# Question 61660

Feb 20, 2017

Here is the answer to 3. (I can't really make out the other questions!)

#### Explanation:

These two exceptions to the usual observation that ionization energy increases across a row of the periodic table occur for two different reasons.

Remember that ionization energy (IE) is the energy required to remove the highest-energy electron from the atom (the one in the orbital of highest energy).

In Mg, this is a $3 s$ electron, while in Al it is a $3 p$ electron. The reduction in IE is strong support for the theoretical prediction that a $p$ orbital is at a higher energy level that the $s$ orbital in the same shell. The higher energy of the $p$ orbital more than compensates for the additional proton in the Al nucleus, and less energy is needed to remove it.

For sulfur, the reason is evidence of Hund's rule, which states that when electrons populate the identical energies of the $p$ orbitals, they do so by having a single electron in each orbital before any pairs are created.

With sulfur, we have the first instance of an electron pair being created in the $3 p$ subshell. With two electrons in the same orbital, the probability that these electrons are close together is much greater than when each electron occupies a different orbital. This results in greater repulsion between these electrons, and their energy is higher. This again compensates for the additional electron in S, and the IE is smaller than for P.

Feb 20, 2017

Here's my answer to Question 1.

#### Explanation: I would expect the energy level diagram to look like the one below. There are only six allowed transitions.

The shortest line (L1) corresponds to the infrared, and the longest line (L6) corresponds to the ultraviolet.

Lines L2 to L4 correspond to increasingly shorter wavelengths in the visible spectrum in the order $\textcolor{red}{\text{R")color(white)(l) bbcolor(orange)("O")color(white)(l) bbcolor(yellow)("Y") color(white)(l)color(green)("G")color(white)(l) bbcolor(blue)("B") color(white)(l)color(indigo)("I")color(white)(l) bbcolor(purple)("V}}$.

Feb 20, 2017

Here's my answer to Question 2.

#### Explanation: We need to refer to a Periodic Table for this question. (From Wikipedia)

a. $\text{Cr}$

$\text{Cr}$ is in Period 4, Group 16.

We would predict its electron configuration to be ${\text{[Ar] 4s"^2 "3d}}^{4}$.

However, because of the stability of half-filled subshells, an electron is "promoted" to the $\text{3d}$ level.

The actual electron configuration is ${\text{[Ar] 4s 3d}}^{5}$.

b. $\text{Si}$

$\text{Si}$ is in Period 3, Group 14.

Its electron configuration is ${\text{[Ne] 3s"^2 "3p}}^{2}$.

c. $\text{Au}$

$\text{Au}$ is in Period 6, Group 11. It contains $\text{4f}$ electrons.

We would predict its electron configuration to be ${\text{[Xe] 6s"^2 "4f"^14 "5d}}^{9}$.

However, because of the extra stability of a filled shell, an electron is "promoted to the "5d level.

The actual electron configuration is ${\text{[Xe] 6s 4f"^14 "5d}}^{10}$.

d. $\text{Ru"^"2+}$

$\text{Ru}$ is in Period 5, Group 8. Its electron configuration is ${\text{[Kr] 5s"^2 "4d}}^{6}$.

When removing electrons to form ions, we remove the $\text{s}$ electrons first.

The electron configuration of $\text{Ru"^"2+}$ is ${\text{[Kr] 4d}}^{6}$.

e. $\text{N"^"3-}$

$\text{N}$ is in Period 2, Group 15. Its electron configuration is ${\text{[He] 2s"^2 "2p}}^{3}$.

The electron configuration of $\text{N"^"3-}$ is ${\text{[He] 2s"^2 "2p}}^{6}$.

f. $\text{Pb"^"4+}$

$\text{Pb}$ is in Period 6, Group 14.

Its electron configuration is ${\text{[Xe] 6s"^2 "4f"^14 "5d"^10 "6p}}^{2}$.

When forming the cation, we remove the $\text{6s}$ and $\text{6p}$ electrons.

The electron configuration of $\text{Pb"^"4+}$ is ${\text{[Xe] 4f"^14 "5d}}^{10}$.

Feb 20, 2017

Here's my answer to Question 3.

#### Explanation: The "dips" in the curve are not because the ionization energies of $\text{Al}$ and $\text{S}$ are anomalously low, but because those of $\text{Mg}$ and $\text{P}$ are anomalously high.

The energy level diagram for $\text{Mg}$ is The $\text{3s}$ subshell is filled, so it is a little more stable than expected.

Hence, more energy is required to remove an electron from this level, and the ionization energy is higher.

The energy level diagram for $\text{P}$ is Here, the $\text{3p}$ sublevel is half-filled, so there is a little extra stabilization.

Hence, more energy is required to remove an electron, and the ionization energy is higher than expected.

Feb 20, 2017

Here's my answer to Question 4.

#### Explanation: a. Energy level diagrams

The electron configuration of $\text{Bi}$ is ${\text{[Xe] 6s"^2 "4f"^14 "5d"^10 "6p}}^{3}$.

In the energy diagram below, all levels are filled up to the $\text{5d}$ level (the red line).

The $\text{6p}$ level (just below the red line) is half-filled. A $\text{Bi}$ atom can lose the three $\text{6p}$ electrons to form a $\text{Bi"^"3+}$ ion.

Remember that $\text{s}$ electrons are removed before $\text{d}$ electrons, so the next most easily removed electrons are in the $\text{6s}$ orbital.

Removing these two electrons forms a $\text{Bi"^"5+}$ ion.

b. Paramagnetism

All electrons are paired in $\text{Bi"^"3+}$ and $\text{Bi"^"5+}$, so these ions will be diamagnetic.

The salts "Bi"("NO"_3)_3 and "Bi"("NO"_3)_5 will be repelled by a magnet.

A $\text{Bi}$ atom has three unpaired electrons, so it will be paramagnetic.

Thus, elemental $\text{Bi}$ will have the strongest attraction to a magnet.