Consider the molecule CIF3. Which of the following best describes the bonding scheme consistent with the molecular geometry predicted by VSEPR? The central atom is dsp3 hybridized; it forms 3 sigma bonds by overlapping with 2p orbitals on F atoms, and the two sets of lone pairs reside in the remaining unhybridized d orbitals. The central atom is dsp3 hybridized; it forms 3 sigma bonds by overlapping with 2p orbitals on F atoms, and the two sets of lone pairs reside in the remaining hybrid orbitals. The central atom is dsp3 hybridized; it forms 1 sigma bond and 2 pi bonds by overlapping with 2p orbitals on F atoms, and the two sets of lone pairs reside in the remaining hybrid orbitals. The central atom is sp2 hybridized; it forms 3 sigma bonds by overlapping with 2p orbitals on F atoms, and there are no lone pairs remaining on the central atom. The central atom is sp2 hybridized; it forms 3 sigma bonds by overlapping with 2p orbitals on F atoms, and the two sets of lone pairs reside in the remaining unhybridized p orbitals.
The Correct Answer and Explanation is :
The correct answer is:
The central atom is dsp3 hybridized; it forms 3 sigma bonds by overlapping with 2p orbitals on F atoms, and the two sets of lone pairs reside in the remaining hybrid orbitals.
Explanation:
- Molecular Geometry and Hybridization:
- The molecular formula ClF3\text{ClF}_3 has chlorine (Cl) as the central atom bonded to three fluorine (F) atoms. Chlorine has 7 valence electrons, and each fluorine contributes one electron for bonding, resulting in 7+3=107 + 3 = 10 total valence electrons around the central atom.
- These 10 electrons form 5 electron domains: 3 bonding pairs (sigma bonds with F atoms) and 2 lone pairs.
- According to VSEPR (Valence Shell Electron Pair Repulsion) theory, 5 electron domains correspond to a trigonal bipyramidal electron geometry. The molecular geometry, considering the lone pairs, becomes T-shaped because the lone pairs occupy equatorial positions to minimize repulsion.
- Hybridization:
- To accommodate 5 electron domains, the chlorine atom must use dsp3^3 hybrid orbitals, which are formed by combining one ss, three pp, and one dd orbital.
- In this hybridization, the 3 bonding pairs occupy three hybrid orbitals, forming sigma bonds with the 2p2p orbitals of fluorine atoms. The remaining two hybrid orbitals contain lone pairs of electrons.
- Lone Pairs:
- The two lone pairs reside in the equatorial positions of the trigonal bipyramidal geometry, which minimizes electron-electron repulsion because equatorial lone pairs are further apart than axial lone pairs.
- Incorrect Options:
- sp2sp^2 hybridization is incorrect because it accommodates only 3 electron domains, not 5.
- Lone pairs do not occupy unhybridized dd-orbitals; all orbitals used are part of the hybridization. dsp3dsp^3 hybrid orbitals include both bonding and lone pair electrons.
This explanation aligns with the bonding and geometry predicted by VSEPR theory and hybridization concepts.