complete the lewis dot structure for the following elements.
state how many electrons will be lost or gained to fulfill the
octet rule.
ne
____ electrons will be (gained / lost)
n
____ electrons will be (gained / lost)
h
____ electrons will be (gained / lost)
c
____ electrons will be (gained / lost)
be
____ electrons will be (gained / lost)

Question

Accepted Answer

### For Ne (Neon)
# Explanation:
## Step1: Determine electron configuration
Neon has an atomic number of 10, so its electron configuration is $1s^2 2s^2 2p^6$. The outermost shell (valence shell) has 8 electrons, which satisfies the octet rule (for main - group elements, having 8 electrons in the valence shell is stable).
## Step2: Decide on electron gain/loss
Since the valence shell already has 8 electrons (a stable octet), Neon will not gain or lose any electrons. So the number of electrons gained or lost is 0. And the choice between gained/lost is neither, but in the context of the question (maybe a bit of a trick here as noble gases are stable), we can say 0 electrons will be (neither gained nor lost, but if we have to choose from the given options, since it's already stable, 0 and no gain or loss. But following the question's format):
# Answer:
0 electrons will be (neither gained nor lost, but in the boxed form for the number: $\boxed{0}$), and for the gain/loss, since it's stable, but if we have to pick, maybe the question expects to recognize it's a noble gas, so 0 electrons will be (gained / lost - but actually neither, but as per the problem's structure, 0 and no gain/loss. But let's proceed with each element:

### For N (Nitrogen)
# Explanation:
## Step1: Determine valence electrons
Nitrogen has an atomic number of 7. Its electron configuration is $1s^2 2s^2 2p^3$. The valence shell (n = 2) has 5 electrons.
## Step2: Calculate electrons needed for octet
To fulfill the octet rule (8 electrons in the valence shell), Nitrogen needs $8 - 5=3$ electrons. So it will gain 3 electrons.
# Answer:
$\boxed{3}$ electrons will be (gained)

### For H (Hydrogen)
# Explanation:
## Step1: Determine valence electrons
Hydrogen has an atomic number of 1. Its electron configuration is $1s^1$. The valence shell (n = 1) follows the duet rule (stable with 2 electrons in the first shell).
## Step2: Calculate electrons needed
It has 1 valence electron, so it needs $2 - 1 = 1$ electron to fulfill the duet rule (similar to octet rule for n = 1). So it will gain 1 electron.
# Answer:
$\boxed{1}$ electrons will be (gained)

### For C (Carbon)
# Explanation:
## Step1: Determine valence electrons
Carbon has an atomic number of 6. Its electron configuration is $1s^2 2s^2 2p^2$. The valence shell (n = 2) has 4 electrons.
## Step2: Calculate electrons needed for octet
To fulfill the octet rule, Carbon can either gain 4 electrons or lose 4 electrons. But it is more energetically favorable to share electrons (covalent bonding), but in the context of ionic bonding (the question might be considering ionic or general octet fulfillment), Carbon needs 4 electrons to reach 8 in the valence shell (or lose 4, but losing 4 is less favorable as Carbon has a moderate electronegativity). However, in terms of the number of electrons to gain to reach octet: $8 - 4=4$. So it will gain 4 electrons (or lose 4, but gaining 4 is more in line with its tendency to form covalent bonds by sharing, but for the octet rule, to reach 8, it needs 4 more).
# Answer:
$\boxed{4}$ electrons will be (gained)

### For Be (Beryllium)
# Explanation:
## Step1: Determine valence electrons
Beryllium has an atomic number of 4. Its electron configuration is $1s^2 2s^2$. The valence shell (n = 2) has 2 electrons.
## Step2: Calculate electrons to lose for stability
To fulfill the octet rule (or the duet rule for the inner shell, but for Be, it is a metal. Metals tend to lose electrons. Be has 2 valence electrons. It is more energetically favorable for Be to lose 2 electrons to have a stable electron configuration like He ($1s^2$) (a stable duet for the new valence shell, which is n = 1 now). So it will lose 2 electrons.
# Answer:
$\boxed{2}$ electrons will be (lost)

So summarizing:

- Ne: $\boxed{0}$ electrons will be (neither gained nor lost, but as per the question's format, 0)
- N: $\boxed{3}$ electrons will be (gained)
- H: $\boxed{1}$ electrons will be (gained)
- C: $\boxed{4}$ electrons will be (gained)
- Be: $\boxed{2}$ electrons will be (lost)

Sovi.AI - AI Math Tutor

QUESTION IMAGE

Question

Explanation:

For Ne (Neon)

Step1: Determine electron configuration

Step2: Decide on electron gain/loss

Step1: Determine valence electrons

Step2: Calculate electrons needed for octet

Step1: Determine valence electrons

Step2: Calculate electrons needed

Answer:

For N (Nitrogen)