Question:
Consider the elements with atomic numbers Z = 8, 9, 11, 19 and 20. The number of ionic compounds possible with the elements having these atomic numbers is
Consider the elements with atomic numbers Z = 8, 9, 11, 19 and 20. The number of ionic compounds possible with the elements having these atomic numbers is
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To solve this quickly, just classify the elements into metals and non-metals. The total number of binary ionic compounds will be (number of metals) \(\times\) (number of non-metals).
Updated On: Jul 1, 2026
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The Correct Option is A
Solution and Explanation
Step 1: Understanding the Question:
We are given a set of atomic numbers and need to find how many unique binary ionic compounds can be formed between them. An ionic compound is typically formed between a metal (cation) and a non-metal (anion).
Step 2: Key Formula or Approach:
1. Identify the elements from their atomic numbers.
2. Classify each element as a metal or a non-metal. Metals tend to lose electrons (form cations), while non-metals tend to gain electrons (form anions).
3. Count the number of possible combinations between one metal and one non-metal.
Step 3: Detailed Explanation:
1.
Identify the elements:
- Z = 8 is Oxygen (O)
- Z = 9 is Fluorine (F)
- Z = 11 is Sodium (Na)
- Z = 19 is Potassium (K)
- Z = 20 is Calcium (Ca)
2.
Classify the elements:
-
Non-metals (form anions): Oxygen (O\(^{2-}\)), Fluorine (F\(^{-}\)) -- [2 non-metals]
-
Metals (form cations): Sodium (Na\(^{+}\)), Potassium (K\(^{+}\)), Calcium (Ca\(^{2+}\)) -- [3 metals]
3.
Count the combinations:
We can form an ionic compound by pairing each metal with each non-metal.
- Combinations with Sodium (Na):
- Na and O \(\rightarrow\) Na\(_2\)O (Sodium oxide)
- Na and F \(\rightarrow\) NaF (Sodium fluoride)
- Combinations with Potassium (K):
- K and O \(\rightarrow\) K\(_2\)O (Potassium oxide)
- K and F \(\rightarrow\) KF (Potassium fluoride)
- Combinations with Calcium (Ca):
- Ca and O \(\rightarrow\) CaO (Calcium oxide)
- Ca and F \(\rightarrow\) CaF\(_2\) (Calcium fluoride)
The total number of possible ionic compounds is the number of metals multiplied by the number of non-metals: \(3 \times 2 = 6\).
Step 4: Final Answer:
The number of possible ionic compounds is 6.
We are given a set of atomic numbers and need to find how many unique binary ionic compounds can be formed between them. An ionic compound is typically formed between a metal (cation) and a non-metal (anion).
Step 2: Key Formula or Approach:
1. Identify the elements from their atomic numbers.
2. Classify each element as a metal or a non-metal. Metals tend to lose electrons (form cations), while non-metals tend to gain electrons (form anions).
3. Count the number of possible combinations between one metal and one non-metal.
Step 3: Detailed Explanation:
1.
Identify the elements:
- Z = 8 is Oxygen (O)
- Z = 9 is Fluorine (F)
- Z = 11 is Sodium (Na)
- Z = 19 is Potassium (K)
- Z = 20 is Calcium (Ca)
2.
Classify the elements:
-
Non-metals (form anions): Oxygen (O\(^{2-}\)), Fluorine (F\(^{-}\)) -- [2 non-metals]
-
Metals (form cations): Sodium (Na\(^{+}\)), Potassium (K\(^{+}\)), Calcium (Ca\(^{2+}\)) -- [3 metals]
3.
Count the combinations:
We can form an ionic compound by pairing each metal with each non-metal.
- Combinations with Sodium (Na):
- Na and O \(\rightarrow\) Na\(_2\)O (Sodium oxide)
- Na and F \(\rightarrow\) NaF (Sodium fluoride)
- Combinations with Potassium (K):
- K and O \(\rightarrow\) K\(_2\)O (Potassium oxide)
- K and F \(\rightarrow\) KF (Potassium fluoride)
- Combinations with Calcium (Ca):
- Ca and O \(\rightarrow\) CaO (Calcium oxide)
- Ca and F \(\rightarrow\) CaF\(_2\) (Calcium fluoride)
The total number of possible ionic compounds is the number of metals multiplied by the number of non-metals: \(3 \times 2 = 6\).
Step 4: Final Answer:
The number of possible ionic compounds is 6.
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