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Answer:
a) ammonium ion
b) amide ion
Explanation:
The order of decreasing bond angles of the three nitrogen species; ammonium ion, ammonia and amide ion is NH4+ >NH3> NH2-. Next we need to rationalize this order of decreasing bond angles from the valence shell electron pair repulsion (VSEPR) theory perspective.
First we must realize that all three nitrogen species contain a central sp3 hybridized carbon atom. This means that a tetrahedral geometry is ideally expected. Recall that the presence of lone pairs distorts molecular structures from the expected geometry based on VSEPR theory.
The amide ion contains two lone pairs of electrons. Remember that the presence of lone pairs causes greater repulsion than bond pairs on the outermost shell of the central atom. Hence, the amide ion has the least H-N-H bond angle of about 105°.
The ammonia molecule contains one lone pair, the repulsion caused by one lone pair is definitely bless than that caused by two lone pairs of electrons hence the bond angle of the H-N-H bond in ammonia is 107°.
The ammonium ion contains four bond pairs and no lone pair of electrons on the outermost nitrogen atom. Hence we expect a perfect tetrahedron with bond angle of 109°.
The answer is in the attachment below:
Ionic bonding is formed in Ionic compounds due to electrostatic force between the oppositely charged ions.
In covalent bonds electrons are shared between the atoms. In case of ionic bond the bond is stronger as there is complete transfer of electrons from one ion to the other.
Since the ionic bonds are more difficult to break than the covalent bonds, ionic compounds have a higher melting point than covalent compounds.
Answer:
The science project or the proposed test using the water as a control element in the experiment determines whether you choose to use deionized water or distilled water. Of the two, distilled water is the purest because the water undergoes boiling that kills off most organic contaminants.
Explanation:
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