`H_2S + 4F_2 to 2HF + B`
The shape of molecule of compound B is

Options

A. Octahedral
B. Tetrahedral
C. Trigonal planar
D. linear

Answer

To determine the shape of the molecule of compound B formed in the reaction \( H_2S + 4F_2 \rightarrow 2HF + B \), we will follow these steps:

### Step 1: Identify Compound B
In the given reaction, we start with \( H_2S \) and \( F_2 \). The \( H_2S \) contains sulfur (S) and the \( F_2 \) contains fluorine (F). The products include \( 2HF \) and an unknown compound B.

### Step 2: Analyze the Reaction
From the reaction, we can see that sulfur reacts with fluorine. The \( H_2S \) provides the sulfur atom, and the \( 4F_2 \) provides fluorine atoms.

### Step 3: Count the Fluorine Atoms
In total, we have 4 molecules of \( F_2 \), which means there are \( 4 \times 2 = 8 \) fluorine atoms available. Since \( 2HF \) is produced, which uses 2 fluorine atoms, we have:
\[ 8 - 2 = 6 \]
This means 6 fluorine atoms remain to combine with sulfur.

### Step 4: Determine Compound B
The remaining 6 fluorine atoms will combine with the sulfur atom to form \( SF_6 \) (sulfur hexafluoride). Therefore, compound B is \( SF_6 \).

### Step 5: Determine the Shape of SF6
The molecular geometry of \( SF_6 \) can be determined by considering the arrangement of the fluorine atoms around the sulfur atom. In \( SF_6 \):
- Sulfur is the central atom.
- There are 6 fluorine atoms bonded to the sulfur atom.

The arrangement of these bonds is such that the molecule adopts an octahedral shape, where the fluorine atoms are positioned at the vertices of an octahedron.

### Conclusion
The shape of the molecule of compound B (SF6) is **octahedral**.

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