"The boron-nitrogen interaction in the studied molecules shows some similarities with the N→B bond in the H3N-BH3 molecule, formally understood as covalent-dative. ... The results show that all the studied BN bonds are triple, since three two-center orbitals have been obtained."
"Formation of a dative bond or coordinate bond between ammonia and boron trifluoride. When the nitrogen donates a pair of electrons to share with the boron, the boron gains an octet. ... In addition, a pair of non-bonding electrons becomes bonding; they are delocalized over two atoms and become lower in energy."
Answer:
Al³⁺, S²⁻; Al₂S₃
Explanation:
Al is a metal, so it loses its three valence electrons in a reaction to form Al³⁺ ions.
S is a nonmetal. It needs two more electrons to complete its octet, so it tends to form S²⁻ ions.
We can use the criss-cross method to work out the formula of the compound.
The steps are
- Write the symbols of the anion and cation.
- Criss-cross the numbers of the charges to become the subscripts of the other ion.
- Write the formula with the new subscripts.
- Divide the subscripts by their highest common factor.
- Omit all subscripts that are 1.
When we use this method with Al³⁺ and S²⁻, the formula for the compound formed becomes Al₂S₃ .