What type of orbital hybridisation is conside | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(A) In $PCl_5$, the central phosphorus atom $(P)$ has $5$ valence electrons. It forms $5$ covalent bonds with $5$ chlorine atoms. Using the formula fo

Vedclass · Accepted Answer

$sp^3d$

Vedclass · Answer

(A) In $PCl_5$, the central phosphorus atom $(P)$ has $5$ valence electrons. It forms $5$ covalent bonds with $5$ chlorine atoms. Using the formula for steric number: $	ext{Steric Number} = \frac{1}{2} (V + M - C + A)$, where $V = 5$ (valence electrons of $P$), $M = 5$ (monovalent atoms), $C = 0$ (cationic charge), and $A = 0$ (anionic charge). $	ext{Steric Number} = \frac{1}{2} (5 + 5) = 5$. $A$ steric number of $5$ corresponds to $sp^3d$ hybridisation, resulting in a trigonal bipyramidal geometry.

What type of orbital hybridisation is considered on $P$ in $PCl_5$?

Similar Questions

The $C-C$ bond length of the following molecules is in the order:

The number of species from the following in which the central atom uses $sp^3$ hybrid orbitals in its bonding is . . . . . . .
$NH_3, SO_2, SiO_2, BeCl_2, CO_2, H_2O, CH_4, BF_3$

As the $p-$character increases,the bond angle in hybrid orbitals formed by $s$ and $p$ atomic orbitals

Describe the hybridization in $PCl_{5}$. Why are the axial bonds longer as compared to equatorial bonds?

The hybridization states of the central atoms of the ions $I_3^-$,$ICl_4^-$ and $ICl_2^-$ are respectively

Vedclass Test Series

Exam Paper Generator

Online Exam Module