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Chapter 4: Problem 18

Among \(\mathrm{KO}_{2}, \mathrm{AlO}_{2}^{-}, \mathrm{BaO}_{2}, \mathrm{NO}_{2}^{+}\) unpaired electron is present in: (a) \(\mathrm{NO}_{2}^{+}\) and \(\mathrm{BaO}_{2}\) (b) \(\mathrm{KO}_{2}\) and \(\mathrm{AlO}_{2}^{-}\) (c) \(\mathrm{KO}_{2}\) only (d) \(\mathrm{BaO}_{2}\) only

Short Answer

Expert verified
Unpaired electron is present in KO鈧� only.

Step by step solution

01

Determine the electron configuration for KO鈧�

KO鈧� (potassium superoxide) contains the superoxide ion O鈧傗伝. In superoxide, each oxygen has 8 electrons, giving O鈧傗倐鈦� a total of 17 electrons. The molecular orbital configuration is similar to O鈧�, but with an additional electron, making it \[ \sigma_{2s}^2 \sigma_{2s}^*2 \sigma_{2p_z}^2 \pi_{2p_x}^2 \pi_{2p_y}^2 \pi_{2p_x}^* \pi_{2p_y}^* \]This configuration results in one unpaired electron.
02

Analyze electron configuration for AlO鈧傗伝

AlO鈧傗伝 is the aluminate ion where aluminum has a +3 oxidation state, and oxygen provides two electrons, and with an extra electron from charge, it results in a closed shell configuration with no unpaired electrons.
03

Examine the electron configuration for BaO鈧�

BaO鈧� consists of a peroxide ion O鈧偮测伝, where each oxygen has 8 electrons. For O鈧偮测伝, each additional electron results in pairing, leading to a complete filling of the molecular orbitals. Thus, there are no unpaired electrons in BaO鈧�.
04

Evaluate electron configuration for NO鈧傗伜

NO鈧傗伜 involves nitrogen and oxygen, where the nitrogen shares coordinates with 5 electrons in its bonds, but loses one to have NO鈧傗伜. Its electronic configuration, when considering MO theory, ends up with closed shell molecular orbitals with no unpaired electrons.
05

Conclusion

The molecule with unpaired electrons, after analyzing all the given compounds, is KO鈧� because it has one unpaired electron due to its superoxide ion configuration.

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Key Concepts

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

Superoxide Ion
A superoxide ion is represented as O鈧傗伝 and is a key component in compounds like KO鈧� (potassium superoxide). In this ion, there is one extra electron compared to the neutral oxygen molecule (O鈧�). Each oxygen atom in the superoxide ion contributes eight electrons, resulting in a total count of 17 electrons for the entire ion. This odd number of electrons means that one electron does not have a pair, leading to an unpaired electron in the system.

Key properties of the superoxide ion include:
  • Appearance of orange-yellow crystals in combination with potassium.
  • Highly reactive nature, particularly with moisture and air.
  • Existence of paramagnetism due to its unpaired electron.
Understanding this concept is crucial for identifying the presence of unpaired electrons in chemical compounds like KO鈧�.
Molecular Orbital Theory
Molecular Orbital Theory (MO Theory) provides a fundamental understanding of how electrons are distributed in molecules. According to MO Theory, atomic orbitals combine to form molecular orbitals, which can be occupied by electrons. These could be of bonding, anti-bonding, or non-bonding character based on the arrangement.

Some highlights of MO Theory are:
  • Electrons are placed in molecular orbitals similar to the filling of atomic orbitals.
  • The theory describes how the arrangement affects the stability and magnetic properties of the molecule.
  • It helps to predict the presence of unpaired electrons by analyzing the occupancy of molecular orbitals.
In the case of superoxide ions, using MO Theory allows us to derive their electron configurations. For O鈧傗伝, it informs us that one more electron is added to the typical O鈧� configuration, affecting bonding characteristics and creating an unpaired electron in antibonding orbitals.
Electron Configuration
Electron configuration refers to the arrangement of electrons in an atom or ion. This concept helps to predict and explain the chemical behavior of elements by showing the distribution of electrons across different energy levels and sublevels.

Key points about electron configurations include:
  • Use of the Aufbau principle to fill orbitals from lower to higher energy levels.
  • The Pauli Exclusion Principle, stating no two electrons can have identical quantum numbers in the same atom.
  • Hund鈥檚 Rule, which optimizes electron configuration within a set of degenerate orbitals for maximum unpaired electrons.
In the context of the given problem, the electron configuration plays a vital role in determining the presence of unpaired electrons. KO鈧� has an additional electron leading to the setup \[\sigma_{2s}^2 \sigma_{2s}^*2 \sigma_{2p_z}^2 \pi_{2p_x}^2 \pi_{2p_y}^2 \pi_{2p_x}^* \pi_{2p_y}^* \]resulting in the unpaired electron that causes its unique chemical properties.

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Most popular questions from this chapter

The species which does not show paramagnetism? (a) \(\mathrm{O}_{2}\) (b) \(\mathrm{O}_{2}^{+}\) (c) \(\mathrm{O}_{2}^{2-}\) (d) \(\mathrm{H}_{2}^{+}\)

In the reaction, $$ \mathrm{HC}_{2} \mathrm{O}_{4}^{-}+\mathrm{PO}_{4}^{3-} \rightleftharpoons \mathrm{HPO}_{4}^{2-}+\mathrm{C}_{2} \mathrm{O}_{4}^{2} $$ the Bronsted bases are : (a) \(\mathrm{HC}_{2} \mathrm{O}_{4}^{-}\) (b) \(\mathrm{PO}_{4}^{3-}\) (c) \(\mathrm{HPO}_{4}^{2-}\) (d) \(\mathrm{C}_{2} \mathrm{O}_{4}^{2-}\)

(A) Auto reduction of ores is used for cinnabar, copper glance and galena ores. (R) The sulphide ores of \(\mathrm{Hg}, \mathrm{Cu}\) and \(\mathrm{Pb}\) react with their corresponding oxides to give respective metals.

An element has the configuration, $$ 1 s^{2}, 2 s^{2} 2 p^{6}, 3 s^{2} 3 p^{6} 3 d^{5}, 4 s^{2} $$ to which block and group of the long form of periodic table, does this belong? (a) \(d\) -block, 7 th group (b) \(d\) -block, 5 th group (c) \(s\) -block, 2 nd group (d) \(p\) -block, 15th group

The first ionisation energy of \(\mathrm{B}\) is less than that of \(\mathrm{Be}\) because : (a) boron has greater nuclear charge than beryllium (b) In beryllium \(s\) -electron is removed and from. boron \(p\) -electron is removed (c) \(p\) -electron of boron is more shielded from nucleus (d) both (b) and (c)
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