Question

fill in the missing information:
table with columns: symbol, atom or ion? (check all that apply), number of protons, number of electrons; rows:

• mg²⁺: checkboxes (neutral atom, cation, anion), empty boxes for protons/electrons
• ru: checkboxes (neutral atom, cation, anion), empty boxes for protons/electrons
• blank symbol: checkboxes (neutral atom, cation, anion), protons=9, electrons=10

Explanation:

For $\boldsymbol{\ce{Mg^{2+}}}$:

Step1: Determine type (cation)

$\ce{Mg^{2+}}$ has a positive charge, so it's a cation (not neutral atom or anion).

Step2: Find protons (atomic number of Mg)

Atomic number of Mg is 12, so protons = 12.

Step3: Calculate electrons (protons - charge)

Electrons = $12 - 2 = 10$.

For $\boldsymbol{\ce{Ru}}$:

Step1: Determine type (neutral atom)

$\ce{Ru}$ is a neutral atom (no charge), so not cation/anion.

Step2: Find protons (atomic number of Ru)

Atomic number of Ru is 44, so protons = 44.

Step3: Electrons in neutral atom (equal to protons)

Electrons = 44.

For the last row:

Step1: Identify element (atomic number 9)

Atomic number 9 is fluorine (F). Since electrons (10) > protons (9), it's an anion ($\ce{F^-}$).

Step2: Confirm type (anion)

Electrons > protons, so anion (not neutral atom/cation).

Filled Table (Row - by - Row):

Symbol	atom or ion? (check)	number of protons	number of electrons
$\ce{Ru}$	$\boldsymbol{\checkmark}$ neutral atom, $\square$ cation, $\square$ anion	44	44
$\ce{F^-}$	$\square$ neutral atom, $\square$ cation, $\boldsymbol{\checkmark}$ anion	9	10

(For the checkboxes, mark the appropriate ones as shown above. The symbol for the last row is $\ce{F^-}$ (or just $\ce{F}$ with anion checked, but $\ce{F^-}$ is more precise).)

Answer:

For $\boldsymbol{\ce{Mg^{2+}}}$:

Step1: Determine type (cation)

$\ce{Mg^{2+}}$ has a positive charge, so it's a cation (not neutral atom or anion).

Step2: Find protons (atomic number of Mg)

Atomic number of Mg is 12, so protons = 12.

Step3: Calculate electrons (protons - charge)

Electrons = $12 - 2 = 10$.

For $\boldsymbol{\ce{Ru}}$:

Step1: Determine type (neutral atom)

$\ce{Ru}$ is a neutral atom (no charge), so not cation/anion.

Step2: Find protons (atomic number of Ru)

Atomic number of Ru is 44, so protons = 44.

Step3: Electrons in neutral atom (equal to protons)

Electrons = 44.

For the last row:

Step1: Identify element (atomic number 9)

Atomic number 9 is fluorine (F). Since electrons (10) > protons (9), it's an anion ($\ce{F^-}$).

Step2: Confirm type (anion)

Electrons > protons, so anion (not neutral atom/cation).

Filled Table (Row - by - Row):

Symbol	atom or ion? (check)	number of protons	number of electrons
$\ce{Ru}$	$\boldsymbol{\checkmark}$ neutral atom, $\square$ cation, $\square$ anion	44	44
$\ce{F^-}$	$\square$ neutral atom, $\square$ cation, $\boldsymbol{\checkmark}$ anion	9	10

(For the checkboxes, mark the appropriate ones as shown above. The symbol for the last row is $\ce{F^-}$ (or just $\ce{F}$ with anion checked, but $\ce{F^-}$ is more precise).)