Chemical Bonding & Molecular Structure

To identify the isostructural pairs among the given species, we first need to understand their molecular geometries and the electron-pair arrangements.

1. NF₃: Nitrogen trifluoride has a trigonal pyramidal structure due to the presence of one lone pair of electrons on nitrogen. It follows the AX₃E steric number formula (where A is the central atom, X represents bonded atoms, and E represents lone pairs).

2. $NO_3^-$: The nitrate ion has a trigonal planar structure. This can be rationalized using VSEPR theory, where it follows the AX₃ steric number formula with no lone pairs on the central nitrogen atom.

3. BF₃: Boron trifluoride is also trigonal planar. It conforms to the AX₃ steric number formula with no lone pairs on boron.

4. H₃O⁺: The hydronium ion has a trigonal pyramidal structure, similar to NF₃, with one lone pair of electrons. It follows the AX₃E formula.

5. HN₃: Hydronitrenium ion has a linear or slightly bent structure, depending on how we look at the resonance structures. Overall, it does not share the same geometric configuration with any of the molecules listed above.

From the above analysis:

• Isostructural Pair 1: NF₃ and H₃O⁺ (both have a trigonal pyramidal structure)
• Isostructural Pair 2: $NO_3^-$ and BF₃ (both have a trigonal planar structure)

Therefore, the correct option is:

Option C:

[NF₃, H₃O⁺] and [$NO_3^-$, BF₃].

Calcium carbide ($CaC_2$) is an ionic compound consisting of ${Ca^{2+}}$ and carbide ions ${C_2^{2-}}$. In this compound, the two carbon atoms are bonded together in the carbide ion ${C_2^{2-}}$. The bonding in the carbide ion involves one sigma ($\sigma$) bond and two pi ($\pi$) bonds, which forms a triple bond between the two carbon atoms.

The reasoning is as follows:

Each carbon atom in the ${C_2^{2-}}$ ion is sp hybridized, having one sigma ($\sigma$) bond from the overlap of sp hybrid orbitals and two pi ($\pi$) bonds from the lateral overlap of p orbitals. Therefore, the carbide ion ${C_2^{2-}}$ has a carbon-carbon triple bond.

Thus, the correct option is:

Option B: one sigma ($\sigma$) and two pi ($\pi$) bonds

To determine the number of paired electrons in an O₂ molecule, we need to understand its electronic configuration and molecular orbital configuration. Each oxygen atom has an atomic number of 8, so its electron configuration is:

$1s^2 2s^2 2p^4$

When two oxygen atoms form a diatomic O₂ molecule, their atomic orbitals combine to form molecular orbitals. For O₂, the relevant molecular orbitals and their filling order are:

$\sigma_{1s}, \sigma_{1s}^*, \sigma_{2s}, \sigma_{2s}^*, \sigma_{2p_z}, \pi_{2p_x} = \pi_{2p_y}, \pi_{2p_x}^* = \pi_{2p_y}^*, \sigma_{2p_z}^*$

Filling these molecular orbitals with the 16 electrons (8 from each oxygen atom) according to the aufbau principle and Hund's rule, we get the following configuration:

$ (\sigma_{1s})^2 (\sigma_{1s}^*)^2 (\sigma_{2s})^2 (\sigma_{2s}^*)^2 (\sigma_{2p_z})^2 (\pi_{2p_x})^2 (\pi_{2p_y})^2 (\pi_{2p_x}^*)^1 (\pi_{2p_y}^*)^1 $

Here’s the breakdown of paired and unpaired electrons:

• $ \sigma_{1s} $ -> 2 paired electrons
• $ \sigma_{1s}^* $ -> 2 paired electrons
• $ \sigma_{2s} $ -> 2 paired electrons
• $ \sigma_{2s}^* $ -> 2 paired electrons
• $ \sigma_{2p_z} $ -> 2 paired electrons
• $ \pi_{2p_x} $ -> 2 paired electrons
• $ \pi_{2p_y} $ -> 2 paired electrons

The antibonding $ \pi_{2p_x}^* $ and $ \pi_{2p_y}^* $ orbitals each have one unpaired electron, but since the question asks for paired electrons, these do not count.

