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Math

Math Snap

PROBLEM

Place the following in order of increasing IE1I E_{1}.
N
F
As
A) N<As<F\mathrm{N}<\mathrm{As}<\mathrm{F}
B) As<N<F\mathrm{As}<\mathrm{N}<\mathrm{F}
C) F<N<As\mathrm{F}<\mathrm{N}<\mathrm{As}
D) F<As<N\mathrm{F}<\mathrm{As}<\mathrm{N}
E) As <F<N<\mathrm{F}<\mathrm{N}

STEP 1

What is this asking?
Which element, Nitrogen (N), Fluorine (F), or Arsenic (As), is easiest to remove an electron from, and which is hardest?
Watch out!
Don't confuse ionization energy with electron affinity.
Ionization energy is about removing an electron, not adding one!

STEP 2

1. Periodic Trends
2. Element Comparison

STEP 3

Ionization energy is the energy needed to remove an electron from an atom.
Think of it like this: how much energy does it take to steal that electron?
A higher ionization energy means it's harder to remove the electron.

STEP 4

As we move left to right across a period (row) in the periodic table, ionization energy generally increases.
This is because the number of protons increases, pulling the electrons in tighter.
It's like adding more magnets to hold onto those electrons!

STEP 5

As we move down a group (column), ionization energy generally decreases.
This is because electrons are further away from the nucleus in larger atoms, and those outer electrons are shielded from the positive charge by the inner electrons.
It's like trying to grab something from the outer layer of a giant onion – easier than grabbing from the core!

STEP 6

Let's find our elements on the periodic table!
Nitrogen (NN) is in period 2, group 15.
Fluorine (FF) is in period 2, group 17.
Arsenic (AsAs) is in period 4, group 15.

STEP 7

Both NN and FF are in the same period (row).
Since FF is to the right of NN, FF has a higher ionization energy.
It's harder to steal an electron from FF than from NN.
So, IE(N)<IE(F)\mathrm{IE}(N) < \mathrm{IE}(F).

STEP 8

Nitrogen (NN) and Arsenic (AsAs) are in the same group (column).
Since AsAs is below NN, AsAs has a lower ionization energy.
It's easier to steal an electron from AsAs than from NN.
So, IE(As)<IE(N)\mathrm{IE}(As) < \mathrm{IE}(N).

STEP 9

We know IE(As)<IE(N)\mathrm{IE}(As) < \mathrm{IE}(N) and IE(N)<IE(F)\mathrm{IE}(N) < \mathrm{IE}(F).
Combining these, we get IE(As)<IE(N)<IE(F)\mathrm{IE}(As) < \mathrm{IE}(N) < \mathrm{IE}(F).

SOLUTION

The correct order of increasing first ionization energy is As<N<FAs < N < F, which corresponds to answer choice B.

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