Problem 31 - 43 Years JEE ADVANCED (1978-2020) + JEE MAIN Chapterwise & Topicwise Solved Papers Chemistry

Question

Among the triatomic molecules/ions, \(\mathrm{BeCl}_{2}, \mathrm{~N}_{3}^{-}, \mathrm{N}_{2} \mathrm{O}, \mathrm{NO}_{2}^{+}, \mathrm{O}_{3}\),\(\mathrm{SCl}_{2}, \quad \mathrm{ICl}_{2}^{-}, \mathrm{I}_{3}^{-}\)and \(\mathrm{XeF}_{2}\), the total number of linear molecule(s)/ion(s) where the hybridization of the central atom does not have contribution from the \(d\)-orbital(s) is [Atomic number : \(\mathrm{S}=16, \mathrm{Cl}=17, \mathrm{I}=53\) and \(\mathrm{Xe}=54]\)

Step-by-Step Solution

Answer

There are 4 such molecules or ions.

Step 1: Identify Linear Molecules/Ions

First, we need to determine which of these triatomic molecules or ions are linear in shape. Linear structures typically have a bond angle of 180 degrees, and they are often molecular species with sp hybridization. Look for common linear molecules like \(BeCl_{2}\), \(N_{3}^{-}\), \(N_{2}O\), \(NO_{2}^{+}\), \(O_{3}\), \(SCl_{2}\), \(ICl_{2}^{-}\), \(I_{3}^{-}\), and \(XeF_{2}\).

Step 2: Check Hybridization for Each

For each linear molecule or ion, identify the hybridization of the central atom. Linear molecules without contributions from d-orbitals can be identified by sp-hybridization, which involves only s and p orbitals.

Step 3: BeCl_2 and N_3^- Analysis

For \(BeCl_2\), beryllium forms two bonds with \(Cl\) and is sp hybridized, which allows for a linear structure. \(N_3^-\) is linear due to its resonance structures and involves sp hybridized orbitals at the terminal nitrogen atoms.

Step 4: Analysis of N_2O and NO_2^+

\(N_{2}O\) is linear due to its bonding pattern, with the central nitrogen using sp hybridization. \(NO_{2}^+\) is linear due to the sp hybridization of the central nitrogen atom.

Step 5: Determine XeF_2 Hybridization

Though having a linear shape, \(XeF_{2}\) features dsp extsuperscript{3} (3-centered 4-electron interaction) hybridization due to xenon's involvement of the d orbitals, so it doesn't fit the criteria of no d orbital contribution.

Step 6: Count Suitable Molecules/Ions

BeCl_2, N_3^-, N_2O, and NO_2^+ are all linear and do not involve d-orbitals in the hybridization of their central atoms. The total number of linear triatomic molecules or ions without d orbital contributions to their hybridization is 4.

Key Concepts

Hybridization

Hybridization is a crucial concept in molecular geometry, describing how atomic orbitals mix to form new hybrid orbitals. These are necessary for creating chemical bonds with specific shapes and orientations. Hybridization explains how atoms can form bonds that don't necessarily follow the expected strict geometric angles given by just their atomic orbitals.
To determine hybridization, you consider the number of atoms connected to the central atom and any lone pairs of electrons, which gives the steric number. Depending on the steric number, different hybridizations occur, such as sp, sp2, sp3, and beyond. For example, a steric number of 2 leads to sp hybridization with linear geometry, like in the molecule \(BeCl_{2}\).
When analyzing molecules, look at the central atom's electron count and bonding to deduce its hybridization. This explains why some molecules, despite having complex atoms, appear simple due to hybrid orbital arrangements.

Linear Molecules

Linear molecules have a distinctive shape characterized by having all their atoms arranged in a straight line. This structure is associated with a bond angle of 180 degrees, which means the atoms are as far apart as possible, minimizing electron pair repulsion.
In the context of triatomic molecules, such as \(BeCl_2\) or \(N_3^-\), linearity occurs when only two points define the geometry, typically due to sp hybridization at the central atom. This hybridization results in two equal sigma bonds and no lone pairs, facilitating a straight-line structure.
Recognizing linear molecules involves understanding electron pair geometry and the hybridization state of the central atom. A molecule will form a linear shape if it has no lone pairs and two equal bond pairs on the central atom, exhibiting sp hybridization, which results in an even distribution of electron density.

Bond Angle

Bond angles are the angles formed between three atoms across at least two bonds. These angles are critical in determining and visualizing the shape of molecules. They arise from the repulsive forces between electron pairs around a central atom, known as VSEPR (Valence Shell Electron Pair Repulsion) theory.
In molecules with sp hybridization like \(N_2O\) and \(NO_2^+\), the bond angle is 180 degrees, signifying a linear arrangement. The lack of lone pairs allows the bonded atoms to spread out evenly, creating maximum distance between electron clouds, which defines a geometric line.
Understanding bond angles provides insight into molecular reactivity and interaction. Molecules with larger bond angles often have less steric hindrance, enabling smoother reactions with other species. Recognizing bond angles in different hybridizations further helps predict physical and chemical properties.

sp Hybridization

sp hybridization occurs when one s orbital mixes with one p orbital in an atom, resulting in two equivalent sp hybrid orbitals. This type of hybridization is typical for linear molecules where the central atom needs to form two sigma bonds in opposite directions, like in \(N_3^-\) and \(BeCl_2\).
Each sp hybrid orbital contains one electron pair used to form a sigma bond, while unhybridized p orbitals may participate in pi bonding if double or triple bonds are present.
In simple molecules, sp hybridization facilitates a straightforward geometry with 180-degree bond angles. This hybridization results in strong, directional bonds that are crucial for maintaining linear structures. Recognizing sp hybridization is fundamental for predicting molecular shapes and is directly related to the linear orientation of the involved atoms.