6.
In the following structure
The ease of breaking phosphate bonds follows the order
a. I > II > III
b. III > II > I
c. II > III> I
d. I = II = III

The correct answer is: a. I > II > III.
In ATP, the phosphoanhydride bonds between the terminal phosphates (I and II) are high‑energy and break more easily than the phosphoester bond linking phosphate III to the ribose sugar, so I and II > III, and among I and II, the outermost terminal bond (I) is cleaved most readily in cells.​

Introduction: ease of breaking phosphate bonds in ATP

In ATP, three phosphate groups are linked in series to a ribose–adenine moiety, forming a triphosphate tail that stores cellular free energy. The ease of breaking phosphate bonds in ATP depends on the type of linkage (phosphoanhydride or phosphoester), the degree of electrostatic repulsion, and the stability of products formed after hydrolysis. Understanding the correct order, I > II > III, is essential for mastering bioenergetics questions in exams such as CSIR NET and NEET.​

Understanding the labeled bonds (I, II, III)

In the given structure (adenosine triphosphate):

• Bond III is between the ribose sugar and the first (α) phosphate; this is a phosphoester bond.​

• Bond II is the phosphoanhydride bond between the α and β phosphates.​

• Bond I is the phosphoanhydride bond between the β and γ (terminal) phosphates and is the bond typically cleaved during ATP → ADP + Pi hydrolysis.​

Phosphoanhydride bonds (I and II) are “high‑energy” because their hydrolysis generates products (ADP/AMP and inorganic phosphate or pyrophosphate) that are more stable and better solvated than ATP, leading to a large negative free energy change.​

Why the order I > II > III is correct

1. Greater electrostatic repulsion at the terminal bond (I):
   The three negatively charged phosphates of ATP are crowded together, and the outermost γ‑phosphate experiences significant repulsion from the inner phosphates, making the β–γ phosphoanhydride bond (I) relatively unstable and easier to cleave. Hydrolysis that removes the γ‑phosphate relieves the greatest amount of charge repulsion, so bond I is cleaved most readily.​

2. Stabilization of products after breaking bond I:
   Hydrolysis of bond I forms ADP and inorganic phosphate (Pi), which show better resonance stabilization and solvation than the intact triphosphate chain in ATP, giving a large negative ΔG. Because the products are much more stable, the reaction proceeds more easily, so bond I is the easiest to break.​

3. Bond II is next, but slightly harder to break:
   When ATP has already lost its γ‑phosphate, ADP now contains two phosphates (α and β), still linked by a phosphoanhydride bond (II), but with less total charge repulsion than in ATP. Thus, hydrolysis of bond II (ADP → AMP + Pi) still releases energy, but generally less than ATP → ADP + Pi, so bond II is easier to break than III but harder than I.​

4. Bond III (phosphoester) is the most stable:
   The linkage between the α‑phosphate and ribose (III) is a phosphoester bond, which is not classified as “high‑energy” and is comparatively stable under physiological conditions. Cleavage of this bond (AMP → adenosine + Pi) is less favorable, requiring more specific enzymatic conditions, so its ease of breaking is lowest among the three.​

Therefore, the ease of breaking phosphate bonds follows:
I (β–γ phosphoanhydride) > II (α–β phosphoanhydride) > III (phosphoester to ribose).

Explanation of each option

Option a: I > II > III (Correct)

• Reflects that both phosphoanhydride bonds (I and II) are high‑energy and easier to hydrolyze than the phosphoester bond (III).​

• Correctly places bond I (terminal β–γ) as easiest to break because its hydrolysis most effectively relieves charge repulsion and yields highly stabilized products (ADP + Pi).​

Option b: III > II > I (Incorrect)

• This option incorrectly suggests the phosphoester bond (III) is easiest to break, but in reality, this bond is more stable and not classed as high‑energy.​

• It also reverses the relative ease of cleavage of the two phosphoanhydride bonds, contradicting the fact that ATP → ADP + Pi (breaking I) is generally more exergonic and more commonly used than ADP → AMP + Pi (breaking II).​

Option c: II > III > I (Incorrect)

• Here, bond II is placed as easiest to break, but experimentally, cells predominantly hydrolyze ATP at the β–γ bond (I), not at the α–β bond (II), indicating that I is more labile in physiological reactions.​

• The ordering III > I here would also mean the phosphoester bond breaks more readily than the terminal anhydride bond, which is inconsistent with the definition of “high‑energy” phosphoanhydride linkages.​

Option d: I = II = III (Incorrect)

• This assumes all three bonds are equally easy to break, ignoring the fundamental chemical distinction between phosphoanhydride (I, II) and phosphoester (III) linkages and their different ΔG of hydrolysis values.​

• Experimental thermodynamic data and cellular metabolism both show that terminal phosphoanhydride hydrolysis (I) is favored, followed by internal phosphoanhydride (II), whereas cleavage of the phosphoester bond (III) is least favored, so equality is not valid.​

If you want, a second article can be created focusing only on “high‑energy bonds of ATP” with additional exam‑style MCQs built around this same concept.