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Of the following sets, which one does not contain isoelectronic species? (a) \(\mathrm{CN}^{-}, \mathrm{N}_{2}, \mathrm{C}_{2}{ }^{2-}\) (b) \(\mathrm{PO}_{4}^{3-}, \mathrm{SO}_{4}^{2-}, \mathrm{ClO}_{4}^{-}\) (c) \(\mathrm{SO}_{3}{ }^{2-}, \mathrm{CO}_{3}^{2-}, \mathrm{NO}_{3}^{-}\) (d) \(\mathrm{BO}_{3}^{3-}, \mathrm{CO}_{3}^{2-}, \mathrm{NO}_{3}^{-}\)

Short Answer

Expert verified
Set (c) does not contain isoelectronic species.

Step by step solution

01

Understand Isoelectronic species

Isoelectronic species are atoms, molecules, or ions that have the same number of electrons. In general, the total number of electrons is found by adding the electrons of each atom involved plus or minus any extra due to the charge on the species.
02

Calculate Electrons for Set (a)

For \(\mathrm{CN}^{-}\): C has 6 electrons, N has 7 electrons, and there is 1 extra electron due to the negative charge. Total = 6 + 7 + 1 = 14 electrons. \(\mathrm{N}_{2}\): Each N has 7 electrons. Total = 7 + 7 = 14 electrons. \(\mathrm{C}_{2}{ }^{2-}\): Each C has 6 electrons, and there are 2 extra electrons due to the 2- charge. Total = 6 + 6 + 2 = 14 electrons.All have 14 electrons.
03

Calculate Electrons for Set (b)

For \(\mathrm{PO}_{4}^{3-}\): P has 15 electrons, each O has 8 electrons, and there are 3 extra electrons due to the 3- charge. Total = 15 + (4脳8) + 3 = 32 electrons. \(\mathrm{SO}_{4}^{2-}\): S has 16 electrons, each O has 8 electrons, and there are 2 extra electrons due to the 2- charge. Total = 16 + (4脳8) + 2 = 32 electrons. \(\mathrm{ClO}_{4}^{-}\): Cl has 17 electrons, each O has 8 electrons, and 1 extra electron due to the negative charge. Total = 17 + (4脳8) + 1 = 32 electrons.All have 32 electrons.
04

Calculate Electrons for Set (c)

For \(\mathrm{SO}_{3}{ }^{2-}\): S has 16 electrons, each O has 8 electrons, and 2 extra electrons due to the 2- charge. Total = 16 + (3脳8) + 2 = 32 electrons. \(\mathrm{CO}_{3}^{2-}\): C has 6 electrons, each O has 8 electrons, and there are 2 extra electrons due to the 2- charge. Total = 6 + (3脳8) + 2 = 24 electrons. \(\mathrm{NO}_{3}^{-}\): N has 7 electrons, each O has 8 electrons, and 1 extra electron due to the negative charge. Total = 7 + (3脳8) + 1 = 24 electrons.SO鈧兟测伝 has 32 electrons, CO鈧兟测伝 and NO鈧冣伝 have 24 electrons.
05

Calculate Electrons for Set (d)

For \(\mathrm{BO}_{3}^{3-}\): B has 5 electrons, each O has 8 electrons, and there are 3 extra electrons due to the 3- charge. Total = 5 + (3脳8) + 3 = 32 electrons. \(\mathrm{CO}_{3}^{2-}\): C has 6 electrons, each O has 8 electrons, and there are 2 extra electrons due to the 2- charge. Total = 6 + (3脳8) + 2 = 24 electrons. \(\mathrm{NO}_{3}^{-}\): N has 7 electrons, each O has 8 electrons, and 1 extra electron due to the negative charge. Total = 7 + (3脳8) + 1 = 24 electrons.BO鈧兟斥伝 has 32 electrons, CO鈧兟测伝 and NO鈧冣伝 have 24 electrons.
06

Conclusion: Determine Non-Isoelectronic Set

Set (c) and Set (d) do not have all species with the same number of electrons. Specifically, the counts differ in electrons between the species within these sets.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Electron Configuration
Electron configuration describes the arrangement of electrons in an atom's orbitals. It is essential in determining how atoms interact with each other. Electrons are filled into orbitals following specific rules, like the Aufbau principle, which states that electrons occupy the lowest energy orbitals first.

For example, the electron configuration for carbon (C) with 6 electrons is written as 1s虏 2s虏 2p虏. This notation tells you the number of electrons in each subatomic orbital. The electron count and distribution significantly influence an atom's chemical behavior.

When two atomic or molecular species have the same electron configuration, they are known as isoelectronic. Let's say you take ext{C}^{4+} ext{ and } ext{Be}, both end up having 2 electrons, with the configuration of 1s虏. This identity in their electron configuration translates to similar chemical properties.
Chemical Species
Chemical species are atoms, ions, molecules, or radicals involved in a chemical process or reaction. Each species has distinct properties and can participate in forming chemical compounds by sharing, losing, or gaining electrons.

