A compound contains atoms \(X, Y, Z\) the oxidation number of \(\mathrm{X}\) is \(+2, \mathrm{Y}\) is \(+5\) and \(\mathrm{Z}\) is \(-2\) The possible form of the compound is: (a) \(\mathrm{XY}_{1} \mathrm{Z}_{2}\) (b) \(\mathrm{Y}_{2}\left(\mathrm{XZ}_{3}\right)_{2}\) (c) \(\mathrm{X}_{3}\left(\mathrm{YZ}_{4}\right)_{2}\) (d) \(\mathrm{X}_{3}\left(\mathrm{Y}_{4} \mathrm{Z}\right)_{2}\)

Short Answer

Expert verified
The compound with the formula \(\mathrm{X}_{3}\left(\mathrm{YZ}_{4}\right)_{2}\) is electrically neutral, making option (c) the correct answer.

Step by step solution

01

Understanding the Principle of Charge Neutrality

For a compound to be stable, the total positive charge must equal the total negative charge. The oxidation numbers indicate the charges atoms would have if the compound were ionic. We need to find a compound where the total sum of the oxidation numbers equals zero.
02

Analyzing Option (a) - XY1Z2

Calculate the net charge: Charge of X (+2) + Charge of Y (+5) + 2 * Charge of Z (-2) = +2 + +5 + (-4) = +3. The compound is not neutral, so (a) cannot be the correct formula.
03

Analyzing Option (b) - Y2(XZ3)2

Calculate the net charge: 2 * Charge of Y (+5) + 2 * (Charge of X (+2) + 3 * Charge of Z (-2)) = 2 * +5 + 2 * (+2 + (-6)) = +10 + 2 * (-4) = +10 - 8 = +2. The compound is not neutral, so (b) cannot be the correct formula.
04

Analyzing Option (c) - X3(YZ4)2

Calculate the net charge: 3 * Charge of X (+2) + 2 * (Charge of Y (+5) + 4 * Charge of Z (-2)) = 3 * +2 + 2 * (+5 + (-8)) = +6 + 2 * (-3) = +6 - 6 = 0. The compound is neutral, so (c) is the correct formula.
05

Analyzing Option (d) - X3(Y4Z)2

For completion, let's verify if option (d) is incorrect. Calculate the net charge: 3 * Charge of X (+2) + 2 * (4 * Charge of Y (+5) + Charge of Z (-2)) = 3 * +2 + 2 * (+20 - 2) = +6 + 2 * 18 = +6 + 36 = +42. The compound is not neutral, so (d) cannot be the correct formula.

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Key Concepts

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

Chemical Neutrality
Understanding chemical neutrality is fundamental in the study of chemistry, especially when examining ionic and covalent compounds. Chemical neutrality refers to the condition where the total positive charge in a compound equals the total negative charge, resulting in a net charge of zero. This balance is not just coincidental but a requirement for the stability of compounds, as nature favors balance and minimization of energy.

Imagine a seesaw with ions on both sides; for the seesaw to be level, both sides must carry the same weight, or charge, in our case. Atoms in a compound 'share' or 'transfer' electrons to achieve a filled outer shell similar to the noble gases, known for their stability because they have a full valence shell. In ionic compounds, which consist of metals and nonmetals, metals tend to lose electrons and become positively charged, while nonmetals tend to gain electrons and become negatively charged.

Using this principle, we can predict and understand the structure of unknown compounds and validate if a proposed formula is feasible. For instance, in the given textbook exercise, the total oxidation numbers must sum up to zero to consider the compound chemically neutral and, thus, likely to be correct. This principal guides us through analyzing which option is the stable form of the compound containing atoms X, Y, and Z.
Oxidation State Calculation
The oxidation state or oxidation number of an atom within a compound provides insight into the electron distribution in that molecule. An oxidation state is a hypothetical charge that an atom would have if all its bonds were purely ionic. While calculating the oxidation state, some rules should be followed:
  • For an atom in its elemental form, the oxidation number is always zero.
  • For a simple monoatomic ion, the oxidation number equals the ion's charge.
  • In compounds, hydrogen typically has an oxidation number of +1, and oxygen has -2, except in peroxides where it is -1.
  • The algebraic sum of oxidation numbers for all atoms in a molecule or polyatomic ion equals the charge on the molecule or ion.

