Question

The trigonal bipyramidal geometry is obtained from the hybridization

• A. ${\mathrm{d}}^2{\mathrm{sp}}^3\mathrm{or}{\mathrm{sp}}^3{\mathrm{d}}^2$
• B. ${\mathrm{dsp}}^2\mathrm{or}{\mathrm{sp}}^2\mathrm{d}$
• C. ${\mathrm{dsp}}^3\mathrm{or}{\mathrm{sp}}^3\mathrm{d}$
• D. None of these

Answer 1

The correct option is C

${\mathrm{dsp}}^3\mathrm{or}{\mathrm{sp}}^3\mathrm{d}$

The explanation for the correct option:-

c) ${\mathrm{dsp}}^3\mathrm{or}{\mathrm{sp}}^3\mathrm{d}$

• When the mixing of two atomic orbitals takes place to give hybrid orbitals having characteristics of both the orbitals, the process is called hybridization
• For eg:- Mixing of s and p orbitals results in the formation of sp hybridized orbitals
• The atomic orbitals which are to be mixed should have comparable energies
• We can assign geometry to a molecule by calculating the steric number of the molecule
• Steric number = No of atoms directly bonded to central atom + No of lone pairs attached to the central atom
• Eg:- For steric number$2$, the hybridization is sp and geometry is linear
• For steric number$5$, the hybridization is ${\mathrm{dsp}}^3\mathrm{or}{\mathrm{sp}}^3\mathrm{d}$ and geometry is trigonal bipyramidal
• Mixing of $1$ s orbital, $3$ p orbitals, and $1$ d orbital takes place in ${\mathrm{sp}}^3\mathrm{d}$ hybridization to form $5$ ${\mathrm{sp}}^3\mathrm{d}$ hybrid orbitals having equal energy.
• An example of a molecule having trigonal bipyramidal geometry is ${\mathrm{PCl}}_5$
• The steric number for ${\mathrm{PCl}}_5$is $5$ as it has $5$ P-Cl bonds and thus hybridization is${\mathrm{sp}}^3\mathrm{d}$.

( Trigonal bipyramidal geometry of ${\mathrm{PCl}}_5$)

The explanation for incorrect options:-

a) ${\mathrm{d}}^2{\mathrm{sp}}^3\mathrm{or}{\mathrm{sp}}^3{\mathrm{d}}^2$

• For steric number$6$, the hybridization is ${\mathrm{d}}^2{\mathrm{sp}}^3\mathrm{or}{\mathrm{sp}}^3{\mathrm{d}}^2$and geometry is octahedral
• Mixing of $1$ s orbital, $3$ p orbitals, and $2$ d orbitals takes place in ${\mathrm{sp}}^3{\mathrm{d}}^2$ hybridization to form $6$ ${\mathrm{sp}}^3{\mathrm{d}}^2$ hybrid orbitals having equal energy
• An example of a molecule having ${\mathrm{sp}}^3{\mathrm{d}}^2$ hybridization is${\mathrm{SF}}_6$.

b) ${\mathrm{dsp}}^2\mathrm{or}{\mathrm{sp}}^2\mathrm{d}$

• For steric number$4$, the hybridization is ${\mathrm{dsp}}^2\mathrm{or}{\mathrm{sp}}^2\mathrm{d}$ and geometry is square planar
• Mixing of $1$ s orbital, $2$ p orbitals, and $1$ d orbital takes place in ${\mathrm{sp}}^2\mathrm{d}$ hybridization to form $4$ ${\mathrm{sp}}^2\mathrm{d}$ hybrid orbitals having equal energy
• An example of a molecule having ${\mathrm{sp}}^2\mathrm{d}$ hybridization is${\mathrm{XeF}}_4$.

Therefore the correct option is (c)${\mathrm{dsp}}^3\mathrm{or}{\mathrm{sp}}^3\mathrm{d}$.