Question

Bromine trifluoride autoionizes to form BrF$_2^+$ and BrF$_4^-$. The shapes of the cation and anion are respectively ________, and ________

• A. bent, square planar
• B. linear, square planar
• C. bent, see-saw
• D. linear, tetrahedral

Answer 1

The Correct Option is A

Step 1: Understanding the Concept:
To find the shape of a molecule or ion, we use VSEPR (Valence Shell Electron Pair Repulsion) theory. We count the number of valence electrons on the central atom, adjust for the ionic charge, and determine the steric number (number of bond pairs + lone pairs) to deduce hybridization and geometry.
Step 2: Key Formula or Approach:
Steric number = $\frac{1}{2} (V + M - C + A)$
where $V$ is valence electrons of central atom, $M$ is number of monovalent surrounding atoms, $C$ is positive charge, $A$ is negative charge.
Step 3: Detailed Explanation:
For the cation $BrF_2^+$:
Central atom is Bromine (Br), Group 17 $\implies V = 7$.
Monovalent atoms (F) $\implies M = 2$.
Cationic charge $C = 1$.
Number of electron pairs = $\frac{1}{2}(7 + 2 - 1) = \frac{8}{2} = 4$.
So, it has $sp^3$ hybridization.
Out of 4 pairs, 2 are bond pairs (with F) and 2 are lone pairs.
The fundamental geometry for 4 pairs is tetrahedral. The presence of 2 lone pairs causes the two bond pairs to form a "bent" or "V-shape".
So, $BrF_2^+$ is bent.
For the anion $BrF_4^-$:
Central atom is Bromine (Br), $V = 7$.
Monovalent atoms (F) $\implies M = 4$.
Anionic charge $A = 1$.
Number of electron pairs = $\frac{1}{2}(7 + 4 + 1) = \frac{12}{2} = 6$.
So, it has $sp^3d^2$ hybridization.
Out of 6 pairs, 4 are bond pairs (with F) and 2 are lone pairs.
The fundamental geometry for 6 pairs is octahedral. According to VSEPR, lone pairs occupy trans (opposite) positions to minimize repulsion at $180^\circ$.
The 4 fluorine atoms will therefore occupy the equatorial plane, creating a "square planar" shape.
So, $BrF_4^-$ is square planar.
Step 4: Final Answer:
The shapes are bent and square planar respectively.