Q. Given below are two statements:
Statement I: $H_2Se$ is more acidic than $H_2Te$
Statement II: $H_2Se$ has higher bond enthalpy for dissociation than $H_2Te$
In the light of the above statements, choose the correct answer:
A) Both Statement I and Statement II are true
B) Statement I is true but Statement II is false
C) Both Statement I and Statement II are false
D) Statement I is false but Statement II is true
Statement I: $H_2Se$ is more acidic than $H_2Te$
Statement II: $H_2Se$ has higher bond enthalpy for dissociation than $H_2Te$
In the light of the above statements, choose the correct answer:
A) Both Statement I and Statement II are true
B) Statement I is true but Statement II is false
C) Both Statement I and Statement II are false
D) Statement I is false but Statement II is true
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Step-by-Step Solution
1
Approach: Evaluate Each Statement Independently
In Statement I and Statement II type questions, we must check each statement separately for correctness, then select the matching option. We need to analyse the acidity trend and bond enthalpy trend for Group 16 hydrides.
Group 16 hydrides in order: $H_2O$, $H_2S$, $H_2Se$, $H_2Te$
Key factors to compare: (1) Acidic strength, (2) Bond enthalpy (H–X bond dissociation energy)
Key factors to compare: (1) Acidic strength, (2) Bond enthalpy (H–X bond dissociation energy)
2
Evaluating Statement I: Is $H_2Se$ more acidic than $H_2Te$?
Acidic strength of binary hydrides in the same group is primarily governed by H–X bond enthalpy. The weaker the H–X bond, the more easily the proton ($H^+$) is released, and the stronger the acid.
As we go down Group 16 (O → S → Se → Te), atomic size increases, H–X bond length increases, and H–X bond enthalpy decreases. This means the H–Te bond is weaker than H–Se bond, so $H_2Te$ releases $H^+$ more easily.
Acidity order: $H_2O < H_2S < H_2Se < H_2Te$
∴ $H_2Te$ is MORE acidic than $H_2Se$, NOT the other way around.
∴ $H_2Te$ is MORE acidic than $H_2Se$, NOT the other way around.
Statement I is FALSE. H₂Se is NOT more acidic than H₂Te. H₂Te is more acidic.
3
Evaluating Statement II: Does $H_2Se$ have higher bond enthalpy than $H_2Te$?
As atomic radius increases down Group 16, the valence orbitals of the central atom become larger and more diffuse. The overlap between these larger orbitals and the small $H\,(1s)$ orbital becomes less effective, making the H–X bond longer and weaker.
Bond enthalpy values for Group 16 H–X bonds (approximate):
H–O bond: ~498 kJ/mol
H–S bond: ~363 kJ/mol
H–Se bond: ~276 kJ/mol
H–Te bond: ~238 kJ/mol
Trend: Bond enthalpy decreases down the group.
H–S bond: ~363 kJ/mol
H–Se bond: ~276 kJ/mol
H–Te bond: ~238 kJ/mol
Trend: Bond enthalpy decreases down the group.
Since $276 > 238$, the H–Se bond has clearly higher bond enthalpy than the H–Te bond.
Statement II is TRUE. H₂Se has higher bond enthalpy (~276 kJ/mol) than H₂Te (~238 kJ/mol).
4
Understanding the Connection Between the Two Statements
Interestingly, the very reason Statement II is true is also the reason Statement I is false. Because $H_2Se$ has a higher bond enthalpy (stronger H–Se bond), it is harder to release the proton from $H_2Se$ — meaning $H_2Se$ is a weaker acid. Conversely, $H_2Te$ has a lower bond enthalpy (weaker H–Te bond), making it easier to release $H^+$ — so $H_2Te$ is a stronger acid.
Higher bond enthalpy → Harder to break → Weaker acid
Lower bond enthalpy → Easier to break → Stronger acid
H₂Se: higher bond enthalpy → weaker acid than H₂Te ✓
H₂Te: lower bond enthalpy → stronger acid than H₂Se ✓
Lower bond enthalpy → Easier to break → Stronger acid
H₂Se: higher bond enthalpy → weaker acid than H₂Te ✓
H₂Te: lower bond enthalpy → stronger acid than H₂Se ✓
5
Why NOT Option A (Both True)?
