To determine which of the given ions has a tetrahedral shape, we need to analyze the electron pair geometry around the central atom in each ion. This can be done using the Valence Shell Electron Pair Repulsion (VSEPR) theory.
Given ions:
A. [tex]\(SO_3^{2-}\)[/tex]
B. [tex]\(PO_4^{3-}\)[/tex]
C. [tex]\(CO_3^{2-}\)[/tex]
D. [tex]\(CN^-\)[/tex]
Let's analyze each one step-by-step:
### A. [tex]\(SO_3^{2-}\)[/tex]
1. Count the valence electrons of the central atom (Sulfur): Sulfur (S) has 6 valence electrons.
2. Add valence electrons from the oxygen atoms: Each oxygen (O) atom has 6 valence electrons, and there are 3 oxygen atoms, contributing 3 valence electrons (one for each bond), total 3.
3. Add electrons for the charge: The ion has a charge of [tex]\(2^-\)[/tex], adding 2 more electrons.
4. Total electron pairs around the sulfur: Thus, the total number of valence electrons is 6 (from S) + 3 (from three oxygen) + 2 (extra for the charge) = 11 valence electrons pairs.
With 11 valence electron pairs, the arrangement does not provide a tetrahedral shape.
### B. [tex]\(PO_4^{3-}\)[/tex]
1. Count valence electrons of the central atom (Phosphorus): Phosphorus (P) has 5 valence electrons.
2. Add valence electrons from the oxygen atoms: Each oxygen contributes one valence electron for bonding purposes (4 1 electron), total 4.
3. Add electrons for the charge: The ion has a charge of [tex]\(3^-\)[/tex], adding 3 more electrons.
4. Total electron pairs around the phosphorus: Thus, the total number of valence electron pairs is 5 (from P) + 4 (for bonding with four oxygens) + 3 (extra for the charge) = 12 valence electrons pairs around phosphorus.
The molecular geometry with 4 bonding pairs (no lone pairs on Phosphorus) arranges itself in a way that results in a tetrahedral shape.
### C. [tex]\(CO_3^{2-}\)[/tex]
1. Count valence electrons of the central atom (Carbon): Carbon (C) has 4 valence electrons.
2. Add valence electrons from the oxygen atoms: Each oxygen contributes one valence electron for bonding (3 1), total 3.
3. Add electrons for the charge: The ion has a charge of [tex]\(2^-\)[/tex], adding 2 more electrons.
4. Total electron pairs around the carbon: Thus, the total number of valence electron pairs is 4 (from C) + 3 (for bonding with three oxygens) + 2 (extra for the charge) = 9 valence electrons pairs.
With 9 valence electron pairs, the arrangement does not provide a tetrahedral shape.
### D. [tex]\(CN^-\)[/tex]
1. Count valence electrons of the central atom (Carbon): Carbon (C) has 4 valence electrons.
2. Add valence electrons from the nitrogen atom: Nitrogen (N) has 5 valence electrons, but we consider 1 for bonding purposes.
3. Add electrons for the charge: The ion has a charge of [tex]\(1^-\)[/tex], adding 1 more electron.
4. Total electron pairs around the carbon: Thus, the total number of valence electron pairs is 4 (from C) + 1 (from N for bonding) + 1 (extra for the charge) = 6 valence electron pairs.
This results in a linear geometry, not a tetrahedral shape.
### Conclusion
Based on the above analyses, the ion with a tetrahedral shape is B. [tex]\(PO_4^{3-}\)[/tex].
