g If attempting to dissolve both silver bromide and silver chloride in aqueous solution through complex ion formation, which dat
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<h2>The compound formed is NaCl and 
</h2>
Explanation:
PART 1:
- Atomic number of sodium is 11.
- Sodium(Na, Z = 11, Group 1A) will lose 1 electron to become
which is isoelectronic to Neon (Ne ,Z = 10).
- The symbol for the sodium ion is
.
- Chlorine atom (
, Z = 17, Group 7A or 17) gains 1 elecron to be isoelectronic to neon.
- The symbol for the compound formed is NaCl.
- PART 2:
- Atomic number of calcium is 20
- Calcium(Ca, Z = 20 ,Group 2A) will lose two electrons to become
which is isoelectronic to argon.
- The symbol for the ion is
- Two chlorine atoms (Cl, Z = 17, Group 7A or 17) each gains one electron to be isoelectronic to argon(Z = 18)
- The symbol for the compound formed is