Formulas for this Timetable
CL- OH- MnO4- NO3- HCO3- SO4^3 - CO3^2- Cr2O7^2- PO4^3-
NH^4+
Na+
CU+
Pb^2+
Au^3+
B^3+
Sn^4+
So what's the question? And note that SO4^2- is correct but SO4^3- is incorrect.
To determine the formulas for the compounds formed between the given cations and anions in the timetable, you need to consider the charges on the ions and balance them to create neutral compounds.
Here's how you can determine the formulas for each compound:
1. NH4+ (Ammonium ion) + Cl- (Chloride ion):
The charges are NH4+ (1+) and Cl- (1-). Since the charges are balanced, the formula will be NH4Cl.
2. Na+ (Sodium ion) + OH- (Hydroxide ion):
The charges are Na+ (1+) and OH- (1-). Since the charges are balanced, the formula will be NaOH.
3. Cu+ (Copper(I) ion) + OH- (Hydroxide ion):
The charges are Cu+ (1+) and OH- (1-). Since the charges are balanced, the formula will be CuOH.
4. Pb^2+ (Lead(II) ion) + OH- (Hydroxide ion):
The charges are Pb^2+ (2+) and OH- (1-). To balance the charges, you need two OH- ions. The formula will be Pb(OH)2.
5. Au^3+ (Gold(III) ion) + OH- (Hydroxide ion):
The charges are Au^3+ (3+) and OH- (1-). To balance the charges, you need three OH- ions. The formula will be Au(OH)3.
6. B^3+ (Boron ion) + OH- (Hydroxide ion):
The charges are B^3+ (3+) and OH- (1-). To balance the charges, you need three OH- ions. The formula will be B(OH)3.
7. Sn^4+ (Tin(IV) ion) + OH- (Hydroxide ion):
The charges are Sn^4+ (4+) and OH- (1-). To balance the charges, you need four OH- ions. The formula will be Sn(OH)4.
It's important to note that some of these compounds may exist as polyatomic ions or complex ions in specific contexts. Additionally, the oxidation state of some elements can vary depending on the compound they form.
To determine the formulas for the ions paired with each element in the given timetable, we need to consider their charges and combine them in a way that balances the charges.
1. NH4+ (Ammonium ion):
The ammonium ion has a charge of +1, which means it needs to combine with an ion that has a charge of -1 to balance the charges. Among the given ions, the OH- ion has a charge of -1. Therefore, the compound NH4OH is formed.
2. Na+ (Sodium ion):
The sodium ion has a charge of +1. It can combine with various negatively charged ions present in the timetable. Let's examine the possibilities:
- NaClO (Sodium hypochlorite): Sodium can combine with the Cl- ion (chloride ion), resulting in NaCl (sodium chloride). The ClO- ion (hypochlorite ion) has a charge of -1, and combining it with Na+ would form NaClO.
- NaMnO4 (Sodium permanganate): Sodium can combine with the MnO4- ion (permanganate ion), which has a charge of -1. Therefore, the compound NaMnO4 is formed.
- NaNO3 (Sodium nitrate): Sodium can combine with the NO3- ion (nitrate ion), which has a charge of -1. Therefore, the compound NaNO3 is formed.
- NaHCO3 (Sodium bicarbonate): Sodium can combine with the HCO3- ion (bicarbonate ion), which has a charge of -1. Therefore, the compound NaHCO3 is formed.
- Na2SO4 (Sodium sulfate): Sodium can combine with the SO4^2- ion (sulfate ion), which has a charge of -2. Therefore, the compound Na2SO4 is formed.
- Na2CO3 (Sodium carbonate): Sodium can combine with the CO3^2- ion (carbonate ion), which has a charge of -2. Therefore, the compound Na2CO3 is formed.
- Na2Cr2O7 (Sodium dichromate): Sodium can combine with the Cr2O7^2- ion (dichromate ion), which has a charge of -2. Therefore, the compound Na2Cr2O7 is formed.
- Na3PO4 (Sodium phosphate): Sodium can combine with the PO4^3- ion (phosphate ion), which has a charge of -3. Therefore, the compound Na3PO4 is formed.
3. Cu+ (Copper(I) ion):
Copper(I) ion has a charge of +1. It can combine with various negatively charged ions in the timetable. Let's examine the possibilities:
- CuCl (Copper(I) chloride): Copper(I) can combine with the Cl- ion (chloride ion). Therefore, the compound CuCl is formed.
- CuNO3 (Copper(I) nitrate): Copper(I) can combine with the NO3- ion (nitrate ion). Therefore, the compound CuNO3 is formed.
4. Pb^2+ (Lead(II) ion):
The Lead(II) ion has a charge of +2. It can combine with various negatively charged ions. Let's examine the possibilities:
- PbCl2 (Lead(II) chloride): Lead(II) can combine with two Cl- ions (chloride ions). Therefore, the compound PbCl2 is formed.
- Pb(NO3)2 (Lead(II) nitrate): Lead(II) can combine with two NO3- ions (nitrate ions). Therefore, the compound Pb(NO3)2 is formed.
5. Au^3+ (Gold(III) ion):
The Gold(III) ion has a charge of +3. It can combine with various negatively charged ions. Let's examine the possibilities:
- AuCl3 (Gold(III) chloride): Gold(III) can combine with three Cl- ions (chloride ions). Therefore, the compound AuCl3 is formed.
6. B^3+ (Boron(III) ion):
The Boron(III) ion has a charge of +3. It can combine with various negatively charged ions. Let's examine the possibilities:
- BCl3 (Boron(III) chloride): Boron(III) can combine with three Cl- ions (chloride ions). Therefore, the compound BCl3 is formed.
7. Sn^4+ (Tin(IV) ion):
The Tin(IV) ion has a charge of +4. It can combine with various negatively charged ions. Let's examine the possibilities:
- SnCl4 (Tin(IV) chloride): Tin(IV) can combine with four Cl- ions (chloride ions). Therefore, the compound SnCl4 is formed.
Note: There is no match for the ion Au^3+ with the given negatively charged ions.