Applied to our exercise, considering the oxidation number of X is +2, Y is +5 and Z is -2, we can calculate the oxidation states and verify if the sum equals zero as it should in a compound that adheres to the rule of chemical neutrality. This systematic approach allows us to determine the correct stoichiometry of the compound by evaluating each given option, ensuring the oxidation states balance out.
Stable Compound Formation
Stable compound formation is the end goal when atoms interact and combine. Stability in chemistry often means having a full valence electron shell, and atoms will share, lose, or gain electrons to achieve this state. A stable compound, therefore, is one where all participating atoms have reached a level of electronic configuration that minimizes their potential energy.

Whether compounds are formed through ionic or covalent bonding, the principle remains the same: atoms will strive for the electronic configuration that provides the most stability. Ionic compounds are typically formed between metals and nonmetals and are primarily held together by the electrostatic attraction between oppositely charged ions. On the other hand, covalent compounds are formed when two nonmetals share electrons to fill their outer shells.

In the context of the exercise, we're looking for a compound that is chemically neutral. The compound will be stable if it has been formed correctly according to the rules of oxidation states and there is an equal balance between the total positive and negative charges. Once identified, the stable form of the compound can be rightfully confirmed, as was shown with option (c) in our step-by-step calculation. The concept of stable compound formation underlines much of what we explore in chemistry, guiding us from individual ion behavior to the structure of complex molecules.

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Most popular questions from this chapter

Which of the following does not represent redox reactions? (a) \(\mathrm{Cr}_{2} \mathrm{O}_{7}^{2-}+2 \mathrm{OH}^{-} \longrightarrow \mathrm{CrO}_{4}^{2-}+\mathrm{H}_{2} \mathrm{O}\) (b) \(\mathrm{SO}_{5}^{2-}+2 \mathrm{I}^{-}+2 \mathrm{H}^{+} \longrightarrow \mathrm{I}_{2}+\mathrm{SO}_{4}^{2-}+\mathrm{H}_{2} \mathrm{O}\) (c) \(2 \mathrm{Ca}(\mathrm{OH})_{2}+2 \mathrm{Cl}_{2} \longrightarrow \mathrm{Ca}(\mathrm{ClO})_{2}+\mathrm{CaCl}_{2}+\) \(2 \mathrm{H}_{2} \mathrm{O}\) (d) \(\mathrm{PCl}_{5} \longrightarrow \mathrm{PCl}_{3}+\mathrm{Cl}_{2}\)

The standard reduction potential for \(\mathrm{Fe}^{2+} / \mathrm{Fe}\) and \(\mathrm{Sn}^{2+} /\) Sn electrodes are \(-0.44\) and \(-0.14\) volts respectively. For the cell reaction: \(\mathrm{Fe}^{2+}+\mathrm{Sn} \longrightarrow \mathrm{Fe}+\mathrm{Sn}^{2+}\) The standard emf is: (a) \(+0.30 \mathrm{~V}\) (b) \(-0.58 \mathrm{~V}\) (c) \(+0.58 \mathrm{~V}\) (d) \(-0.300 \mathrm{~V}\)

The equivalent weight of \(\mathrm{Na}_{2} \mathrm{~S}_{2} \mathrm{O}_{3}\) in the reaction \(2 \mathrm{Na}_{2} \mathrm{~S}_{2} \mathrm{O}_{3}+\mathrm{I}_{2} \longrightarrow \mathrm{Na}_{2} \mathrm{~S}_{4} \mathrm{O}_{6}+2 \mathrm{NaI}\) is (M = molecular weight):

The equivalent conductance at infinite dilution of \(\mathrm{HCl}\) and \(\mathrm{NaCl}\) are \(426.15\) and \(126.15 \mathrm{mho} \mathrm{cm}^{2} \mathrm{~g} \mathrm{eq}^{-1}\) respectively. It can be said that the mobility of: (a) \(\mathrm{H}^{+}\) ions is much more than that of \(\mathrm{Cl}\) " ions (b) \(\mathrm{Cl}^{-}\) ions is much more than that of \(\mathrm{H}^{+}\) ions (c) \(\mathrm{H}^{+}\) ions is much more than that of \(\mathrm{Na}^{+}\) ions (d) \(\mathrm{Na}^{+}\) ions is much more than that of \(\mathrm{H}^{+}\) ions

Conductivity (Seimens \(S\) ) is directly proportional to area of the vessel and the concentration of the solution in it and is inversely proportional to the length of the vessel, then constant of proportionality is expressed in: (a) \(\mathrm{S} \mathrm{m} \mathrm{mol}^{-1}\) (b) \(\mathrm{S}^{2} \mathrm{~m}^{2} \mathrm{~mol}^{-2}\) (c) \(\mathrm{S} \mathrm{m}^{2} \mathrm{~mol}^{-1}\) (d) \(\mathrm{S}^{2} \mathrm{~m}^{2} \mathrm{~mol}\)

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