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

V) N₂O₅ is the unstable compound.

The presence of N₂ is the hint of its instability. N₂ is a dinitrogen molecule and is an unstable compound because nitrogen a triple bond with itself.

Answer 2

Answer: The correct answer is Option (ii)

Explanation:

Nitrogen is the 7th element of the periodic table having electronic configuration of $1s^22s^22p^3$

Number of valence electrons for this element are 5.

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

Formula used to calculate the number of atomic orbitals around central metal atom is:

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

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

• 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 will be $sp^3$

• 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 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.

• 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 will be $sp^2$

• 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 will be $sp$

• 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 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.