The oxidation number of phosphorous in ATP (adenosine triphosphate) is: (a) 2 (b) 3 (c) 4 (d) 5

Short Answer

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The oxidation number of phosphorous in ATP is (d) 5.

Step by step solution

01

Identify the Compound Structure

ATP or adenosine triphosphate is a nucleotide that consists of adenosine (adenine ring and ribose sugar) and three phosphate groups linked together. The oxidation number we are interested in is of the phosphorous in the phosphate groups.
02

Understand the Oxidation State Rules

The oxidation state of an atom is the charge that an atom would have if all bonds were ionic. Oxygen has an oxidation state of -2 in most of its compounds, and hydrogen has an oxidation state of +1 when bonded to non-metals.
03

Calculate the Oxidation State of Phosphorus

In the phosphate ion, \(PO_4^{3-}\), phosphorus is bonded to four oxygen atoms. If we consider all bonds to be ionic, the combined oxidation state for the oxygens is \(4 \times (-2) = -8\). Since the phosphate ion carries a -3 charge overall, then the oxidation state of phosphorus must be such that when added to the -8 from oxygen equals the total charge: \(x + (-8) = -3\), solving for x gives us the oxidation state of phosphorus, \(x = +5\).

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Key Concepts

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

Understanding Oxidation States
When it comes to mastering redox processes and predicting the behavior of elements in compounds, oxidation states play a pivotal role. The term 'oxidation state' refers to a hypothetical charge that an atom would possess if all of its bonds were purely ionic. This concept helps us understand how electrons are shared or transferred in chemical reactions.

By adhering to certain guidelines such as oxygen usually having an oxidation state of -2 and hydrogen being +1 when bonded to non-metals, students can deduce the overall charge of complex molecules. It's like piecing together a puzzle to find out how much charge each piece (atom) must have to achieve the established charge of the larger structure. For example, in a phosphate group, where the net charge is -3, the oxidation state of phosphorous can be calculated by balancing the charge contributed by the oxygen atoms in the structure.
Exploring Phosphorous Chemistry
Phosphorous is an incredibly versatile element, central to many biochemical compounds, including ATP. It has multiple oxidation states, spanning from -3 to +5, which enable it to form a wide range of compounds. Phosphorus forms phosphates when it reacts with oxygen; these are key constituents of DNA, RNA, and ATP, dictating the chemical energy currency of the cell.

In ATP, phosphorus primarily exhibits the +5 oxidation state, allowing for the formation of a stable triphosphate group. This stability is crucial for its role in energy transfer. Moreover, understanding the chemical properties of phosphorus and how it forms bonds is essential for grasping its function in biological molecules and technological applications.
Chemical Bonding Concepts
Chemical bonding is the force that holds atoms together in molecules and compounds. The main types of chemical bonds include ionic, covalent, and metallic. Ionic bonds form between metals and non-metals where electrons are transferred, while covalent bonds occur when atoms share electrons. In organic and biochemistry contexts, like in ATP, covalent bonds are more prevalent.

In the context of ATP and its phosphate groups, the bonds are predominantly covalent, but we can use the concept of oxidation states by imagining these bonds as ionic, allowing for easier computation of an element's formal charge within the molecule. This understanding not only aids in recognizing how molecules hold together but also in predicting the products of chemical reactions.
NEET Chemistry Preparation
NEET aspirants need to develop a strong foundational understanding of concepts like oxidation states and chemical bonding, as they are crucial for answering various chemistry questions in the exam. However, beyond theory, NEET requires the ability to apply concepts effectively to solve problems efficiently.

Preparing for questions on phosphorous chemistry and oxidation numbers, for example, will involve practicing with compounds like ATP. Aspiring students should not only memorize the rules but also engage in solving numerous practice problems to become proficient. It is through consistent practice and a clear conceptual understanding that one can ace the NEET chemistry section.

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Most popular questions from this chapter

The value of \(E^{\circ}\) at \(298 \mathrm{~K}\) for the given reaction is \(1.5\) \(\mathrm{V}\) find the \(\Delta \mathrm{G}\) value here? \(\mathrm{Cr}+3 \mathrm{Ag}^{+}(0.1 \mathrm{M}) \rightarrow 3 \mathrm{Ag}^{\prime}+\mathrm{Cr}^{3+}\) (a) \(-422.83 \mathrm{KJ}\) (b) \(-211.415 \mathrm{KJ}\) (c) \(+211.415 \mathrm{KJ}\) (d) \(+422.83 \mathrm{KJ}\)

A solution of sodium sulphate in water is electrolyzed using inert electrodes. The products at the cathode and anode are respectively: (a) \(\mathrm{O}_{2}, \mathrm{H}_{2}\) (b) \(\mathrm{O}_{2}, \mathrm{Na}\) (c) \(\mathrm{O}_{2}, \mathrm{SO}_{2}\) (d) \(\mathrm{H}_{2}, \mathrm{O}_{2}\)

The oxidation number of sulphur in \(\mathrm{S}_{8}, \mathrm{~S}_{2} \mathrm{~F}_{2}, \mathrm{H}_{2} \mathrm{~S}\) respectively, are: (a) \(0,+1\) and \(-2\) (b) \(+2,+1\) and \(-2\) (c) \(0,+1\) and \(+2\) (d) \(-2,+1\) and \(-2\).

\(\mathrm{Cu}^{+}\) (aq) is unstable in solution and undergoes simultaneous oxidation and reduction according to the reaction, \(2 \mathrm{Cu}^{+}\) (aq) \(\rightleftharpoons \mathrm{Cu}^{2+}(\mathrm{aq})+\mathrm{Cu}(\mathrm{s})\) Choose correct \(E^{\circ}\) for above reaction if \(\mathrm{E}^{\circ}\left(\mathrm{Cu}^{2+} / \mathrm{Cu}\right)=0.34 \mathrm{~V}\) and \(\mathrm{E}^{\circ}\left(\mathrm{Cu}^{2+} / \mathrm{Cu}^{+}\right)=0.15 \mathrm{~V}\) (a) \(-0.38 \mathrm{~V}\) (b) \(+0.49 \mathrm{~V}\) (c) \(+0.38 \mathrm{~V}\) (d) \(-0.19 \mathrm{~V}\)

In the electrolysis of a fused salt, the weight of the deposit on the electrode, will not depend on: (a) Current intensity (b) Electrochemical equivalent of the ions (c) Time for electrolysis (d) Temperature of the bath

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