Question

Identify the set containing isoelectronic species.

• A. \( \text{N}_2 \), \( \text{O}_2^- \), \( \text{NO}^+ \)
• B. \( \text{N}_2 \), \( \text{CO} \), \( \text{NO}^+ \)
• C. \( \text{F}_2 \), \( \text{O}_2^- \), \( \text{N}_2 \)
• D. \( \text{N}_2 \), \( \text{O}_2^+ \), \( \text{C}_2 \)

Hint:

Isoelectronic species have the same number of electrons. To identify them, calculate the total number of electrons for each species in the set.

Answer 1

The Correct Option is B

Step 1: Known Information.
Isoelectronic species are atoms or molecules that have the same number of electrons.
We need to identify the set of species with the same number of electrons.
Step 2: Analyze Each Option. 1. Option 1: \( \text{N}_2 \), \( \text{O}_2^- \), \( \text{NO}^+ \)
\( \text{N}_2 \):
Nitrogen (\( \text{N} \)) has 7 electrons.
\( \text{N}_2 \) has \( 7 + 7 = 14 \) electrons.
\( \text{O}_2^- \):
Oxygen (\( \text{O} \)) has 8 electrons.
\( \text{O}_2 \) has \( 8 + 8 = 16 \) electrons.
Adding one extra electron (\( \text{O}_2^- \)) gives \( 16 + 1 = 17 \) electrons.
\( \text{NO}^+ \):
Nitrogen (\( \text{N} \)) has 7 electrons.
Oxygen (\( \text{O} \)) has 8 electrons.
\( \text{NO} \) has \( 7 + 8 = 15 \) electrons.
Removing one electron (\( \text{NO}^+ \)) gives \( 15 - 1 = 14 \) electrons.
Electron Count: \( \text{N}_2 \) (14), \( \text{O}_2^- \) (17), \( \text{NO}^+ \) (14).
Not all species have the same number of electrons.
2. Option 2: \( \text{N}_2 \), \( \text{CO} \), \( \text{NO}^+ \)
\( \text{N}_2 \):
As calculated earlier, \( \text{N}_2 \) has 14 electrons.
\( \text{CO} \):
Carbon (\( \text{C} \)) has 6 electrons.
Oxygen (\( \text{O} \)) has 8 electrons.
\( \text{CO} \) has \( 6 + 8 = 14 \) electrons.
\( \text{NO}^+ \):
As calculated earlier, \( \text{NO}^+ \) has 14 electrons.
Electron Count: \( \text{N}_2 \) (14), \( \text{CO} \) (14), \( \text{NO}^+ \) (14).
All species have the same number of electrons.
3. Option 3: \( \text{F}_2 \), \( \text{O}_2^- \), \( \text{N}_2 \)
\( \text{F}_2 \):
Fluorine (\( \text{F} \)) has 9 electrons.
\( \text{F}_2 \) has \( 9 + 9 = 18 \) electrons.
\( \text{O}_2^- \):
As calculated earlier, \( \text{O}_2^- \) has 17 electrons.
\( \text{N}_2 \):
As calculated earlier, \( \text{N}_2 \) has 14 electrons.
Electron Count: \( \text{F}_2 \) (18), \( \text{O}_2^- \) (17), \( \text{N}_2 \) (14).
Not all species have the same number of electrons.
4. Option 4: \( \text{N}_2 \), \( \text{O}_2^+ \), \( \text{C}_2 \)
\( \text{N}_2 \):
As calculated earlier, \( \text{N}_2 \) has 14 electrons.
\( \text{O}_2^+ \):
Oxygen (\( \text{O} \)) has 8 electrons.
\( \text{O}_2 \) has \( 8 + 8 = 16 \) electrons.
Removing one electron (\( \text{O}_2^+ \)) gives \( 16 - 1 = 15 \) electrons.
\( \text{C}_2 \):
Carbon (\( \text{C} \)) has 6 electrons.
\( \text{C}_2 \) has \( 6 + 6 = 12 \) electrons.
Electron Count: \( \text{N}_2 \) (14), \( \text{O}_2^+ \) (15), \( \text{C}_2 \) (12).
Not all species have the same number of electrons.
Step 3: Correct Answer. The set containing isoelectronic species is:
\[ \boxed{\text{N}_2, \text{CO}, \text{NO}^+} \]