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Lemur [1.5K]
3 years ago
13

Write the empirical formula of at least four binary ionic compounds that could be formed from the following ions: Zn2+, Ni4+, F-

, O2-

Chemistry
2 answers:
hram777 [196]3 years ago
8 0

Ionic compounds are formed between oppositely charged ions.

A binary ionic compound is composed of ions of two different elements - one of which is a positive ion(metal), and the other is negative ion (nonmetal).

To write the empirical formula of binary ionic compound we must remember that one ion should be positive and other ion should be negative, then only the correct formula should be written. To write the empirical formula the charges of opposite ions should be criss-crossed.

First empirical formula of binary ionic compound is written betweenZn^{2+} (Positive ion)and F^{-} (Negative ion)

First Formula would be ZnF_{2}

Second empirical formula is between Zn^{2+}(Positive ion) and O^{2-}(Negative ion)

Second Formula would be Zn_{2}O_{2}

Note : When the subscript are same they get cancel out, so Zn_{2}O_{2} would be written as ZnO

Third empirical formula is between Ni^{4+}(Positive ion) and F^{-}(Negative ion)

Third Formula would be :NiF_{4}

Forth empirical formula is between Ni^{4+}(Positive ion)and O^{2-}(negative ion)

Forth Formula would be : Ni_{2}O_{4} or NiO_{2}

Note- The subscript will be simplified and the formula will be written as NiO_{2}.

The empirical formula of four binary ionic compounds are : ZnF_{2}, ZnO, NiF_{4},NiO_{2}


7nadin3 [17]3 years ago
5 0

The empirical formula of at least four binary ionic compounds :

1. ZnF₂

2. Ni₂O₄ ---.>NiO₂

3. ZnO

4. NiF₄

<h3>Further explanation</h3>

Ions are atoms that have a charge

Ions are formed because events are capturing or releasing electrons.

Ions can be differentiated based on the number of constituent elements they become

  • 1. monatomic ion, if it comes from one element, such as K⁺, Mg²⁺, Cl⁻
  • 2. polyatomic ion, if it comes from two or more elements such as PO₄³⁻, SO₄²⁻, ClO₃⁻

Based on the charge, ions can be divided into:

  • 1. positive ion (cation)

happens if the atom releases electrons, for example

Al³⁺, ---> Al⇒ Al³⁺ + 3e⁻

Pb²⁺, ----> Pb⇒ Pb²⁺ + 2e-

  • 2. negative ion (anion)

happens if the atom captures electrons, like

S²⁻, S + 2e⁻⇒ S²⁻

F⁻, F + e⁻⇒ F⁻

For ionic compounds consisting of cations (ions +) and anions (ions -)

The naming is preceded by the cation then the anion with the -ide suffix added

Generally for this ion compound consists of metal elements as cations and non-metals as anions

Binary ionic compounds are formed from 2 different atomic elements, namely metal and non-metal

Metals as cations are positively charged ions.

Nonmetals as anions are negatively charged ions.

Cation and anion charge numbers are placed crosswise.

From some of these ions

Possible compounds from the incorporation of these ions are ions

1. ZnF₂

2. Ni₂O₄ ---.>NiO₂

3. ZnO

4. NiF₄

<h3>Learn more</h3>

The net ionic equation

brainly.com/question/8885824

ion stability

brainly.com/question/1781595

ionic bonding

brainly.com/question/1603987

Keywords: binary ionic compounds, elements, cations, anions, polyatomic, monoatomic

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<h3>Answer:</h3>

4 g AgCl

<h3>General Formulas and Concepts:</h3>

<u>Math</u>

<u>Pre-Algebra</u>

Order of Operations: BPEMDAS

  1. Brackets
  2. Parenthesis
  3. Exponents
  4. Multiplication
  5. Division
  6. Addition
  7. Subtraction
  • Left to Right

<u>Chemistry</u>

<u>Stoichiometry</u>

  • Reading a Periodic Table
  • Using Dimensional Analysis
<h3>Explanation:</h3>

<u>Step 1: Define</u>

[RxN]   2AgNO₃ + BaCl₂ → 2AgCl + Ba(NO₃)₂

[Given]   5.0 g AgNO₃

<u>Step 2: Identify Conversions</u>

[Reaction - Stoich] 2AgNO₃ → 2AgCl

Molar Mass of Ag - 107.87 g/mol

Molar Mass of N - 14.01 g/mol

Molar Mass of O - 16.00 g/mol

Molar Mass of Cl - 35.45 g/mol

Molar Mass of AgNO₃ - 107.87 + 14.01 + 3(16.00) = 169.88 g/mol

Molar Mass of AgCl - 107.87 + 35.45 = 143.32 g/mol

<u>Step 3: Stoichiometry</u>

  1. Set up:                              \displaystyle 5.0 \ g \ AgNO_3(\frac{1 \ mol \ AgNO_3}{169.88 \ g \ AgNO_3})(\frac{2 \ mol \ AgCl}{2 \ mol \ AgNO_3})(\frac{143.22 \ g \ AgCl}{1 \ mol \ AgCl})
  2. Multiply/Divide:                                                                                                  \displaystyle 4.21533 \ g \ AgCl

<u>Step 4: Check</u>

<em>Follow sig fig rules and round. We are given 1 sig fig.</em>

4.21533 g AgCl ≈ 4 g AgCl

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