Determine the number of protons and neutrons for Sodium (23/11 Na) and Fluorine (19/9 F). Also, describe the electron configuration for Na+ and F- ions.

Q2.a.i:

• Sodium (\text{^{23}_{11}Na})
  • A) Protons: The atomic number (subscript) indicates the number of protons. Number of protons = $\boxed{\text{11}}$
  • B) Neutrons: The number of neutrons is the mass number (superscript) minus the atomic number. Number of neutrons = $23 - 11 = \boxed{\text{12}}$
• Fluorine (\text{^{19}_{9}F})
  • A) Protons: The atomic number (subscript) indicates the number of protons. Number of protons = $\boxed{\text{9}}$
  • B) Neutrons: The number of neutrons is the mass number (superscript) minus the atomic number. Number of neutrons = $19 - 9 = \boxed{\text{10}}$

Q2.a.ii:

• A) Sodium ion ($\text{Na}^+$) A sodium atom ($\text{Na}$) has 11 electrons (2, 8, 1). A sodium ion ($\text{Na}^+$) loses one electron, so it has 10 electrons (2, 8).

  $$\text{Nucleus: 11p, 12n}$$ $$\text{1st shell: 2e}^-$$ $$\text{2nd shell: 8e}^-$$

  (Diagram showing a central nucleus with 11 protons and 12 neutrons, surrounded by two electron shells: the first with 2 electrons and the second with 8 electrons.)

• B) Fluorine ion ($\text{F}^-$) A fluorine atom ($\text{F}$) has 9 electrons (2, 7). A fluorine ion ($\text{F}^-$) gains one electron, so it has 10 electrons (2, 8).

  $$\text{Nucleus: 9p, 10n}$$ $$\text{1st shell: 2e}^-$$ $$\text{2nd shell: 8e}^-$$

  (Diagram showing a central nucleus with 9 protons and 10 neutrons, surrounded by two electron shells: the first with 2 electrons and the second with 8 electrons.)

Q2.b.i: Two reasons why salt is important in the diet of humans are:

• It helps maintain fluid balance and blood pressure in the body.
• It is essential for nerve impulse transmission and muscle contraction.

Q2.b.ii: Two examples of an abiotic factor which is non-climatic in a terrestrial ecosystem are:

• Soil pH
• Soil mineral composition

Q2.c.i: An electrical conductor is a material that allows electric current to flow through it easily due to the presence of free electrons or ions.

Q2.c.ii: