Given below are two statements:
Statement I: SO\(_2\) and H\(_2\)O both possess V-shaped structures.
Statement II: The bond angle of SO\(_2\) is less than that of H\(_2\)O. In the light of the above statements, choose the most appropriate answer from the options given below:
Given below are two statements:
Statement I: SO\(_2\) and H\(_2\)O both possess V-shaped structures.
Statement II: The bond angle of SO\(_2\) is less than that of H\(_2\)O. In the light of the above statements, choose the most appropriate answer from the options given below:
- Both Statements I and Statement II are incorrect
- Both Statement I and Statement II are correct
- Statement I is correct but Statement II is incorrect
- Statement I is incorrect but Statement II is correct
The Correct Option is C
Solution and Explanation
1. Statement I: Both SO\(_2\) and H\(_2\)O possess V-shaped (bent) structures due to the presence of lone pairs on the central atom. - SO\(_2\): The central sulfur atom forms two double bonds with oxygen and has one lone pair of electrons. - H\(_2\)O: The central oxygen atom forms two single bonds with hydrogen and has two lone pairs of electrons. Thus, this statement is correct.
2. Statement II: The bond angle of SO\(_2\) is approximately 119.5° (sp\(^2\) hybridization), while the bond angle of H\(_2\)O is 104.5° (sp\(^3\) hybridization). - The larger bond angle in SO\(_2\) is due to the greater repulsion caused by the double bonds in SO\(_2\), compared to the single bonds in H\(_2\)O. Thus, this statement is incorrect. \[ \text{SO}_2: \angle \text{O-S-O} \approx 119.5^\circ \quad \text{H}_2\text{O}: \angle \text{H-O-H} \approx 104.5^\circ \]
Conclusion: Statement I is correct, but Statement II is incorrect.
Top Questions on p -Block Elements
- $\textbf{Choose the correct statements about the hydrides of group 15 elements.}$
A. The stability of the hydrides decreases in the order \(\text{NH}_3 > \text{PH}_3 > \text{AsH}_3 > \text{SbH}_3 > \text{BiH}_3\)
B. The reducing ability of the hydrides increases in the order \(\text{NH}_3 < \text{PH}_3 < \text{AsH}_3 < \text{SbH}_3 < \text{BiH}_3\)
C. Among the hydrides, \(\text{NH}_3\) is a strong reducing agent while \(\text{BiH}_3\) is a mild reducing agent.
D. The basicity of the hydrides increases in the order \(\text{NH}_3 < \text{PH}_3 < \text{AsH}_3 < \text{SbH}_3 < \text{BiH}_3\)
Choose the most appropriate from the option given below:- JEE Main - 2024
- Chemistry
- p -Block Elements
- $\textbf{Reduction potential of ions are given below:}$
\[ \begin{array}{ccc} \text{ClO}_4^- & \text{IO}_4^- & \text{BrO}_4^- \\ E^\circ = 1.19 \, \text{V} & E^\circ = 1.65 \, \text{V} & E^\circ = 1.74 \, \text{V} \\ \end{array} \]
The correct order of their oxidising power is:- JEE Main - 2024
- Chemistry
- p -Block Elements
- Identify the correct statements about p-block elements and their compounds.
(A) Non metals have higher electronegativity than metals.
(B) Non metals have lower ionisation enthalpy than metals.
(C) Compounds formed between highly reactive nonmetals and highly reactive metals are generally ionic.
(D) The non-metal oxides are generally basic in nature.
(E) The metal oxides are generally acidic or neutral in nature.- JEE Main - 2024
- Chemistry
- p -Block Elements
- In which of the following species, the ratio of s-electrons to p-electrons is the same?
- TS EAMCET - 2024
- Chemistry
- p -Block Elements
The number of oxygen atoms present in chemical formula of fuming sulphuric acid is _______.
- JEE Main - 2024
- Chemistry
- p -Block Elements
Questions Asked in JEE Main exam
- Which of the following molecules hydrolyzes fast?
- JEE Main - 2025
- Organic Chemistry
- The distance of the point \( P(1, 2, 3) \) from the plane \( 3x - 4y + 12z - 7 = 0 \) is:
- JEE Main - 2025
- Geometry and Vectors
- In group 17, which property does not follow the regular trend?
- JEE Main - 2025
- Periodicity of Elements
- The ratio of the magnetic field at the center of a circular coil to the magnetic field at a distance \( x \) from the center of the circular coil is:
- JEE Main - 2025
- Electromagnetism
- The moment of inertia of a uniform rod of mass \( m \) and length \( l \) is \( \alpha \) when rotated about an axis passing through the centre and perpendicular to the length. If the rod is broken into equal halves and arranged as shown, then the moment of inertia about the given axis is:
- JEE Main - 2025
- rotational motion