What is the electron configuration of Beryllium (Z=4) and its Be2+ ion? Why does Beryllium prefer to lose…

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What is the electron configuration of Beryllium (Z=4) and its Be2+ ion? Why does Beryllium prefer to lose electrons?

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14. Beryllium (Z=4)

a) The Beryllium atom (Z=4) has 4 electrons.

Fill the lowest energy orbital, 1s, with 2 electrons.
Fill the next lowest energy orbital, 2s, with the remaining 2 electrons.
The s, p configuration for the Beryllium atom is:
$\boxed{1s^2 2s^2}$
b) Beryllium forms a Be $^{2+}$ ion by losing 2 electrons.
The neutral Beryllium atom has 4 electrons ( $1s^2 2s^2$ ).
The Be $^{2+}$ ion has $4 - 2 = 2$ electrons.
The electron configuration of Be $^{2+}$ is $1s^2$ .
The Noble Gas with 2 electrons and an electron configuration of $1s^2$ is Helium.
The Noble Gas with the same electronic configuration as Be $^{2+}$ is:
$\boxed{\text{Helium (He)}}$
c) Beryllium prefers to lose 2 electrons rather than gain 6 to fill its outer shell because losing 2 electrons requires significantly less energy. By losing its two valence electrons from the 2s orbital, Beryllium achieves a stable electron configuration identical to that of the Noble Gas Helium ( $1s^2$ ). Gaining 6 electrons would be energetically unfavorable due to the large energy input required and increased electron-electron repulsion.

15. Aluminum Phosphate (AlPO $_4$ )

a) Aluminum (Al) has an atomic number Z=13. It typically forms an Al $^{3+}$ ion by losing its 3 valence electrons.
A neutral Aluminum atom has 13 protons and 13 electrons.
The Al $^{3+}$ ion has 13 protons (in the nucleus) and $13 - 3 = 10$ electrons.
These 10 electrons are arranged in shells.
The first shell (K-shell) holds 2 electrons, and the second shell (L-shell) holds the remaining 8 electrons.
The structure for the Aluminum ion (Al $^{3+}$ ) can be described as:
- A nucleus containing 13 protons.
- An inner electron shell (K-shell) containing 2 electrons.
- An outer electron shell (L-shell) containing 8 electrons. (Note: Neutrons are not included as the mass number is not provided.)
b) The formula for Aluminum Phosphate is AlPO $_4$ and not Al $_3$ (PO $_4$ ) $_3$ due to the principle of charge balance and the convention of writing ionic formulas in their simplest whole-number ratio.
Aluminum (Al) is in Group 13 and typically forms an ion with a charge of +3, so it is Al $^{3+}$ .
The Phosphate radical (PO $_4$ ) has a charge of -3, so it is PO $_4^{3-}$ .
For a neutral ionic compound, the total positive charge must balance the total negative charge.
$(\text{Charge of Al}^{3+}) \times (\text{Number of Al}^{3+} \text{ ions}) + (\text{Charge of PO}_4^{3-}) \times (\text{Number of PO}_4^{3-} \text{ ions}) = 0$ $(+3) \times (1) + (-3) \times (1) = 0$
A 1:1 ratio of Al $^{3+}$ ions to PO $_4^{3-}$ ions results in a neutral compound.
Ionic compound formulas are always written using the simplest whole-number ratio of the ions.
Therefore, one Al $^{3+}$ ion combines with one PO $_4^{3-}$ ion to form AlPO $_4$ . While Al $_3$ (PO $_4$ ) $_3$ represents the same 1:1 ratio, it is not the simplest formula and thus not the correct chemical formula.
The formula is $\boxed{\text{AlPO}_4}$ because one Al $^{3+}$ ion balances the charge of one PO $_4^{3-}$ ion, and the formula represents the simplest whole-number ratio of the ions.

What is the electron configuration of Beryllium (Z=4) and its Be2+ ion? Why does Beryllium prefer to lose electrons?

Step-by-step solution

Fill the lowest energy orbital, 1s, with 2 electrons.

Fill the next lowest energy orbital, 2s, with the remaining 2 electrons.

The neutral Beryllium atom has 4 electrons ( $1s^2 2s^2$ ).

The Be $^{2+}$ ion has $4 - 2 = 2$ electrons.

The electron configuration of Be $^{2+}$ is $1s^2$ .

The Noble Gas with 2 electrons and an electron configuration of $1s^2$ is Helium.

A neutral Aluminum atom has 13 protons and 13 electrons.

The Al $^{3+}$ ion has 13 protons (in the nucleus) and $13 - 3 = 10$ electrons.

These 10 electrons are arranged in shells.

Aluminum (Al) is in Group 13 and typically forms an ion with a charge of +3, so it is Al $^{3+}$ .

The Phosphate radical (PO $_4$ ) has a charge of -3, so it is PO $_4^{3-}$ .

For a neutral ionic compound, the total positive charge must balance the total negative charge.

A 1:1 ratio of Al $^{3+}$ ions to PO $_4^{3-}$ ions results in a neutral compound.

Ionic compound formulas are always written using the simplest whole-number ratio of the ions.

Still stuck on this one?

More Chemistry Questions

What is the electron configuration of Beryllium (Z=4) and its Be2+ ion? Why does Beryllium prefer to lose electrons?

Step-by-step solution

Fill the lowest energy orbital, 1s, with 2 electrons.

Fill the next lowest energy orbital, 2s, with the remaining 2 electrons.

The neutral Beryllium atom has 4 electrons (1s22s21s^2 2s^21s22s2).

The Be2+^{2+}2+ ion has 4−2=24 - 2 = 24−2=2 electrons.

The electron configuration of Be2+^{2+}2+ is 1s21s^21s2.

The Noble Gas with 2 electrons and an electron configuration of 1s21s^21s2 is Helium.

A neutral Aluminum atom has 13 protons and 13 electrons.

The Al3+^{3+}3+ ion has 13 protons (in the nucleus) and 13−3=1013 - 3 = 1013−3=10 electrons.

These 10 electrons are arranged in shells.

Aluminum (Al) is in Group 13 and typically forms an ion with a charge of +3, so it is Al3+^{3+}3+.

The Phosphate radical (PO4_44​) has a charge of -3, so it is PO43−_4^{3-}43−​.

For a neutral ionic compound, the total positive charge must balance the total negative charge.

A 1:1 ratio of Al3+^{3+}3+ ions to PO43−_4^{3-}43−​ ions results in a neutral compound.

Ionic compound formulas are always written using the simplest whole-number ratio of the ions.

Still stuck on this one?

More Chemistry Questions

The neutral Beryllium atom has 4 electrons ( $1s^2 2s^2$ ).

The Be $^{2+}$ ion has $4 - 2 = 2$ electrons.

The electron configuration of Be $^{2+}$ is $1s^2$ .

The Noble Gas with 2 electrons and an electron configuration of $1s^2$ is Helium.

The Al $^{3+}$ ion has 13 protons (in the nucleus) and $13 - 3 = 10$ electrons.

Aluminum (Al) is in Group 13 and typically forms an ion with a charge of +3, so it is Al $^{3+}$ .

The Phosphate radical (PO $_4$ ) has a charge of -3, so it is PO $_4^{3-}$ .

A 1:1 ratio of Al $^{3+}$ ions to PO $_4^{3-}$ ions results in a neutral compound.