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The oxidation state(s) of \(\mathrm{Cl}\) in \(\mathrm{CaOCl}_{2}\) (bleaching powder) is/are: (a) \(+1\) only (b) \(-1\) only (c) \(+1\) and \(-1\) (d) none of these

Short Answer

Expert verified
The oxidation states of \(\mathrm{Cl}\) in \(\mathrm{CaOCl}_{2}\) (bleaching powder) are \(+1\) and \(-1\).

Step by step solution

01

Understand the compound structure

Recognize that bleaching powder, or \(\mathrm{CaOCl}_{2}\), is a mixed salt made of \(\mathrm{CaCl}_{2}\) and \(\mathrm{Ca}(\mathrm{OCl})_{2}\). Thus, it contains both \(\mathrm{Cl^{-}}\) ions and \(\mathrm{OCl^{-}}\) ions.
02

Assign oxidation states

For the \(\mathrm{CaCl}_{2}\) part, chloride ions are in the \(\mathrm{-1}\) oxidation state. For the \(\mathrm{Ca}(\mathrm{OCl})_{2}\) part, chlorine has a \(\mathrm{+1}\) oxidation state because oxygen is \(\mathrm{-2}\) and calcium is \(\mathrm{+2}\), maintaining charge neutrality.
03

Compare with the options given

Based on the assignments, chlorine has both \(\mathrm{+1}\) and \(\mathrm{-1}\) oxidation states in bleaching powder. Therefore, the correct answer is that both oxidation states are present for chlorine.

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Key Concepts

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

Oxidation Number Calculation
Understanding the oxidation number of an element within a compound is crucial for grasping many chemical concepts, such as redox reactions and compound formation. The oxidation number, often known as the oxidation state, signifies the degree of oxidation of an atom in a chemical compound. It is an indicator of the hypothetical charge an atom would have if all bonds to atoms of different elements were completely ionic.

When calculating the oxidation states, remember that the sum of oxidation states for all atoms in a neutral molecule must be zero, while in an ion, it should equal the ion's charge. Rules for assigning oxidation numbers include recognizing that elements in their elemental form have an oxidation state of zero and that for monoatomic ions, the oxidation state is equal to the ion's charge. In compounds, typically oxygen has an oxidation state of (-2), hydrogen is usually (+1), and other elements have varying states depending on the substance.

In the case of bleaching powder (CaOCl_{2}), the individual oxidation states of chlorine can be calculated by first understanding the compound's structure and then applying these rules to determine that chlorine exists in both (+1) and (-1) states within the different parts of the compound.
Chemical Compound Structure
A chemical compound's structure provides vital information about the way atoms are bonded to each other and how the molecule functions within chemical reactions. In the example of bleaching powder (CaOCl_{2}), its mixed salt structure can be seen as deriving from calcium chloride (CaCl_{2}) and calcium hypochlorite (Ca(OCl)_{2}). This kind of mixed salt is formed when calcium hydroxide (Ca(OH)_{2}) reacts with chlorine gas (Cl_{2}).

The distinct parts of the structure can be mentally visualized by separating the anions associated with the calcium cation (Ca^{2+}). We have the chloride anion (Cl^{-}) with a (-1) charge, and the hypochlorite anion (OCl^{-}), where the bond between oxygen and chlorine is more polar. It's important to note that the chlorine in the chloride anion has a different environment—and thus a different oxidation state—than the chlorine in the hypochlorite anion. This compound structure is essential when calculating the overall charge and assigning oxidation numbers appropriately.
Bleaching Powder Composition
Bleaching powder, chemically represented as CaOCl_{2}), is not a single simple compound but rather a mixture composed of calcium chloride (CaCl_{2}) and calcium hypochlorite (Ca(OCl)_{2}). It's vital to identify the function of bleaching powder; it's widely used as a disinfectant and oxidizing agent.

Upon breaking down its composition, the bleaching effect is primarily due to the presence of hypochlorite ions which release chlorine when in contact with water, leading to the formation of hypochlorous acid—a powerful bleaching agent. Meanwhile, the calcium chloride part remains relatively inert in terms of bleaching but contributes to the overall stability of the powder.

The dual existence of calcium chloride and calcium hypochlorite within bleaching powder is the reason behind its unique chemical properties, including the different oxidation states of chlorine. Understanding this composition allows chemists to exploit its properties for various practical applications, especially in sanitation and the treatment of water.

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Most popular questions from this chapter

A sample of calcium carbonate \(\left(\mathrm{CaCO}_{3}\right)\) has the following percentage composition : \(\mathrm{Ca}=40 \%\), \(\mathrm{C}=12 \%, \mathrm{O}=48 \%\). If the law of constant proportions is true, then the weight of calcium in \(4 \mathrm{~g}\) of a sample of calcium carbonate obtained from another source will be : (a) \(0.016 \mathrm{~g}\) (b) \(0.16 \mathrm{~g}\) (c) \(1.6 \mathrm{~g}\) (d) \(16 \mathrm{~g}\)

\(50 \mathrm{~mL}\) of \(20.8 \%(\mathrm{w} / \mathrm{V}) \mathrm{BaCl}_{2}(a q)\) and \(100 \mathrm{~mL}\) of \(9.8 \%(\mathrm{w} / \mathrm{V}) \mathrm{H}_{2} \mathrm{SO}_{4}(a q)\) solutions are mixed. Molarity of \(\mathrm{Cl}^{-}\) ions in the resulting solution is: (At. wt. of \(\mathrm{Ba}=137\) ) (a) \(0.333 \mathrm{M}\) (b) \(0.666 \mathrm{M}\) (c) \(0.1 \mathrm{M}\) (d) \(1.33 M\)

Balance the following equation and choose the quantity which is the sum of the coefficients of the products: $$ \ldots \ldots \mathrm{CS}_{2}+\ldots \ldots \mathrm{Cl}_{2} \longrightarrow \mathrm{CCl}_{4}+\ldots \ldots \mathrm{S}_{2} \mathrm{Cl}_{2} $$ (a) 5 (b) 3 (c) 6 (d) 2

Caffiene has a molecular weight of 194 . If it contains \(28.9 \%\) by mass of nitrogen, number of atoms of nitrogen in one molecule of caffeine is : (a) 4 (b) 6 (c) 2 (d) 3

\(32 \mathrm{~g}\) of a sample of \(\mathrm{FeSO}_{4} \cdot 7 \mathrm{H}_{2} \mathrm{O}\) were dissolved in dilute sulphuric acid and water and it's volume was made up to 1 litre, \(25 \mathrm{~mL}\) of this solution required \(20 \mathrm{~mL}\) of \(0.02 \mathrm{M} \mathrm{KMnO}_{4}\) solution for complete oxidation. Calculate the weight \(\%\) of \(\mathrm{FeSO}_{4} \cdot 7 \mathrm{H}_{2} \mathrm{O}\) in the sample. (a) \(34.75\) (b) \(69.5\) (c) \(89.5\) (d) None of these

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