Total number of paired electrons:

$2 ( \sigma_{1s} ) + 2 ( \sigma_{1s}^* ) + 2 ( \sigma_{2s} ) + 2 ( \sigma_{2s}^* ) + 2 ( \sigma_{2p_z} ) + 2 ( \pi_{2p_x} ) + 2 ( \pi_{2p_y} ) = 14$

Therefore, the number of paired electrons in O₂ molecules is:

Option D: 14

To determine which molecule or ion assumes a linear structure, we need to consider the molecular geometry, which is often predicted by the VSEPR (Valence Shell Electron Pair Repulsion) theory.

Let’s analyze each option:

Option A: SnCl₂

Tin(II) chloride, SnCl₂, has a bent structure rather than a linear one due to the presence of a lone pair of electrons on the tin atom, which repels the bonds between Sn and Cl.

Option B: NCO^-

The cyanate ion, NCO^-, has a linear structure because the central atom carbon forms a double bond with nitrogen and a triple bond with oxygen, aligning all atoms in a straight line.

Option C: CS₂

Carbon disulfide, CS₂, has a linear structure. The central carbon atom forms double bonds with each sulfur atom, allowing the molecule to maintain a straight line.

Option D: $NO_2^+$

The nitronium ion, $NO_2^+$, also has a linear structure. It has a central nitrogen atom bonded to two oxygen atoms with no lone pairs on nitrogen, thus the molecule assumes a linear shape.

So, the linear structures are assumed by Options B, C, and D.

Isostructural compounds are those that have the same shape and similar bond angles. To determine which compound is isostructural with carbon dioxide (CO₂), we need to analyze the molecular geometry of each of the options.

1. CO₂ has a linear structure with the carbon atom in the center, double-bonded to two oxygen atoms, giving it a bond angle of 180°. The structure can be represented as:

$ \text{O} = \text{C} = \text{O} $

2. HgCl₂ also has a linear structure. The mercury atom in the center is single-bonded to two chlorine atoms, giving it a bond angle of 180°. The structure can be represented as:

$ \text{Cl} - \text{Hg} - \text{Cl} $

3. SnCl₂ has a bent (or V-shaped) structure due to lone pairs on the tin atom. Hence, it is not isostructural with CO₂.

4. C₂H₂ (acetylene) has a linear structure with a carbon-carbon triple bond and hydrogen atoms at each end, giving it a bond angle of 180°. The structure can be represented as:

$ \text{H} - \text{C} \equiv \text{C} - \text{H} $

5. NO₂ has a bent structure due to the presence of a lone electron on the nitrogen atom, which causes repulsion and bends the molecule. Therefore, it is not isostructural with CO₂.

Given this analysis, CO₂ is isostructural with both HgCl₂ and C₂H₂.

The statement "SnCl₂ is a non-linear molecule" is TRUE.

To understand why SnCl₂ (Tin(II) chloride) is a non-linear molecule, we need to look at its molecular geometry. Tin (Sn) is the central atom and it has a lone pair of electrons along with two bonding pairs of electrons with the chlorine (Cl) atoms. This creates a V-shaped or bent molecular geometry.

According to the Valence Shell Electron Pair Repulsion (VSEPR) theory, the lone pair of electrons on the tin atom will repel the bonding pairs, causing the molecule to take on a shape that minimizes repulsion. Therefore, the electron geometry of SnCl₂ is trigonal planar, but because of the lone pair, the molecular shape is bent.

The bond angle in a bent molecule like SnCl₂ deviates from the typical 180 degrees and is approximately 120 degrees. This bent shape confirms that SnCl₂ is non-linear.

Hence, the correct answer is:

Option A: TRUE