Answered

Dive into the world of knowledge and get your queries resolved at IDNLearn.com. Our platform offers reliable and comprehensive answers to help you make informed decisions quickly and easily.

Write the empirical formula for at least four ionic compounds that could be formed from the following ions:

[tex] CN^{-}, Fe^{2+}, CrO_4^{2-}, NH_4^{+} [/tex]

A. [tex] (Fe^{2+})(CN^{-})_2 [/tex]
B. [tex] (NH_4^{+})_2(CrO_4^{2-}) [/tex]
C. [tex] (NH_4^{+})(CN^{-}) [/tex]
D. [tex] (Fe^{2+})(CrO_4^{2-}) [/tex]

Sagot :

GinnyAnsweridn
Sure! To write the empirical formulas for ionic compounds formed from the given ions [tex]\(CN^-\)[/tex], [tex]\(Fe^{2+}\)[/tex], [tex]\(CrO_4^{2-}\)[/tex], and [tex]\(NH_4^+\)[/tex], we need to combine these ions to form neutral compounds. Let's determine the possible combinations step-by-step:

### 1. Combination of [tex]\(Fe^{2+}\)[/tex] and [tex]\(CN^-\)[/tex]:

The iron ion ([tex]\(Fe^{2+}\)[/tex]) has a charge of [tex]\(2+\)[/tex], while the cyanide ion ([tex]\(CN^{-}\)[/tex]) has a charge of [tex]\(1-\)[/tex]. To form a neutral compound, we need two cyanide ions to balance the charge of one iron ion.

- Charges: [tex]\(Fe^{2+}\)[/tex] and [tex]\(2 \times CN^{-}\)[/tex]
- Formula: [tex]\(Fe(CN)_2\)[/tex]

The empirical formula is [tex]\( \text{Fe(CN)}_2 \)[/tex].

### 2. Combination of [tex]\(NH_4^+\)[/tex] and [tex]\(CrO_4^{2-}\)[/tex]:

The ammonium ion ([tex]\(NH_4^+\)[/tex]) has a charge of [tex]\(1+\)[/tex], while the chromate ion ([tex]\(CrO_4^{2-}\)[/tex]) has a charge of [tex]\(2-\)[/tex]. To balance the charges, we need two ammonium ions for every one chromate ion.

- Charges: [tex]\(2 \times NH_4^+\)[/tex] and [tex]\(CrO_4^{2-}\)[/tex]
- Formula: [tex]\((NH_4)_2CrO_4\)[/tex]

The empirical formula is [tex]\((\text{NH}_4)_2\text{CrO}_4\)[/tex].

### 3. Combination of [tex]\(Fe^{2+}\)[/tex] and [tex]\(CrO_4^{2-}\)[/tex]:

Both the iron ion ([tex]\(Fe^{2+}\)[/tex]) and the chromate ion ([tex]\(CrO_4^{2-}\)[/tex]) have charges of [tex]\(2+\)[/tex] and [tex]\(2-\)[/tex] respectively. They can combine in a 1:1 ratio to form a neutral compound.

- Charges: [tex]\(Fe^{2+}\)[/tex] and [tex]\(CrO_4^{2-}\)[/tex]
- Formula: [tex]\(FeCrO_4\)[/tex]

The empirical formula is [tex]\( \text{FeCrO}_4 \)[/tex].

### 4. Combination of [tex]\(NH_4^+\)[/tex] and [tex]\(CN^-\)[/tex]:

The ammonium ion ([tex]\(NH_4^+\)[/tex]) has a charge of [tex]\(1+\)[/tex], and the cyanide ion ([tex]\(CN^{-}\)[/tex]) has a charge of [tex]\(1-\)[/tex]. These ions can combine in a 1:1 ratio to form a neutral compound.

- Charges: [tex]\(NH_4^+\)[/tex] and [tex]\(CN^-\)[/tex]
- Formula: [tex]\(NH_4CN\)[/tex]

The empirical formula is [tex]\( \text{NH}_4\text{CN} \)[/tex].

In summary, the empirical formulas for the four ionic compounds that could be formed from the given ions are:

1. [tex]\( \text{Fe(CN)}_2 \)[/tex]
2. [tex]\( (\text{NH}_4)_2\text{CrO}_4 \)[/tex]
3. [tex]\( \text{FeCrO}_4 \)[/tex]
4. [tex]\( \text{NH}_4\text{CN} \)[/tex]
Thank you for using this platform to share and learn. Don't hesitate to keep asking and answering. We value every contribution you make. IDNLearn.com is dedicated to providing accurate answers. Thank you for visiting, and see you next time for more solutions.