Question

What are the excited states and the hybridization of \[As{O_4}^{3 - }\] and \[P{F_4}^ - \] ?

Answer 1

Hint: Quantum mechanics is an important branch of science. It deals with the energy’s states and excitations. The quantum state in a molecule that has higher energy than the ground state can be known as the excited state. There will be an infinite number of excited states of a molecule.

Complete answer:
Given molecules are \[As{O_4}^{3 - }\] and \[P{F_4}^ - \]
These both molecules have an infinite number of excited states.
In \[As{O_4}^{3 - }\] , the central metal atom is arsenic. Arsenic has the atomic number of \[33\] , the valence electrons in arsenic are five. Whereas oxygen has an atomic number of \[8\] , it has \[6\] valence electrons. as there were four oxygen atoms in \[As{O_4}^{3 - }\] , the total number of valence electrons will be equal to \[24\] and there is a negative charge of \[ - 3\] . thus, the total number of electrons in \[As{O_4}^{3 - }\] will be equal to \[35\]

These \[35\] electrons are arranged as follows:

This molecule has steric number \[4\] , and also has \[A{X_4}\] system which has a geometry of tetrahedral and has hybridization of \[s{p^3}\]

In \[P{F_4}^ - \] , the central metal atom is phosphorous, with five valence electrons and each fluorine has \[7\] valence electrons as four fluorine atoms are there, the total valence electrons of four fluorine atoms will be \[28\] and this molecule has one negative charge. Thus, the total valence electrons will be equal to \[34\] .

These \[34\] electrons are arranged as follows:

This molecule has a steric number of \[5\] and has the hybridization of \[s{p^3}d\] with \[A{X_4}E\] system.

Note: The steric number can be given by the sum of the bonding pair of electrons and lone pair of electrons on the central metal atoms but not on terminal atoms. The atom with low electronegativity occupies the central position of a molecule.