3
{Step 1:} Analyzing the tetrahedral structure of each compound:
\( H_3BO_3 \):
This molecule has a trigonal planar structure, not tetrahedral.
\( [B(OH)_4]^- \):
The structure of this ion is tetrahedral, as boron is surrounded by four OH groups in a tetrahedral arrangement.
\( BH_4^- \):
This molecule also has a tetrahedral structure, as boron is surrounded by four hydrogen atoms in a tetrahedral shape.
\( [BCI_3 \cdot NH_3] \):
The boron atom in this compound is surrounded by three chlorine atoms and one ammonia molecule, resulting in a tetrahedral geometry.
\( SiO_4^{2-}\):
The silicate ion has a tetrahedral shape as the silicon atom is surrounded by four oxygen atoms in a tetrahedral arrangement.Thus, the number of compounds/ions with tetrahedral shape is four.
| Molisch's lest | Barfoed Test | Biuret Test | |
|---|---|---|---|
| A | Positive | Negative | Negativde |
| B | Positive | Positive | Negative |
| C | Negative | Negative | Positive |
The set of elements which obey the general electronic configuration \( (n-1)d^{10} ns^2 \) is:}