Write the electron configuration for Mg2+ and Cl- ions and the valency of the hypochlorite ion in Ca(ClO)2.

Here are the solutions to the questions:

Question 11:

a) Write the resulting configuration of an $\text{Mg}^{2+}$ ion.
Magnesium ($\text{Mg}$) has an atomic number of 12. A neutral $\text{Mg}$ atom has 12 electrons.
The electron configuration for neutral $\text{Mg}$ is $1s^2 2s^2 2p^6 3s^2$.
When $\text{Mg}$ forms an $\text{Mg}^{2+}$ ion, it loses 2 electrons from its outermost shell ($3s^2$).
The resulting configuration for $\text{Mg}^{2+}$ is:

$$\boxed{1s^2 2s^2 2p^6}$$

b) Identify the valency of the hypochlorite anion ($\text{ClO}^-$) in this compound.
The compound is Calcium Hypochlorite ($\text{Ca(ClO)}_2$).
Calcium ($\text{Ca}$) is in Group 2 and forms a $\text{Ca}^{2+}$ ion.
Since the compound is electrically neutral, the total negative charge from the hypochlorite anions must balance the $+2$ charge from the calcium ion.
There are two $\text{ClO}$ anions for every one $\text{Ca}^{2+}$ ion.
Let the charge of the hypochlorite anion be $x$.

$$(+2) + 2(x) = 0$$ $$2x = -2$$ $$x = -1$$

The valency of the hypochlorite anion ($\text{ClO}^-$) is:

$$\boxed{1}$$

c) If calcium ($\text{Z}=20$) reacts with chlorine gas ($\text{Cl}_2$), it forms calcium chloride ($\text{CaCl}_2$).

i) Write the s,p configuration for the $\text{Cl}^-$ ion ($\text{Z}=17$).
Chlorine ($\text{Cl}$) has an atomic number of 17. A neutral $\text{Cl}$ atom has 17 electrons.
The electron configuration for neutral $\text{Cl}$ is $1s^2 2s^2 2p^6 3s^2 3p^5$.
When $\text{Cl}$ forms a $\text{Cl}^-$ ion, it gains 1 electron to complete its outermost p-orbital.
The s,p configuration for the $\text{Cl}^-$ ion is:

$$\boxed{1s^2 2s^2 2p^6 3s^2 3p^6}$$

ii) Write a balanced equation and include state symbols.
Calcium ($\text{Ca}$) is a solid metal. Chlorine gas ($\text{Cl}_2$) is a gas. Calcium chloride ($\text{CaCl}_2$) is an ionic solid.
The balanced chemical equation is:

$$\boxed{\text{Ca}(s) + \text{Cl}_2(g) \to \text{CaCl}_2(s)}$$

Question 12:
A chemist is synthesizing a drug that requires Ammonium Carbonate.

a) Write the formula for the Ammonium radical and the Carbonate radical including their charges.

• Ammonium radical: $\boxed{\text{NH}_4^+}$
• Carbonate radical: $\boxed{\text{CO}_3^{2-}}$

b) Write out the final chemical formula for Ammonium Carbonate.
To form a neutral compound, two ammonium ions ($\text{NH}_4^+$) are needed to balance the charge of one carbonate ion ($\text{CO}_3^{2-}$).
The final chemical formula for Ammonium Carbonate is:

$$\boxed{(\text{NH}_4)_2\text{CO}_3}$$

Question 13:
A forensic chemist in Nairobi is analyzing a gold bar suspected to be a fake alloy. The bar contains a rare element Gallium (Ga) which has 2 isotopes: $\text{}^{69}\text{Ga}$ and $\text{}^{71}\text{Ga}$. The reported relative atomic mass ($\text{Ar}$) of the certificate is 69.72.

a) If the abundance of $\text{}^{69}\text{Ga}$ is approximately 60%, write an equation for the abundance of $\text{}^{71}\text{Ga}$ and calculate the exact percentage abundance of the isotopes.

Let $x$ be the percentage abundance of $\text{}^{69}\text{Ga}$ and $y$ be the percentage abundance of $\text{}^{71}\text{Ga}$.
The sum of the abundances must be 100%:

$$x + y = 100$$

The equation for the abundance of $\text{}^{71}\text{Ga}$ is:

$$\boxed{y = 100 - x}$$

The relative atomic mass ($\text{Ar}$) is calculated using the formula:

$$\text{Ar} = \frac{(\text{mass of isotope 1} \times \text{abundance 1}) + (\text{mass of isotope 2} \times \text{abundance 2})}{100}$$

Substitute the given values:

$$69.72 = \frac{(69 \times x) + (71 \times y)}{100}$$

Substitute $y = 100 - x$ into the equation:

$$69.72 = \frac{(69x) + (71(100 - x))}{100}$$

Multiply both sides by 100:

$$6972 = 69x + 7100 - 71x$$ $$6972 = 7100 - 2x$$

Rearrange to solve for $x$:

$$2x = 7100 - 6972$$ $$2x = 128$$ $$x = \frac{128}{2}$$ $$x = 64\%$$

Now calculate $y$:

$$y = 100 - x = 100 - 64$$ $$y = 36\%$$

b) Calculate the exact percentage abundance of the isotopes.
The exact percentage abundance of the isotopes are:

• $\text{}^{69}\text{Ga}$: $\boxed{64\%}$
• $\text{}^{71}\text{Ga}$: $\boxed{36\%}$

Question 14:

a) Write the s,p configuration for the Beryllium atom ($1s^2 2s^2$).
Beryllium ($\text{Be}$) has an atomic number of 4.
The s,p configuration for the Beryllium atom is:

$$\boxed{1s^2 2s^2}$$

b) Beryllium often reacts to form a $\text{Be}^{2+}$ ion which always has the same electronic configuration as the Helium atom.

c) Why does Beryllium prefer to lose 2 electrons rather than gain 6 electrons to fill its outer shell?
Beryllium prefers to lose 2 electrons because it has 2 valence electrons in its outermost $2s$ orbital. Losing these 2 electrons results in a stable electron configuration identical to that of Helium ($1s^2$), which is a noble gas. This process requires significantly less energy than gaining 6 electrons to fill its $2p$ orbital and achieve a Neon-like configuration ($1s^2 2s^2 2p^6$). Atoms tend to react in ways that achieve stability with the minimum energy input.

Question 15:
A new solid state battery uses Aluminum Phosphate ($\text{AlPO}_4$) as an electrolyte.

a) Draw the structure of Aluminum ion ($\text{Z}=13$).
Aluminum ($\text{Al}$) has an atomic number of 13. A neutral $\text{Al}$ atom has 13 electrons.
Its electron configuration is $1s^2 2s^2 2p^6 3s^2 3p^1$.
When Aluminum forms an ion, it typically loses its 3 valence electrons to become $\text{Al}^{3+}$.
The electron configuration for $\text{Al}^{3+}$ is $1s^2 2s^2 2p^6$.
This means the $\text{Al}^{3+}$ ion has 2 electrons in the first shell and 8 electrons in the second shell.

Here is a Bohr model representation of the Aluminum ion ($\text{Al}^{3+}$):

$$(Nucleus: 13 protons)$$