Q14.168CP - Chemistry: Molecular Nature Of Matter And Change

Question

Element E forms an oxide of general structure A and a chloride of general structure B:

                                       

                      A                                                                                       B

For the anion $\mathbf{E}{{\mathbf{F}}_{\mathbf{5}}}^{\mathbf{-}}$ , what is 

(a) the molecular shape; 

(b) the hybridization of E; 

(c) the O.N. of E?

Step-by-Step Solution

Answer

1. Shape is square pyramidical
2. Hybridization is  $s{p}^3{d}^2$
3. Oxidation number is +4

Step 1: Molecular shape and geometry of anion

As given, the anion $\mathbf{E}{{\mathbf{F}}_{\mathbf{5}}}^{\mathbf{-}}$ has five bonds and one lone pair so that the shape is square pyramidical and the geometry is octahedral and having hybridisation is $\mathit{s}{\mathit{p}}^{\mathbf{3}}{\mathit{d}}^{\mathbf{2}}$because E has seven valence electrons and the formula for calculating 

hybridization is:

$H=\frac{1}{2}\left(V+M-C+A\right)$

V= Valence electrons 

M=no. of atoms to which ion attached 

C= charge of the cation

A= charge of the anion

So,

It has a valence of 6 electrons, and M=5 charge is -1

So,

$\begin{array}{rcl}H& =& \frac{1}{2}\left(6+5-1\right)\\ & =& \frac{10}{2}\\ & =& 5\\ & & \end{array}$

So, $s{p}^3{d}^2$

Step 2: Oxidation number of anion

The oxidation number is calculated:

As fluoride has a -1 charge and a total of 5 Fluoride are attached so the charge is -5, and the total charge on a molecule is -1

Let the charge on E is x:

 $\begin{array}{rcl}& & E{F_5}^{-}\\ x+\left(5\times -1\right)& =& -1\\ x-5& =& -1\\ x& =& -1+5\\ x& =& +4\end{array}$           

So, the Oxidation number is +4