To determine the nature of a chemical species, one examines its electron configuration, charge, and atomic composition. For instance, a negatively charged ion or an anion, such as ext{Cl}^{-}, is a chemical species with electrons added to the neutral chlorine atom. This gain creates new chemical behavior, allowing it to bond differently than its neutral counterpart.

Isoelectronic chemical species have identical electron numbers, which makes understanding chemical reactions more intuitive. Remember that isoelectronic species don鈥檛 necessarily share the same chemical properties, even though they have the same electronic configurations. Imagining them as puzzle pieces with the same shape but different colors can help discern the interactions.
Molecular Ions
Molecular ions are charged molecules formed when a molecule gains or loses electrons. These ions result typically from molecules interacting in a chemical reaction, leading to charged species that play vital roles in chemical processes.

A common example of a molecular ion is the sulfate ion ( ext{SO}_{4}^{2-} ext{). In this ion, sulfur is bonded to four oxygen atoms, with the entire molecule carrying a 2- charge. Molecular ions like these maintain the overall geometry and bond arrangement of the molecule, altering only with the loss or gain of electrons.

Calculating the total number of electrons in molecular ions is crucial in identifying their properties and reactivity. This analysis often requires considering the atomic components and the overall charge, leading to a comprehensive understanding of how such ions behave in reactions.

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Most popular questions from this chapter

When a white hygroscopic powder \(X\) is treated with cold water, a compound \(Y\) is formed which forms a white precipitate with \(\mathrm{BaCl}_{2}\) in acidic medium. \(\mathrm{Y}\) also gives a white amorphous precipitate with neutral \(\mathrm{AgNO}_{3}\). Compound \(\mathrm{Y}\) can also be obtained by heating \(\mathrm{H}_{3} \mathrm{PO}_{4}\) here \(\mathrm{X}\) and \(\mathrm{Y}\) are respectively (a) \(\mathrm{P}_{4} \mathrm{O}_{10}\) and \(\mathrm{H}_{3} \mathrm{PO}_{2}\) (b) \(\mathrm{P}_{4} \mathrm{O}_{10}\) and \(\mathrm{HPO}_{3}\) (c) \(\mathrm{P}_{4} \mathrm{O}_{6}\) and \(\mathrm{H}_{3} \mathrm{PO}_{2}\) (d) \(\mathrm{N}_{2} \mathrm{O}_{5}\) and \(\mathrm{HPO}_{3}\)

For \(\mathrm{H}_{3} \mathrm{PO}_{3}\) and \(\mathrm{H}_{3} \mathrm{PO}_{4}\) the correct choice is (a) \(\mathrm{H}_{3} \mathrm{PO}_{3}\) is dibasic and reducing (b) \(\mathrm{H}_{3} \mathrm{PO}_{3}\) is dibasic and non-reducing (c) \(\mathrm{H}_{3} \mathrm{PO}_{4}\) is tribasic and reducing (d) \(\mathrm{H}_{3} \mathrm{PO}_{4}\) is tribasic and non-reducing

\(\mathrm{PCl}_{3}\) and \(\mathrm{PCl}_{5}\) both exists; \(\mathrm{NCl}_{3}\) exists but \(\mathrm{NCl}_{5}\) does not exist. It is due to (a) lower electronegativity of \(\mathrm{P}\) than \(\mathrm{N}\) (b) lower tendency of \(\mathrm{N}\) to form covalent bond (c) availability of vacant d orbital in P but not in \(\mathrm{N}\) (d) statement is itself incorrect

A yellow metallic powder is burnt in a stream of fluorine to obtain a colourless gas A which is thermally stable and chemically inert. Its molecule has octahedral geometry. Another colourless gas \(\mathrm{B}\) with same constituent atoms as that of \(\mathrm{A}\) is obtained when sulphur dichloride is heated with sodium fluoride. Its molecule has trigonal bipyramidal geometry. Gases A and \(\mathrm{B}\) are respectively (a) \(\mathrm{SF}_{6}\) and \(\mathrm{SF}_{4}\) (b) \(\mathrm{SF}_{4}\) and \(\mathrm{SF}_{6}\) (c) \(\mathrm{SF}_{4}\) and \(\mathrm{S}_{2} \mathrm{~F}_{2}\) (d) \(\mathrm{KCl}\) and \(\mathrm{KBr}\)

Among the following molecules: (1) \(\mathrm{XeO}_{3}\) (2) \(\mathrm{XeOF}_{4}\) (3) \(\mathrm{XeF}_{6}\) Those having same number of lone pairs on \(\mathrm{Xe}\) are (a) (2) and (3) only (b) (1) and (2) only (c) \((1),(2)\) and (3) only (d) (1) and (3) only

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