Option A would require Statement I to be true, i.e., $H_2Se$ is more acidic than $H_2Te$. This is incorrect. The acidity order within the same group for hydrides is determined by bond enthalpy, not electronegativity. $H_2Te$ is unambiguously more acidic than $H_2Se$.
6
Why NOT Option B (Statement I true, II false)?
Option B would require Statement II to be false, meaning $H_2Se$ has lower bond enthalpy than $H_2Te$. This contradicts the well-established trend of decreasing bond enthalpy down the group. H–Se bond enthalpy (~276 kJ/mol) is definitively higher than H–Te bond enthalpy (~238 kJ/mol).
7
Final Answer
Statement I: $H_2Se$ is more acidic than $H_2Te$ → FALSE (H₂Te is more acidic)
Statement II: $H_2Se$ has higher bond enthalpy than $H_2Te$ → TRUE (~276 vs ~238 kJ/mol)
Correct Answer: Option D — Statement I is false but Statement II is true
Statement II: $H_2Se$ has higher bond enthalpy than $H_2Te$ → TRUE (~276 vs ~238 kJ/mol)
Correct Answer: Option D — Statement I is false but Statement II is true
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Theory
1. Acidity of Group 16 Hydrides — Periodic Trend
The acidity of group 16 hydrides follows a well-defined periodic trend: $H_2O < H_2S < H_2Se < H_2Te$. Water is by far the least acidic, and acidity increases progressively down the group. The primary factor controlling this trend is the H–X bond dissociation enthalpy. As we descend Group 16 from oxygen to tellurium, the central atom’s atomic radius increases significantly (O: 73 pm, S: 103 pm, Se: 119 pm, Te: 142 pm). This increase in size causes the H–X bond to become progressively longer. A longer bond means less effective orbital overlap between the hydrogen 1s orbital and the central atom’s valence orbital, resulting in a weaker bond. Since acid strength is essentially the ease of releasing a proton ($H^+$), a weaker H–X bond directly translates to greater acid strength. It is important to note that electronegativity, while higher for elements at the top of the group, does not override the bond enthalpy effect for within-group comparisons of hydrides. This is why $H_2O$ with its very electronegative oxygen is still less acidic than $H_2S$, $H_2Se$, or $H_2Te$.
2. Bond Enthalpy Trend in Group 16 Hydrides
Bond enthalpy (bond dissociation energy) is defined as the enthalpy change when one mole of a particular bond is broken homolytically in the gaseous phase under standard conditions. For the H–X bonds in group 16 hydrides, the bond enthalpy decreases systematically down the group: H–O (~498 kJ/mol) $>$ H–S (~363 kJ/mol) $>$ H–Se (~276 kJ/mol) $>$ H–Te (~238 kJ/mol). This decrease occurs because the central atom’s valence orbitals (2p for O, 3p for S, 4p for Se, 5p for Te) become progressively larger, more diffuse, and more spread out as the principal quantum number increases. The small, compact hydrogen 1s orbital cannot overlap effectively with these increasingly large orbitals, leading to weaker sigma bonds. This is why the H–Se bond with bond enthalpy ~276 kJ/mol is definitively stronger than the H–Te bond at ~238 kJ/mol — a direct consequence of selenium being smaller than tellurium. In JEE, students must be able to cite these approximate bond enthalpy values and correctly apply the trend.
3. Why Bond Enthalpy Dominates Over Electronegativity for Within-Group Acidity
A common misconception among students is that higher electronegativity of the central atom should always mean greater acid strength (since more electronegative atoms stabilise the conjugate base anion better). While this reasoning works when comparing acids across different groups or periods, it does NOT apply to within-group comparisons of binary hydrides. For group 16 hydrides, oxygen is the most electronegative (3.44) followed by sulfur (2.58), selenium (2.55), and tellurium (2.10). If electronegativity were the dominant factor, $H_2O$ would be the strongest acid — which is opposite to the actual trend. The reason bond enthalpy dominates here is thermodynamic: breaking the H–X bond is the rate-determining step for proton donation in the gas phase and contributes significantly to the free energy of ionisation in solution. The very large differences in bond enthalpy (498 for O-H vs 238 for Te-H) dwarf the electronegativity effect on the energetics of proton release. This principle is tested directly in JEE Main and Advanced.
