Question

In each of the following pairs of substances, one is stable and known, whereas the other is unstable. For each pair, choose the stable substance, and explain why the other compound is unstable.

a. ${\mathbf{\text{NF}}}_{\mathbf{\text{5}}}$or${\mathbf{\text{PF}}}_{\mathbf{\text{5}}}$

b.${\mathbf{\text{AsF}}}_{\mathbf{\text{5}}}$or${\mathbf{\text{AsI}}}_{\mathbf{\text{5}}}$

c.${\mathbf{\text{NF}}}_{\mathbf{3}}$or${\mathbf{\text{NBr}}}_{\mathbf{\text{3}}}$

Short answer

(a) ${\text{NF}}_{\text{5}}$will become unstable as it does not have aorbital and${\text{PF}}_{\text{5}}$will become stable as it has aorbital.

(b)${\text{AsF}}_{\text{5}}$will become stable as atomic size of fluorine is smaller and${\text{AsI}}_{\text{5}}$will become unstable as atomic size of iodine is larger.

(c) ${\text{NF}}_3$will become stable as atomic size of fluorine is smaller and ${\text{NBr}}_3$will become unstable as atomic size of bromine is larger.

Step-by-step solution

Concept Introduction

Electrons in an atom's outermost shell are known as valence electrons. The arrangement of electrons around the nucleus of an atom is known as electron configuration.

Electronic configuration of NF5and PF5

(a)

The given substances are ${\text{NF}}_{\text{5}}$or${\text{PF}}_{\text{5}}$.

To determine the stability of both substances, use the concept of the electronic configurations.

Here, the substance${\text{PF}}_{\text{5}}$, is said to be stable since phosphorus which is the central atom has an electron configuration of ${\text{1s}}^{\text{2}}{\text{2s}}^{\text{2}}{\text{2p}}^{\text{6}}{\text{3s}}^{\text{2}}{\text{3p}}_{\text{x}}^{\text{1}}{\text{3p}}_{\text{y}}^{\text{1}}{\text{3p}}_{\text{z}}^{\text{1}}$.

On the other hand, the central atom of${\text{NF}}_{\text{5}}$which is nitrogen has an electron configuration of${\text{1s}}^{\text{2}}{\text{2s}}^{\text{2}}{\text{2p}}_{\text{x}}^{\text{1}}{\text{2p}}_{\text{y}}^{\text{1}}{\text{2p}}_{\text{z}}^{\text{1}}$.

Hence, in phosphorus, there will be five orbitals that will end up being half filled since the electrons from the$\text{3s}$orbitalwill be excited to$\text{3d}$ orbital. But this phenomenon cannot occur in the nitrogen central atom since there is no$d$orbitalhere.

Therefore, ${\text{NF}}_{\text{5}}$will become unstable and ${\text{PF}}_{\text{5}}$will become stable.

Electronic configuration of  AsF5and AsI5

(b)

The given substances are${\text{AsF}}_{\text{5}}$or${\text{AsI}}_{\text{5}}$.

To determine the stability of both substances, use the concept of atomic size.

Here, the substance${\text{AsF}}_{\text{5}}$, is said to be stable since fluorinewhich is the constituent atom of the arsenic has an smaller atomic size.

On the other hand, the constituent atom of${\text{AsI}}_{\text{5}}$which is iodine has a larger atomic size.

Hence, in${\text{AsF}}_{\text{5}}$, arsenic will be able to hold the fluorine atoms at a stable condition since arsenic is larger than fluorine.But the arsenic in${\text{AsI}}_{\text{5}}$cannot hold the iodine atoms in a large number since an iodine atom is larger than arsenic.

Therefor${\text{AsF}}_{\text{5}}$e, will become stable and ${\text{AsI}}_{\text{5}}$will become unstable.

Electronic configuration of NF3 and NBr3

(c)

The given substances are${\text{NF}}_3$ or${\text{NBr}}_3$ .

To determine the stability of both substances, use the concept of atomic size.

Here, the substance${\text{NF}}_3$is said to be stable since nitrogen which is the central atom has lesser difference in atomic size.

Nitrogen has an atomic number of$7$, Fluorine gas$9$, and Bromine has$35$which is relatively large.

Since bromine is really large, nitrogen cannot be stable in holding them together. And since fluorine is smaller, there willbe a stronger force in nitrogen holding them together, making them stable.

Therefore, ${\text{NF}}_3$will become stable and ${\text{NBr}}_3$will become unstable.