Question

Which of the following pair of compounds consists equal number of lone pair of electrons in the valence shell of central atom?

• A. $BrF_5$ and $XeF_6$
• B. ICl and $H_2S$
• C. $ClF_3$ and $XeF_2$
• D. $IF_7$ and $XeF_4$

Hint:

To prevent counting errors, always remember the valence electron counts based on the periodic group: Group 15 = 5, Group 16 = 6, Group 17 (Halogens) = 7, Group 18 (Noble Gases) = 8.

Answer 1

The Correct Option is A

Step 1: Understanding the Question:
The question asks us to identify the option where both molecules possess the exact same number of non-bonding electron pairs (lone pairs) specifically residing on their central atom.

Step 2: Key Formula or Approach:
The number of lone pairs (LP) on a central atom can be quickly calculated using:
$$LP = \frac{(\text{Valence } e^- \text{ on central atom}) - (\text{Number of bonding electrons utilized})}{2}$$
Note: Halogens and Hydrogen utilize exactly 1 electron from the central atom to form a single bond.

Step 3: Detailed Explanation:
Let's analyze the pairs one by one:
(a) $BrF_5$ and $XeF_6$:
- $BrF_5$: Bromine (group 17) has 7 valence $e^-$. It forms 5 single bonds with 5 F atoms. Remaining $e^- = 7 - 5 = 2$. This is 1 lone pair.
- $XeF_6$: Xenon (group 18) has 8 valence $e^-$. It forms 6 single bonds with 6 F atoms. Remaining $e^- = 8 - 6 = 2$. This is 1 lone pair.
Both have 1 lone pair. This is a match.
(b) ICl and $H_2S$:
- ICl: Iodine has 7 valence $e^-$, forms 1 bond. Remaining $e^- = 6 \rightarrow$ 3 lone pairs.
- $H_2S$: Sulphur has 6 valence $e^-$, forms 2 bonds. Remaining $e^- = 4 \rightarrow$ 2 lone pairs. (Mismatch)
(c) $ClF_3$ and $XeF_2$:
- $ClF_3$: Chlorine has 7 valence $e^-$, forms 3 bonds. Remaining $e^- = 4 \rightarrow$ 2 lone pairs.
- $XeF_2$: Xenon has 8 valence $e^-$, forms 2 bonds. Remaining $e^- = 6 \rightarrow$ 3 lone pairs. (Mismatch)
(d) $IF_7$ and $XeF_4$:
- $IF_7$: Iodine has 7 valence $e^-$, forms 7 bonds. Remaining $e^- = 0 \rightarrow$ 0 lone pairs.
- $XeF_4$: Xenon has 8 valence $e^-$, forms 4 bonds. Remaining $e^- = 4 \rightarrow$ 2 lone pairs. (Mismatch)

Step 4: Final Answer:
The pair that shares an equal number of lone pairs (1 each) is $BrF_5$ and $XeF_6$, which corresponds to option (a).