Your answer is B: The outer electron of Adam B has moved to a higher energy state
In general chemistry, isotopes are a group of substances that belong to the same element. An element is characterized in the periodic table by their atomic number, which is the number of protons in an atom. Therefore, these substances have the same atomic numbers, but differ in mass numbers. Mass number is the sum of the number of protons and neutrons in the nucleus of an atom.
To determine the atomic weight of an element, you take the average weight of all the existent isotopes of that said element. The calculation would require to multiply the exact mass of the isotope to its abundance. Then, sum them all up.
Atomic weight = 98(0.18) + 112(0.82)
Atomic weight = 109.48 amu
Answer:
<h3>1)</h3>
Structure One:
Structure Two:
Structure Three:
Structure Number Two would likely be the most stable structure.
<h3>2)</h3>
- All five C atoms: 0
- All six H atoms to C: 0
- N atom: +1.
The N atom is the one that is "likely" to be attracted to an anion. See explanation.
Explanation:
When calculating the formal charge for an atom, the assumption is that electrons in a chemical bond are shared equally between the two bonding atoms. The formula for the formal charge of an atom can be written as:
.
For example, for the N atom in structure one of the first question,
- N is in IUPAC group 15. There are 15 - 10 = 5 valence electrons on N.
- This N atom is connected to only 1 chemical bond.
- There are three pairs, or 6 electrons that aren't in a chemical bond.
The formal charge of this N atom will be
.
Apply this rule to the other atoms. Note that a double bond counts as two bonds while a triple bond counts as three.
<h3>1)</h3>
Structure One:
Structure Two:
Structure Three:
In general, the formal charge on all atoms in a molecule or an ion shall be as close to zero as possible. That rules out Structure number one.
Additionally, if there is a negative charge on one of the atoms, that atom shall preferably be the most electronegative one in the entire molecule. O is more electronegative than N. Structure two will likely be favored over structure three.
<h3>2)</h3>
Similarly,
- All five C atoms: 0
- All six H atoms to C: 0
- N atom: +1.
Assuming that electrons in a chemical bond are shared equally (which is likely not the case,) the nitrogen atom in this molecule will carry a positive charge. By that assumption, it would attract an anion.
Note that in reality this assumption seldom holds. In this ion, the N-H bond is highly polarized such that the partial positive charge is mostly located on the H atom bonded to the N atom. This example shows how the formal charge assumption might give misleading information. However, for the sake of this particular problem, the N atom is the one that is "likely" to be attracted to an anion.
Answer:
a)CH₄, BH₃, and CCl₄
Explanation:
<u>London dispersion forces:-
</u>
The bond for example, in the molecule is F-F, which is non-polar in nature because the two fluorine atoms have same electronegativity values.
The intermolecular force acting in the molecule are induced dipole-dipole forces or London Dispersion forces / van der Waals forces which are the weakest intermolecular force.
Out of the given options, H₂O , NH₃ exhibits hydrogen bonding which is:-
<u>Hydrogen bonding:-
</u>
Hydrogen bonding is a special type of the dipole-dipole interaction and it occurs between hydrogen atom that is bonded to highly electronegative atom which is either fluorine, oxygen or nitrogen atom.
Thus option B and C rules out.
<u>Hence, the correct option which represents the molecules which would exhibit only London forces is:- a)CH₄, BH₃, and CCl₄</u>