Question

Nitrogen has a reasonable oxidation number in all of the following compounds, and yet one of them is not stable. Which one and why? i) NF3 ii) NF5 iii) HNO3 iv) NO v) N2O5

Answer 1

NF3 has a reasonable oxidation number

Answer 2

Answer: The correct answer for this question is Option (ii)

Explanation:

Nitrogen is the 7th element of the periodic table. Number of electrons that this element contains are 7 and the electronic configuration of this element is $1s^22s^22p^3$

Number of valence electrons for this element are 5.

To know the stability of the compound, we first need to find out the hybridization of nitrogen in each compound. Hybridization is calculated by looking at the Lewis dot structures for all the compounds.

Formula that is used to calculate the number of atomic orbitals around central metal atom (which is nitrogen) is:

Number of atomic orbitals around central metal atom = Number of bond pairs + Number of lone pairs

Bond pairs for a double and triple bond are taken as 1 only.

Option (i): Hybridization of nitrogen in $NF_3$

Number of bond pairs = 3

Number of lone pairs = 1

Number of atomic orbitals around Nitrogen atom = 3 + 1 = 4

So, hybridization of nitrogen here will be $sp^3$

Option (ii): Hybridization of nitrogen in $NF_5$

Number of bond pairs = 5

Number of lone pairs = 0

Number of atomic orbitals around Nitrogen atom = 5 + 0 = 5

So, hybridization of nitrogen here will be $sp^3d$. It is visible from the configuration that nitrogen atom do not have any d-orbitals. Thus, this hybridization is not possible and hence, this compound is not possible. Therefore, it is considered as an unstable molecule.

Option (iii): Hybridization of nitrogen in $HNO_3$

Number of bond pairs = 3

Number of lone pairs = 0

Number of atomic orbitals around Nitrogen atom = 3 + 0 = 3

So, hybridization of nitrogen here will be $sp^2$

Option (iv): Hybridization of nitrogen in $NO$

Number of bond pairs = 1

Number of lone pairs = 1

Number of atomic orbitals around Nitrogen atom = 1 + 1 = 1

So, hybridization of nitrogen here will be $sp$

Option (v): Hybridization of nitrogen in $N_2O_5$

Number of bond pairs = 3

Number of lone pairs = 0

Number of atomic orbitals around Nitrogen atom = 3 + 0 = 3

So, hybridization of nitrogen here will be $sp^2$ and this hybridization will be same for both the nitrogen atoms that present in the given compound.

From the above information, the correct answer is Option (ii).

The lewis dot structures of the compounds are shown in the image attached below.