Q.Arrange the following bonds in order of increasing ionic character giving reason.
N—H, F—H, C—H and O—H

The question asks for the oxide of nitrogen with a linear molecular structure. To decide, you need to draw the Lewis structure for each molecule and then apply VSEPR theory to predict the shape around the central atom(s). A linear structure arises when the central atom has a steric number of 2 (two bonding pairs and no lone pairs) or when the molecule is symmetric with two terminal atoms.

Let’s examine each option.

1. $\mathrm{N_2O_5}$ (dinitrogen pentoxide)

   In the solid state, $\mathrm{N_2O_5}$ exists as $\mathrm{NO_2^+ \, NO_3^-}$ (nitronium nitrate). The $\mathrm{NO_2^+}$ ion is linear (steric number 2 on N, no lone pairs), but the $\mathrm{NO_3^-}$ ion is trigonal planar. In the gas phase, the molecule has a bent $\mathrm{N-O-N}$ bridge with $\mathrm{NO_2}$ groups. Overall, the molecule is not linear — it has a central oxygen with two $\mathrm{NO_2}$ groups attached at an angle.

2. $\mathrm{N_2O_4}$ (dinitrogen tetroxide)

   This is a dimer of $\mathrm{NO_2}$. The structure has a central $\mathrm{N-N}$ single bond, with each nitrogen bonded to two oxygens. Each nitrogen is $\mathrm{sp^2}$ hybridized (steric number 3), giving a trigonal planar geometry around each N. The molecule as a whole is planar but not linear — the $\mathrm{N-N}$ bond is flanked by $\mathrm{NO_2}$ groups at about 134°.

3. $\mathrm{N_2O}$ (nitrous oxide)

   The Lewis structure is best represented as $\mathrm{N \equiv N^+ - O^-}$ (resonance hybrid). The central nitrogen is bonded to one N and one O, with a triple bond to the terminal N and a single bond to O. The central N has no lone pairs and two bonding regions — steric number 2, giving a linear geometry. The molecule is $\mathrm{N \equiv N-O}$, and it is indeed linear. This matches the requirement. …

Concept — coordination number of cubic lattices. The number of nearest neighbours (coordination number) is a fixed geometric property of each lattice type:

• Simple cubic: 6
• Body-centred cubic (BCC): 8
• Face-centred cubic / CCP and HCP: 12

Step 1 — geometry of BCC. In BCC, atoms sit at the 8 corners and one at the body centre. Touching occurs along the body diagonal: the centre atom is in contact with all 8 corner atoms, each at distance …

The hybridization of the central nitrogen atom is determined by counting the number of electron domains around it—both bonding pairs and lone pairs (or unpaired electrons). This count, called the steric number, tells us how many atomic orbitals must mix to accommodate all the electron density.

For nitrogen in these oxides, we need to examine the Lewis structure of each species and count what surrounds the nitrogen.

1. NO₂ (nitrogen dioxide)

Nitrogen has 5 valence electrons, and each oxygen contributes 6. Total valence electrons: $5 + 2(6) = 17$ electrons—an odd number, so one electron must remain unpaired.

The Lewis structure places nitrogen in the center with two N=O double bonds (using resonance, each bond is intermediate between single and double). After forming two bonds and distributing electrons, nitrogen has one unpaired electron (a radical).

• σ-bonds: 2 (one to each oxygen)
• Lone pairs on N: 0
• Unpaired electrons: 1

The unpaired electron occupies an orbital, so the steric number is $2 + 1 = 3$. Three hybrid orbitals are needed → $sp^2$ hybridization.

The geometry is bent (the unpaired electron and two bonds arrange themselves trigonal planar, but the molecular shape is bent).

2. NO₂⁺ (nitronium ion)

Removing one electron from NO₂ gives NO₂⁺. Now we have $5 + 2(6) - 1 = 16$ valence electrons.

The Lewis structure shows nitrogen forming two N=O double bonds with no lone pairs and no unpaired electrons (all electrons are paired in bonds).

• σ-bonds: 2
• Lone pairs on N: 0
• Unpaired electrons: 0

Steric number = $2 + 0 = 2$. Two hybrid orbitals → $sp$ hybridization.

The geometry is linear (O=N=O at 180°).

3. NO₂⁻ (nitrite ion)

Adding one electron to NO₂ gives NO₂⁻. Total valence electrons: $5 + 2(6) + 1 = 18$.

The Lewis structure has nitrogen forming two N–O bonds (with resonance, each is between single and double) and now carries one lone pair of electrons.

• σ-bonds: 2
• Lone pairs on N: 1
• Unpaired electrons: 0

Steric number = $2 + 1 = 3$. Three hybrid orbitals → $sp^2$ hybridization. …

The key idea here is that bond length depends on two main factors: bond order (triple bonds are shorter than double bonds, which are shorter than single bonds) and the size of the atoms involved. For bonds of the same order, larger atoms give longer bonds.

Let’s examine each bond:

1. $\mathrm{C \equiv C}$ — a triple bond between two carbon atoms. Triple bonds are the shortest type. Carbon is a relatively small atom, so this bond is very short.

2. $\mathrm{C \equiv N}$ — a triple bond between carbon and nitrogen. Nitrogen is slightly smaller than carbon (atomic radius: N ≈ 75 pm, C ≈ 77 pm), so this bond is actually a bit shorter than $\mathrm{C \equiv C}$. However, the difference is tiny, and in many contexts they are nearly equal. For exam purposes, $\mathrm{C \equiv N}$ is often considered slightly shorter than $\mathrm{C \equiv C}$.

3. $\mathrm{C=O}$ — a double bond between carbon and oxygen. Double bonds are longer than triple bonds. Oxygen is smaller than carbon, but the bond order effect dominates: a double bond is clearly longer than a triple bond.

4. $\mathrm{N-O}$ — a single bond between nitrogen and oxygen. Single bonds are the longest. Both atoms are small, but the bond order of 1 makes this bond significantly longer than any double or triple bond. …