4. Statement I and Statement II Question Strategy for JEE Main
Statement I and Statement II questions (also called Assertion-Reason or Two-Statement type) are a regular feature of JEE Main Chemistry. The strategy for solving these efficiently is: (1) Evaluate Statement I completely on its own — determine if it is factually correct or incorrect, without being influenced by Statement II. (2) Evaluate Statement II completely on its own — determine its factual correctness independently. (3) Match the combination (both true, I true and II false, I false and II true, both false) to the given options. The most common traps in these questions are: (a) a statement that is partially correct but not entirely, (b) a statement that seems related to the other but is actually independent, and (c) a correct fact stated with the wrong direction of comparison (as in Statement I here — acidity does increase down the group, but the direction of the comparison between Se and Te is reversed). Careful, independent evaluation of each statement prevents these errors.
❓
Frequently Asked Questions
1
Which is more acidic — H₂Se or H₂Te, and why?
H₂Te is more acidic than H₂Se. The H–Te bond enthalpy (~238 kJ/mol) is lower than the H–Se bond enthalpy (~276 kJ/mol). A weaker H–X bond is broken more easily, releasing H⁺ more readily. Since acid strength depends on ease of proton donation, H₂Te with its weaker H–Te bond is the stronger acid.
2
What is the complete acidity order of group 16 hydrides?
The acidity order of group 16 hydrides is: H₂O < H₂S < H₂Se < H₂Te. This increasing acidity down the group is due to the decreasing H–X bond enthalpy, which makes proton donation progressively easier as we go from oxygen to tellurium.
3
What are the approximate bond enthalpy values for group 16 H–X bonds?
Approximate H–X bond enthalpy values: H–O: ~498 kJ/mol, H–S: ~363 kJ/mol, H–Se: ~276 kJ/mol, H–Te: ~238 kJ/mol. These decrease down the group due to increasing atomic radius and decreasing orbital overlap effectiveness. These values are important to remember for JEE.
4
Why is H₂O less acidic than H₂S despite oxygen being more electronegative?
Although oxygen is more electronegative (3.44) than sulfur (2.58), H₂O is less acidic than H₂S because the O–H bond enthalpy (~498 kJ/mol) is far higher than S–H (~363 kJ/mol). For within-group comparisons, bond dissociation enthalpy dominates over electronegativity in determining acid strength of binary hydrides.
5
Why does bond enthalpy decrease down Group 16?
As we go down Group 16, the central atom’s valence orbitals (2p → 3p → 4p → 5p) become progressively larger, more diffuse, and more spread out. The small hydrogen 1s orbital cannot overlap effectively with these increasingly large orbitals, making the H–X bond progressively longer and weaker with decreasing bond enthalpy.
6
What is the pKa trend for group 16 hydrides?
The pKa values (lower = stronger acid): H₂O (~15.7) > H₂S (~7.0) > H₂Se (~3.9) > H₂Te (~2.6). This confirms that acidic strength increases down the group, with H₂Te being the strongest acid among common group 16 hydrides. A lower pKa value directly corresponds to greater acid strength.
7
How should I approach Statement I and Statement II questions in JEE?
Evaluate each statement completely independently: (1) Is Statement I factually correct on its own? (2) Is Statement II factually correct on its own? (3) Match the combination to options. Never let your assessment of one statement influence the other. Common trap: a statement with the correct concept but the wrong direction of comparison — exactly as in Statement I of this question.
8
Does electronegativity affect the acidity of group 16 hydrides?
Electronegativity does contribute to acid strength by stabilising the conjugate base anion, but for within-group comparisons of binary hydrides, the bond enthalpy effect dominates. The differences in bond enthalpy across the group are so large that they override the electronegativity effect. This is why the most electronegative element (O) gives the least acidic hydride (H₂O) in this group.
9
Is H₂Te stable under normal conditions?
H₂Te is a colourless, toxic gas at room temperature that is less thermally stable than H₂S or H₂Se. Its instability is consistent with its low H–Te bond enthalpy (~238 kJ/mol). In fact, the thermal stability of group 16 hydrides also decreases down the group (H₂O > H₂S > H₂Se > H₂Te), which is directly related to the decreasing bond enthalpy.
10
What is the reducing character trend of group 16 hydrides?
Reducing character of group 16 hydrides increases down the group: H₂O < H₂S < H₂Se < H₂Te. A stronger reducing agent donates electrons or hydrogen more easily, which again correlates with the decreasing H–X bond enthalpy. H₂Te with the weakest H–Te bond is the strongest reducing agent and the strongest acid in this series.
🔗
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