SiO4^(4-)`, which is known as the silicate ion. It consists of one silicon (Si) atom and four oxygen (O) atoms.

**Hint:** Look for the components of the ion and their charges.

### Step 2: Count the Valence Electrons
Silicon has 4 valence electrons, and each oxygen atom has 6 valence electrons. Since there are four oxygen atoms, the total number of valence electrons contributed by oxygen is \(4 \times 6 = 24\). The overall charge of the ion is -4, which means we need to add 4 more electrons to the total count.

Total valence electrons = \(4 + 24 + 4 = 32\) electrons.

**Hint:** Remember to account for the charge of the ion when calculating total valence electrons.

### Step 3: Determine the Central Atom and Bonding
In `SiO4^(4-)`, silicon is the central atom, and it forms bonds with the four oxygen atoms. Each silicon-oxygen bond can be considered a single bond.

**Hint:** Identify the central atom and how many atoms it bonds with.

### Step 4: Analyze the Structure
The arrangement of the four oxygen atoms around the silicon atom leads to a specific geometric shape. Since there are four bonds and no lone pairs on the silicon atom, the geometry will be based on the arrangement of these four bonding pairs.

**Hint:** Consider the VSEPR (Valence Shell Electron Pair Repulsion) theory to predict the geometry based on bonding pairs.

### Step 5: Determine the Geometry
According to VSEPR theory, four bonding pairs of electrons around a central atom will arrange themselves in a tetrahedral geometry to minimize repulsion. Therefore, the geometry of the `SiO4^(4-)` ion is tetrahedral.

**Hint:** Recall the common geometries associated with different numbers of bonding pairs and lone pairs.

### Final Answer
The geometry of the `SiO4^(4-)` ion is tetrahedral.

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