Match List-I with List-II.List-I
(Molecule) List-II
(Shape) (A) \(NH_3\) (I) Square pyramid (B) \(BrF_5\) (II) Tetrahedral (C) \(PCL_5\) (III) Trigonal pyramidal (D) \(CH_4\) (IV) Trigonal bipyramidal
Choose the correct answer from the options given below:
| List-I (Molecule) | List-II (Shape) |
|---|---|
| (A) \(NH_3\) | (I) Square pyramid |
| (B) \(BrF_5\) | (II) Tetrahedral |
| (C) \(PCL_5\) | (III) Trigonal pyramidal |
| (D) \(CH_4\) | (IV) Trigonal bipyramidal |
- A-IV, B-III, C-I, D-II
- A-II, B-IV, C-I, D-III
- A-III, B-I, C-IV, D-II
- A-III, B-IV, C-I, D-II
The Correct Option is C
Solution and Explanation
NH$_3$: Trigonalpyramidal ({sp}$^3$ hybridization with one lone pair on N).
BrF$_5$: Squarepyramidal ({sp}$^3${d}$^2$ hybridization with one lone pair on Br).
PCl$_5$: Trigonalbipyramidal ({sp}$^3${d} hybridization, no lone pairs).
CH$_4$: Tetrahedral ({sp}$^3$ hybridization, no lone pairs).
Matching: A-III, B-I, C-IV, D-II.
Final Answer: A-III, B-I, C-IV, D-II.
Top Questions on Chemical bonding and molecular structure
- Given below are two statements:
Statement I: A hypothetical diatomic molecule with bond order zero is quite stable.
Statement II: As bond order increases, the bond length increases.
In the light of the above statements, choose the most appropriate answer from the options given below:- NEET (UG) - 2025
- Chemistry
- Chemical bonding and molecular structure
Identify the correct orders against the property mentioned:
A. H$_2$O $>$ NH$_3$ $>$ CHCl$_3$ - dipole moment
B. XeF$_4$ $>$ XeO$_3$ $>$ XeF$_2$ - number of lone pairs on central atom
C. O–H $>$ C–H $>$ N–O - bond length
D. N$_2$>O$_2$>H$_2$ - bond enthalpy
Choose the correct answer from the options given below:- NEET (UG) - 2025
- Chemistry
- Chemical bonding and molecular structure
- Name the type of bond in the formation of chlorine molecule.
- Maharashtra Class X Board - 2025
- Science & Technology
- Chemical bonding and molecular structure
- Show the formation of magnesium chloride by electron transfer. Write the name of the cation and anion present in the compound formed. (Atomic Number of Mg = 12, Cl = 17)
- CBSE Class X - 2025
- Science
- Chemical bonding and molecular structure
- Total number of non-bonded electrons present in \( \text{NO}_2 \); ion based on Lewis theory is:
- JEE Main - 2025
- Chemistry
- Chemical bonding and molecular structure
Questions Asked in JEE Main exam
Let $ P_n = \alpha^n + \beta^n $, $ n \in \mathbb{N} $. If $ P_{10} = 123,\ P_9 = 76,\ P_8 = 47 $ and $ P_1 = 1 $, then the quadratic equation having roots $ \alpha $ and $ \frac{1}{\beta} $ is:
- JEE Main - 2025
- Relations and functions
- Density of 3 M NaCl solution is 1.25 g/mL. The molality of the solution is:
- JEE Main - 2025
- Solutions
In the given circuit the sliding contact is pulled outwards such that the electric current in the circuit changes at the rate of 8 A/s. At an instant when R is 12 Ω, the value of the current in the circuit will be A.- JEE Main - 2025
- Electrical Circuits
- Let \( \vec{a} = \hat{i} + 2\hat{j} + \hat{k} \) and \( \vec{b} = 2\hat{i} + 7\hat{j} + 3\hat{k} \). Let \( L_1 : \vec{r} = (-\hat{i} + 2\hat{j} + \hat{k}) + \lambda \vec{a}, \lambda \in {R} \) and \( L_2 : \vec{r} = (\hat{j} + \hat{k}) + \mu \vec{b}, \mu \in \mathbb{R} \) be two lines. If the line \( L_3 \) passes through the point of intersection of \( L_1 \) and \( L_2 \), and is parallel to \( \vec{a} + \vec{b} \), then \( L_3 \) passes through the point:
- JEE Main - 2025
- Vectors
For $ \alpha, \beta, \gamma \in \mathbb{R} $, if $$ \lim_{x \to 0} \frac{x^2 \sin \alpha x + (\gamma - 1)e^{x^2} - 3}{\sin 2x - \beta x} = 3, $$ then $ \beta + \gamma - \alpha $ is equal to: