Problem 123
Question
The incorrect statement/s among the following is /are (a) \(\mathrm{NCl}_{5}\) does not exist while \(\mathrm{PCl}_{5}\) does (b) Lead prefers to form tetravalent compounds (c) The three \(\mathrm{C}-\mathrm{O}\) bonds are not equal in the carbonate ion (d) Both \(\mathrm{O}_{2}^{+}\)and \(\mathrm{NO}\) are paramagnetic
Step-by-Step Solution
Verified Answer
Statements (b) and (c) are incorrect.
1Step 1: Understanding the Problem
We need to identify which statements among the given options are incorrect.
2Step 2: Analyzing Statement (a)
Statement (a) says "\(\mathrm{NCl}_{5}\) does not exist while \(\mathrm{PCl}_{5}\) does." This is correct because nitrogen cannot expand its valence shell to accommodate 5 chlorine atoms due to the absence of d-orbitals, whereas phosphorus can.
3Step 3: Analyzing Statement (b)
Statement (b) is "Lead prefers to form tetravalent compounds." This is incorrect because lead typically prefers to form divalent (\(\text{Pb}^{2+}\)) compounds due to the inert pair effect.
4Step 4: Analyzing Statement (c)
Statement (c) says "The three \(\mathrm{C}-\mathrm{O}\) bonds are not equal in the carbonate ion." In the carbonate ion (\(\mathrm{CO}_3^{2-}\)), all three \(\mathrm{C}-\mathrm{O}\) bonds are equivalent due to resonance, making this statement incorrect.
5Step 5: Analyzing Statement (d)
Statement (d) claims "Both \(\mathrm{O}_{2}^{+}\) and \(\mathrm{NO}\) are paramagnetic." This is true because both molecules have unpaired electrons in their molecular orbitals.
Key Concepts
PCl5 StructureInert Pair EffectCarbonate Ion ResonanceParamagnetism in Molecular Orbitals
PCl5 Structure
The molecule \(PCl_5\) has a unique and interesting structure that allows it to exist due to phosphorus's ability to utilize d-orbitals. This ability to expand its valence shell enables it to form bonds with five chlorine atoms, hence creating a stable compound.
Phosphorus forms \(sp^3d\) hybridized orbitals in \(PCl_5\), resulting in a trigonal bipyramidal arrangement.
This geometric configuration helps evenly distribute electron pairs around the phosphorus atom, minimizing electron-electron repulsion. Thus, it results in a stable and efficient molecular structure.
Phosphorus forms \(sp^3d\) hybridized orbitals in \(PCl_5\), resulting in a trigonal bipyramidal arrangement.
This geometric configuration helps evenly distribute electron pairs around the phosphorus atom, minimizing electron-electron repulsion. Thus, it results in a stable and efficient molecular structure.
- The molecule consists of two axial and three equatorial chlorine atoms.
- The axial bonds are slightly longer than the equatorial ones due to spatial arrangement and electron repulsion differences.
Inert Pair Effect
The inert pair effect is a key concept in understanding the chemistry of heavier p-block elements. It explains why elements like lead prefer to form compounds in lower oxidation states than expected.
Lead tends to form stable \(\text{Pb}^{2+}\) ions rather than \(\text{Pb}^{4+}\) ions. This inclination arises because the s-electrons (the 'inert pair') are less often involved in bonding as the atomic size increases down the group.
Lead tends to form stable \(\text{Pb}^{2+}\) ions rather than \(\text{Pb}^{4+}\) ions. This inclination arises because the s-electrons (the 'inert pair') are less often involved in bonding as the atomic size increases down the group.
- The inert pair effect becomes more pronounced with larger and heavier atoms.
- This effect affects the stability of oxidation states across the periodic table.
Carbonate Ion Resonance
Resonance is a powerful concept that explains the structure of molecules like the carbonate ion \(\mathrm{CO}_3^{2-} \).
In reality, the carbonate ion exhibits resonance, where the electrons are delocalized across the three oxygen atoms. This delocalization allows all \(\mathrm{C}-\mathrm{O}\) bonds to be equivalent in length and strength, unlike what some models initially suggest.
In reality, the carbonate ion exhibits resonance, where the electrons are delocalized across the three oxygen atoms. This delocalization allows all \(\mathrm{C}-\mathrm{O}\) bonds to be equivalent in length and strength, unlike what some models initially suggest.
- Resonance structures show possible electron locations but do not depict the actual molecule.
- The actual structure is a hybrid of these resonance forms, contributing to bond equivalency.
Paramagnetism in Molecular Orbitals
Paramagnetism in molecules arises from the presence of unpaired electrons in the molecular orbitals. Molecules such as \(\mathrm{O}_2^+\) and \(\mathrm{NO}\) exhibit paramagnetic behavior due to these unpaired electrons.
In the molecular orbital theory framework, each molecular orbital can accommodate two electrons with opposite spins. When one or more orbitals have unpaired electrons, the molecule exhibits paramagnetism, attracting magnetic fields.
In the molecular orbital theory framework, each molecular orbital can accommodate two electrons with opposite spins. When one or more orbitals have unpaired electrons, the molecule exhibits paramagnetism, attracting magnetic fields.
- In \(\mathrm{O}_2^+\), one electron is removed from the otherwise paired electrons of \(\mathrm{O}_2\), leaving an unpaired electron in its highest occupied orbital.
- The molecule \(\mathrm{NO}\) naturally has an unpaired electron in its molecular orbital configuration, contributing to its paramagnetic nature.
Other exercises in this chapter
Problem 121
Which of the following order is incorrect? (1) \(\mathrm{PH}_{3}\mathrm{PH}_{3}>\mathrm{AsH}_{3}>\mathrm{SbH}_{3}>\mathrm{BiH}_{3}\) (Dipole moment) (a) only 1
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Match the following: List I 1\. most electronegative 2\. most electron affinity 3\. liquid halogen 4\. radio active halogen List II (i) \(\mathrm{I}_{2}\) (ii)
View solution Problem 124
Which of the following halogens does not turn starch iodine paper blue? (a) \(\mathrm{I}_{2}\) (b) \(\mathrm{F}_{2}\) (c) \(\mathrm{Cl}_{2}\) (d) \(\mathrm{Br}_
View solution Problem 126
Which of these is/are not true about the oxoacids of halogens? (a) All of them are good reducing agents (b) All of them are monobasic (c) They may be monobasic
View solution