Question

station #8

1. fill in the following table.

atom/ion atomic # p⁺ n⁰ e⁻ charge atom, cation, or anion
li⁺
n
i

Explanation:

Step1: Recall atomic number and basic particle - charge relationships

For an element, the atomic number ($Z$) is equal to the number of protons ($p^+$). In a neutral atom, the number of electrons ($e^-$) is equal to the number of protons. In an ion, the number of electrons is adjusted according to the charge. The number of neutrons ($n^0$) can be found if the mass - number ($A$) is known using $n^0=A - Z$. The charge of a particle is determined by the difference between the number of protons and electrons.

Step2: Analyze $Li^+$

Lithium ($Li$) has an atomic number of $3$. So, for $Li^+$, the number of protons $p^+=3$. The mass - number of common lithium is approximately $7$. Then the number of neutrons $n^0 = 7-3 = 4$. Since it is a $1+$ ion, the number of electrons $e^-=3 - 1=2$. The charge is $+ 1$, and it is a cation.

Atom/Ion	Atomic #	$p^+$	$n^0$	$e^-$	Charge	Atom, Cation, or Anion

Step3: Analyze $N$ (nitrogen)

Nitrogen has an atomic number of $7$. So, $p^+=7$. The common mass - number of nitrogen is $14$, so $n^0=14 - 7 = 7$. In a neutral nitrogen atom, $e^- = 7$, the charge is $0$, and it is an atom.

Atom/Ion	Atomic #	$p^+$	$n^0$	$e^-$	Charge	Atom, Cation, or Anion
$N$	$7$	$7$	$7$	$7$	$0$	Atom

Step4: Analyze $I^-$

Iodine ($I$) has an atomic number of $53$. So, $p^+=53$. A common mass - number of iodine is $127$, so $n^0=127 - 53 = 74$. Since it is a $1 -$ ion, $e^-=53 + 1=54$, the charge is $-1$, and it is an anion.

Atom/Ion	Atomic #	$p^+$	$n^0$	$e^-$	Charge	Atom, Cation, or Anion
$N$	$7$	$7$	$7$	$7$	$0$	Atom
$I^-$	$53$	$53$	$74$	$54$	$-1$	Anion

Answer:

Atom/Ion	Atomic #	$p^+$	$n^0$	$e^-$	Charge	Atom, Cation, or Anion
$N$	$7$	$7$	$7$	$7$	$0$	Atom
$I^-$	$53$	$53$	$74$	$54$	$-1$	Anion