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 Concept:
The number of lone pairs is calculated using the formula: $LP = \frac{V - (N \times B)}{2}$, where $V$ is valence electrons, $N$ is number of bonds, and $B$ is the bond type (usually 1 for halogens).

Step 2: Formula Application:
$BrF_5$: Br has 7 valence $e^-$, 5 bonds. $LP = (7-5)/2 = 1$. $XeF_6$: Xe has 8 valence $e^-$, 6 bonds. $LP = (8-6)/2 = 1$.

Step 3: Explanation:
Both $BrF_5$ and $XeF_6$ have exactly one lone pair on the central atom. Comparing other pairs: - $ClF_3$ (2 LP) vs $XeF_2$ (3 LP). - $IF_7$ (0 LP) vs $XeF_4$ (2 LP). - $H_2S$ (2 LP) vs ICl (3 LP on Iodine).

Step 4: Final Answer:
The pair with equal lone pairs is $BrF_5$ and $XeF_6$.