To determine the correct statement, let's examine each option:
Ammonia (NH3) has a trigonal pyramidal molecular geometry, which leads to a net dipole moment. In this molecule, the lone pair of electrons on nitrogen creates an orbital dipole, and the bond dipoles from the N-H bonds point towards the nitrogen, aligning with the orbital dipole. Hence, the resultant dipole is significant and points in the same general direction as the orbital dipole.
Both O2 and H2 are homonuclear diatomic molecules with nonpolar covalent bonds. Since atoms of the same element share electrons equally, no bond dipole or polarization exists.
Dipole moment is a vector quantity, not scalar. It has both magnitude and direction.
Boron trifluoride (BF3) has a trigonal planar structure. Although each B-F bond has a polarity, the symmetrical arrangement causes the bond dipoles to cancel out, resulting in a net dipole moment of zero.
Thus, the correct statement is: Dipole moment of ammonia is due to orbital dipole and resultant dipole in same direction.
To determine the correct statement among the given options, we need to analyze the concept of dipole moment and its formation in different molecules:
The correct statement is: Dipole moment of ammonia is due to orbital dipole and resultant dipole in same direction.
Regarding the molecular orbital (MO) energy levels for homonuclear diatomic molecules, the INCORRECT statement(s) is (are):
Which of the following statement is true with respect to H\(_2\)O, NH\(_3\) and CH\(_4\)?
(A) The central atoms of all the molecules are sp\(^3\) hybridized.
(B) The H–O–H, H–N–H and H–C–H angles in the above molecules are 104.5°, 107.5° and 109.5° respectively.
(C) The increasing order of dipole moment is CH\(_4\)<NH\(_3\)<H\(_2\)O.
(D) Both H\(_2\)O and NH\(_3\) are Lewis acids and CH\(_4\) is a Lewis base.
(E) A solution of NH\(_3\) in H\(_2\)O is basic. In this solution NH\(_3\) and H\(_2\)O act as Lowry-Bronsted acid and base respectively.
A solid cylinder of mass 2 kg and radius 0.2 m is rotating about its own axis without friction with angular velocity 5 rad/s. A particle of mass 1 kg moving with a velocity of 5 m/s strikes the cylinder and sticks to it as shown in figure.
The angular velocity of the system after the particle sticks to it will be: