Letter d, because they are both alkali metals (group one)
Hi, you have not provided structure of the aldehyde and alkoxide ion.
Therefore i'll show a mechanism corresponding to the proton transfer by considering a simple example.
Explanation: For an example, let's consider that proton transfer is taking place between a simple aldehyde e.g. acetaldehyde and a simple alkoxide base e.g. methoxide.
The hydrogen atom attached to the carbon atom adjacent to aldehyde group are most acidic. Hence they are removed by alkoxide preferably.
After removal of proton from aldehyde, a carbanion is generated. As it is a conjugated carbanion therefore the negative charge on carbon atom can conjugate through the carbonyl group to form an enolate which is another canonical form of the carbanion.
All the structures are shown below.
2 Ionic bonds form between metal atoms and nonmetal atoms.
4 The less electronegative atoms transfers one or more electrons to the more electronegative atom.
5 The metal atom forms a cation and the nonmetal atom forms an anion.
7 The attraction between ions with an opposite charge forms an ionic bond.
Answer:
The correct answer is Option A (There is no magnetic flux through the wire loop.)
Explanation:
Magnetic flux measures the entire magnetic field that passes through the wire loop.
The right hand rule can be demonstrated on how magnetic flux is generated through the moving current in the wire loop. The magnetic flux through the wire loop will decrease as it moves upward through the magnetic field region.
If the direction of the vector area of the wire loop is to the right, and the switch is closed, it will push the magnetic flux to the right which will now be increased due to an equal increase in the current in the wire loop. But, when the switch is open, this will halt the movement of current through the wire loop thus affecting the generation of magnetic field. This would make the magnetic flux to be zero.
