Understanding Isostructural Molecules

The question asks us to identify which molecule among the given options is isostructural with carbon dioxide (\(\rm CO_2\)). Isostructural means having the same structure. For simple molecules, this usually implies having the same molecular geometry and often a similar arrangement of atoms, though not necessarily the same atoms.

Structure of \(\rm CO_2\)

Let's first determine the structure of \(\rm CO_2\). The central atom is Carbon (C), which is bonded to two Oxygen (O) atoms. Carbon has 4 valence electrons, and Oxygen has 6. Total valence electrons = \(4 + 2 \times 6 = 16\). The Lewis structure shows double bonds between C and each O, with 2 lone pairs on each O.

\(\rm O=C=O\)

The central Carbon atom has two double bonds and no lone pairs. According to VSEPR theory, this corresponds to an AX\(_2\) electron geometry and molecular geometry. Therefore, \(\rm CO_2\) is a linear molecule.

The hybridization of the central Carbon atom in \(\rm CO_2\) is sp.

Analyzing the Structures of Options

Let's analyze the structure of each given option:

Option 1: \(\rm ICl^-_2\)

• The central atom is Iodine (I). Iodine has 7 valence electrons. The ion has a -1 charge, adding 1 electron. So, Iodine has 7 + 1 = 8 valence electrons effectively available for bonding and lone pairs around it.
• Iodine is bonded to two Chlorine (Cl) atoms. This uses 2 electrons for two single bonds (\(\rm I-Cl\)).
• Remaining electrons around I = 8 - 2 = 6 electrons, which form 3 lone pairs.
• VSEPR notation: AX\(_2\)E\(_3\) (2 bonding pairs, 3 lone pairs).
• The molecular geometry is linear. The atoms are arranged as Cl-I-Cl in a straight line.
• Hybridization of the central Iodine atom is sp\(_3\)d.

Option 2: \(\rm SnCl_2\)

• The central atom is Tin (Sn). Tin is in Group 14, so it has 4 valence electrons.
• Tin is bonded to two Chlorine (Cl) atoms. This uses 2 electrons for two single bonds (\(\rm Sn-Cl\)).
• Remaining electrons around Sn = 4 - 2 = 2 electrons, which form 1 lone pair.
• VSEPR notation: AX\(_2\)E\(_1\) (2 bonding pairs, 1 lone pair).
• The molecular geometry is bent or V-shaped. The atoms are not in a straight line.
• Hybridization of the central Tin atom is sp\(_2\).

Option 3: \(\rm C_2H_2\) (Acetylene)

• This molecule contains two Carbon atoms and two Hydrogen atoms. Its structure is H-C≡C-H.
• It is a linear molecule overall. The atoms are arranged in a straight line: H-C-C-H.
• Each Carbon atom is bonded to one Hydrogen atom and one Carbon atom via a triple bond.
• Let's look at the environment around each Carbon atom. Each C is bonded to one H and one C (via a triple bond), representing 2 bonding groups and 0 lone pairs.
• VSEPR around each C: AX\(_2\).
• The geometry around each C is linear, resulting in an overall linear molecule.
• Hybridization of each Carbon atom is sp.

Comparing Structures with \(\rm CO_2\)

Let's compare the structures:

Both \(\rm CO_2\) and \(\rm C_2H_2\) are linear molecules. While \(\rm ICl^-_2\) is also linear and has the same number of atoms as \(\rm CO_2\), \(\rm C_2H_2\) is also a linear molecule. In some contexts, "isostructural" can refer primarily to the overall shape, especially linearity for simple molecules.

Based on the options provided and commonly encountered examples, \(\rm C_2H_2\) is considered isostructural with \(\rm CO_2\) due to both molecules having a linear shape.

Conclusion

Comparing the structures, \(\rm CO_2\) is linear, \(\rm ICl^-_2\) is linear, \(\rm SnCl_2\) is bent, and \(\rm C_2H_2\) is linear. Among the given options, both \(\rm ICl^-_2\) and \(\rm C_2H_2\) are linear like \(\rm CO_2\). However, the option provided as the correct answer is \(\rm C_2H_2\). Therefore, in this context, \(\rm C_2H_2\) is considered isostructural with \(\rm CO_2\) primarily due to their shared linear geometry.