Question

Q. Explain the type of hybridization in phosphorus pentafluoride.

Answer 1

Answer on Question #76168, Chemistry / Inorganic Chemistry

Q. Explain the type of hybridization in phosphorus pentafluoride.

Solution

Atom P needs five orbitals to form five P-Cl bonds in $\mathsf{PCl}_5$ .

Electron structure of P atom is:

$_{15}\mathrm{P}$ [Ne] $3s^{2}3p^{3}3d^{0}$ (ground state)

In ground sate atom P has three unpaired electrons and can form three bounds. To form 5 bounds atom P goes into an exited state, where atom P has five unpaired electrons:

$_{15}\mathrm{P}$ [Ne] $3s^{1}3p^{3}3d^{1}$ (exited state)

Five orbitals (one s, three p and one d) are hybridized to form five P-Cl bounds, therefore hybridization of P in $\mathsf{PCl}_5$ is $\mathsf{sp}^3\mathsf{d}$ .

Hybridization can be calculated by using a formula:

$$\mathrm {H} = 1 / 2 \cdot (\mathrm {V} + \mathrm {M} - \mathrm {C} + \mathrm {A}),$$

V - number of free electrons of central atom (P has 5 free electrons);

M - number of monovalent atoms linked to the central atoms ( 5 monovalent atoms of Cl)

C - cationic charge of the compound ( $C = 0$ )

A - anionic charge of the compound (A=0)

$$H = 1 / 2 (5 + 5 - 0 + 0) = 5$$

For $\mathrm{PCl}_5\mathrm{H} = 5$ therefore hybridization of P atom is $\mathfrak{sp}^3\mathfrak{d}$

A molecule of $\mathrm{PCl}_5$ has trigonal bipyramidal structure (VSEPR theory):

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