Which one of the following molecules is polar?
- $XeF_4$
- $IF_5$
- $SbF_5$
- $CF_4$
The Correct Option is B
Solution and Explanation
The geometry of \(IF_5\) is square pyramidal with an unsymmetric charge distribution therefore this molecule is polar.
So, the correct option is (B): \(IF_5\).
Top Questions on Chemical bonding and molecular structure
- A group 15 element forms \( d\pi - d\pi \) bond with transition metals. It also forms a hydride, which is the strongest base among the hydrides of other group members that form \( d\pi - d\pi \) bonds. The atomic number of the element is ________________________.
- JEE Main - 2025
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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
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- Which of the following has $\text{sp}^3\text{d}^2$ hybridisation?
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Questions Asked in JEE Main exam
In the first configuration (1) as shown in the figure, four identical charges \( q_0 \) are kept at the corners A, B, C and D of square of side length \( a \). In the second configuration (2), the same charges are shifted to mid points C, E, H, and F of the square. If \( K = \frac{1}{4\pi \epsilon_0} \), the difference between the potential energies of configuration (2) and (1) is given by:
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- Electromagnetic Field (EMF)
- The value of \( (\sin 70^\circ)(\cot 10^\circ \cot 70^\circ - 1) \) is:
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- Trigonometric Identities
- Ice at \( -5^\circ C \) is heated to become vapor with temperature of \( 110^\circ C \) at atmospheric pressure. The entropy change associated with this process can be obtained from:
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- Thermodynamics
- Let C be the circle of minimum area enclosing the ellipse E: \( \frac{x^2}{a^2} + \frac{y^2}{b^2} = 1 \) with eccentricity \( \frac{1}{2} \) and foci \( (\pm 2, 0) \). Let PQR be a variable triangle, whose vertex P is on the circle C and the side QR of length 29 is parallel to the major axis and contains the point of intersection of E with the negative y-axis. Then the maximum area of the triangle PQR is:
- JEE Main - 2025
- Coordinate Geometry
- Let circle $C$ be the image of
$$ x^2 + y^2 - 2x + 4y - 4 = 0 $$
in the line
$$ 2x - 3y + 5 = 0 $$
and $A$ be the point on $C$ such that $OA$ is parallel to the x-axis and $A$ lies on the right-hand side of the centre $O$ of $C$.
If $B(\alpha, \beta)$, with $\beta < 4$, lies on $C$ such that the length of the arc $AB$ is $\frac{1}{6}$ of the perimeter of $C$, then $\beta - \sqrt{3}\alpha$ is equal to:
Concepts Used:
Chemical Bonding and Molecular Structure
Such a group of atoms is called a molecule. Obviously, there must be some force that holds these constituent atoms together in the molecules. The attractive force which holds various constituents (atoms, ions, etc.) together in different chemical species is called a chemical bond.
Types of Chemical Bonds:
There are 4 types of chemical bonds which are formed by atoms or molecules to yield compounds.
- Ionic Bonds - Ionic bonding is a type of chemical bonding which involves a transfer of electrons from one atom or molecule to another.
- Covalent Bonds - Compounds that contain carbon commonly exhibit this type of chemical bonding.
- Hydrogen Bonds - It is a type of polar covalent bonding between oxygen and hydrogen wherein the hydrogen develops a partial positive charge
- Polar Bonds - In Polar Covalent chemical bonding, electrons are shared unequally since the more electronegative atom pulls the electron pair closer to itself and away from the less electronegative atom.
Factors Affecting Bond Enthalpy in Chemical Bonding:
- Size of the Atom
- Multiplicity of Bonds
- Number of Lone Pair of Electrons Present
- Bond Angle