Solved: CHEMISTRY Question for NEET

NEET CHEMISTRYMedium

The correct shape and hybridisation of BrF $_5$ is:

Square pyramidal and sp $^3$ d $^2$

Square pyramidal and d $^2$ sp $^3$

Trigonal bipyramidal and sp $^3$ d $^2$

Trigonal bipyramidal and d $^2$ sp $^3$

Determine Hybridisation: The central atom is Bromine (Br), which has 7 valence electrons. It forms single bonds with 5 Fluorine (F) atoms. Using the formula $H = \frac{1}{2}(V + M - C + A)$ , where $V=7$ and $M=5$ : $H = \frac{1}{2}(7 + 5) = 6$ . A steric number of 6 corresponds to sp $^3$ d $^2$ hybridisation .
Determine Bond Pairs and Lone Pairs: There are 5 bonding pairs (Br-F). The number of lone pairs is $H - \text{Bond Pairs} = 6 - 5 = 1$ .
Determine Shape: The electron geometry for sp $^3$ d $^2$ is octahedral. However, with 5 bond pairs and 1 lone pair (AB $_5$ E type), the molecular shape becomes Square Pyramidal .

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