Choose the correct option
Molecule | Shape | ||
---|---|---|---|
A | \(BrF_5\) | i | T-shape |
B | \(H_2O\) | ii | See-saw |
C | \(ClF_3\) | iii | Bent |
D | \(SF_4\) | iv | Square Pyramidal |
Analyze Each Molecule Based on VSEPR Theory:
BrF5: The molecule has five bonded pairs and one lone pair around bromine, leading to a square pyramidal shape.
H2O: Water has two bonded pairs and two lone pairs, giving it a bent shape.
ClF3: Chlorine trifluoride has three bonded pairs and two lone pairs, resulting in a T-shape.
SF4: Sulfur tetrafluoride has four bonded pairs and one lone pair, leading to a see-saw shape.
Match Each Molecule with the Correct Shape:
(A) BrF5 - Square pyramidal
(B) H2O - Bent
(C) ClF3 - T-shape
(D) SF4 - See-saw
Conclusion: Based on the shapes identified above, the correct answer is Option (1).
Let \( y = f(x) \) be the solution of the differential equation
\[ \frac{dy}{dx} + 3y \tan^2 x + 3y = \sec^2 x \]
such that \( f(0) = \frac{e^3}{3} + 1 \), then \( f\left( \frac{\pi}{4} \right) \) is equal to:
Find the IUPAC name of the compound.
If \( \lim_{x \to 0} \left( \frac{\tan x}{x} \right)^{\frac{1}{x^2}} = p \), then \( 96 \ln p \) is: 32