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Chapter 3: Problem 81

The isoelectronic pair is: a. \(\mathrm{ClO}_{2}^{-}, \mathrm{ClF}_{2}^{+}\) b. \(\mathrm{IF}_{2}^{+}, \mathrm{I}_{3}^{-}\) c. \(\mathrm{Cl}_{2} \mathrm{O}_{3}, \mathrm{ICl}_{2}^{-}\) d. \(\mathrm{ICl}_{2}-\mathrm{ClO}_{2}\)

Short Answer

Expert verified
None of the options are isoelectric pairs.

Step by step solution

01

Understand Isoelectric Species

Isoelectric species are atoms, ions, or molecules that have the same number of electrons. When comparing these species, we need to calculate the total number of electrons for each and confirm that they match.
02

Calculate Electrons for Option A

For - \(\mathrm{ClO}_{2}^{-}\): Cl = 17, O = 16 each, plus an extra electron because of the negative charge gives a total of \(17 + 2(16) + 1 = 50\) electrons.- \(\mathrm{ClF}_{2}^{+}\): Cl = 17, F = 9 each, minus one electron for the positive charge gives a total of \(17 + 2(9) - 1 = 34\) electrons. These numbers do not match.
03

Calculate Electrons for Option B

For - \(\mathrm{IF}_{2}^{+}\): I = 53, F = 9 each, minus one electron for the positive charge gives a total of \(53 + 2(9) - 1 = 70\) electrons.- \(\mathrm{I}_{3}^{-}\): I = 53 each, plus one electron for the negative charge gives a total of \(3(53) + 1 = 160\) electrons. These numbers do not match.
04

Calculate Electrons for Option C

For - \(\mathrm{Cl}_{2} \mathrm{O}_{3}\): Cl = 17 each, O = 16 each gives a total of \(2(17) + 3(16) = 82\) electrons.- \(\mathrm{ICl}_{2}^{-}\): I = 53, Cl = 17 each, plus one electron for the negative charge gives a total of \(53 + 2(17) + 1 = 88\) electrons. These numbers do not match.
05

Calculate Electrons for Option D

For - \(\mathrm{ICl}_{2}^{-}\): I = 53, Cl = 17 each, plus one electron for the negative charge gives a total of \(53 + 2(17) + 1 = 88\) electrons.- \(\mathrm{ClO}_{2}\): Cl = 17, O = 16 each gives a total of \(17 + 2(16) = 49\) electrons. These numbers do not match.
06

Conclusion: Identify the Isoelectric Pair

None of the given options present a pair of species that have the same number of electrons.

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Key Concepts

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

Electron Count
Calculating the electron count is an essential step when working with isoelectronic species, which are atoms, ions, or molecules that share the same number of electrons. To determine the electron count:
  • Animals, ions, or molecules must have their atomic number paired with the correct charge contribution.
  • Neutral atoms use their atomic number to determine their electron count, as the number of protons and electrons are equal.
  • For ions, add an electron if the charge is negative, or subtract one if positive, since charges indicate a gain or loss in electrons.
For example, in the molecule \(\mathrm{ClO}_{2}^{-} \), Chlorine (Cl) contributes 17 electrons, and Oxygen (O) contributes 16 electrons each. The negative charge means an additional electron, so the total is \( 17 + 2 \times 16 + 1 = 50 \) electrons. Getting the electron count right is crucial for comparing different species.
Isoelectronic Pairs
Isoelectronic pairs involve comparing species to identify pairs with identical numbers of electrons. This is significant because:
  • Isoelectronic pairs often share similar chemical and physical properties due to the identical electron cloud configuration.
  • The process involves calculating each species' electron count, as discussed before, and determining if they match.
The absence of an isoelectronic pair in the original exercise emphasizes the importance of precise calculations and understanding this concept effectively. If two species, like \( \mathrm{Ne} \) and \( \mathrm{Mg}^{2+} \), share an electron count, they are considered isoelectronic, both having 10 electrons.
Ionic Charges
Ionic charges play a pivotal role in determining electron count and, afterward, identifying isoelectronic pairs. An ion's charge signifies either a surplus or deficit of electrons compared to its neutral state:
  • Positive charges (cations) signify electron loss, which lowers the total electron count from the element's neutral state.
  • Negative charges (anions) mean electron gain, increasing the electron count from the element's neutral state.
For instance, \( \mathrm{ClF}_{2}^{+} \) loses an electron due to the +1 charge, thereby changing its total electron count to \( 34 \) instead of an uncharged \( 35 \). Understanding ionic charges helps in predicting the behavior of the species and their possible interactions.
Species Comparison
Comparing species is a necessary step in determining isoelectronic pairs. After calculating electron counts and considering ionic charges, comparisons allow us to discern similarities in electron configurations among species:
  • Start by listing all species under examination.
  • Calculate and list electron counts, adjusting for any charge translations.
  • Compare these counts in pairs to identify any matches.
None of the options in the exercise met the criteria for isoelectronic pairs due to varied electron counts, highlighting the necessity for diligence in computation. This forms the foundation for further chemical analysis.

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Most popular questions from this chapter

Photoelectric emission is observed from a surface for frequencies \(\mathrm{v}_{1}\) and \(\mathrm{v}_{2}\) of the incident radiation \(\left(\mathrm{v}_{1}>\mathrm{v}_{2}\right)\). If the maximum kinetic energies of the photoelectrons in the two cases are in the ratio \(1: \mathrm{k}\) then the threshold frequency \(\mathrm{v}_{0}\) is given by a. \(\frac{\mathrm{kv}_{2}-\mathrm{v}_{\mathrm{l}}}{\mathrm{k}-1}\) b. \(\frac{\mathrm{v}_{2}-\mathrm{v}_{1}}{\mathrm{k}}\) c. \(\frac{\mathrm{v}_{2}-\mathrm{v}_{1}}{\mathrm{k}-1}\) d. \(\frac{\mathrm{kv}_{1}-\mathrm{v}_{2}}{\mathrm{k}-1}\)

The electronic configuration of four elements \(\mathrm{P}\), \(\mathrm{Q}, \mathrm{R}\) and \(\mathrm{S}\) given below. Choose the element that would most readily form a diatomic molecule. \(\mathrm{P}=1 \mathrm{~s}^{2} 2 \mathrm{~s}^{2} 2 \mathrm{p}^{1}, \mathrm{Q}=1 \mathrm{~s}^{2} 2 \mathrm{~s}^{2} 2 \mathrm{p}^{5}\) \(\mathrm{R}=1 \mathrm{~s}^{2} 2 \mathrm{~s}^{2} 2 \mathrm{p}^{6} 3 \mathrm{~s}^{1}, \mathrm{~S}=1 \mathrm{~s}^{2} 2 \mathrm{~s}^{2} 2 \mathrm{p}^{2}\) a. \(\mathrm{Q}\) b. \(\underline{S}\) c. \(\mathrm{P}\) d. \(\mathrm{R}\)

Which of the following statement(s) are correct? 1\. the electronic configuration of \(\mathrm{Cr}\) is \([\mathrm{Ar}] 3 \mathrm{~d}^{5}\) \(4 \mathrm{~s}^{1}\) (atomic number of \(\mathrm{Cr}=24\) ) 2\. the magnetic quantum number may have a negative value 3\. in silver atom, 23 electrons have a spin of one type and 24 of the opposite type (atomic number of \(\mathrm{Ag}=47\) ) 4\. the oxiation state of nitrogen in \(\mathrm{HN}_{3}\) is \(-3\). a. \(1,2,3\) b. \(2,3,4\) c. 3,4 d. \(1,2,4\)

The ionization energy of \(\mathrm{He}^{+}\)is \(19.6 \times 10^{-18} \mathrm{~J}\) atom \(^{1}\). Calculate the energy of the first stationary state of \(\mathrm{Li}^{2+}\) a. \(19.6 \times 10^{-18} \mathrm{~J}\) atom \(^{-1}\) b. \(4.41 \times 10^{-18} \mathrm{~J}\) atom-1 c. \(19.6 \times 10^{-19} \mathrm{~J}\) atom \(^{-1}\) d. \(4.41 \times 10^{-17} \mathrm{~J}\) atom-1

The kinetic energy of photoelectron emitted on irradiating a metal surface with frequency \(v\) is related by \(\mathrm{KE}=\mathrm{hv}-\mathrm{IE}\). The plots of \(\mathrm{KE}\) vs. incidented frequency \(\mathrm{v}\) shows: a. A straight line with slope equal to Planck's constant. b. A straight line with intercept on \(x\)-axis equal to the product of threshold frequency and Planck's constant. c. A straight line with extrapolated intercept on \(\mathrm{y}\) axis equal to ionization energy d. A straight line with intercept on \(x\)-axis equal to threshold frequency
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