Question

version b
name
period
chemistry
quiz: atomic notation, ions, & valence electrons
unit 3 atomic structure & theory
1
2

1. what charge would nitrogen (n) form based on its location on the periodic table?

a. +1
b. -1
c. 0
d. -3
e. none of these answers are correct

1. what charge would aluminum (al) form based on its location on the periodic table?

a. +1
b. +2
c. -1
d. +3
e. none of these answers are correct
5.
6.

1. how many valence electrons does the element carbon (c) have?

a. 1
b. 4
c. 7
d. 3

1. how many valence electrons does the element neon (ne) have?

a. 6
b. 2
c. 5
d. 1

Explanation:

Step1: Atomic Notation for \(^{80}\text{Br}\)

The atomic number (\(Z\)) of bromine (\(Br\)) is \(35\) (from the periodic table). The mass number (\(A\)) is given as \(80\). The number of protons (\(p^{+}\)) is equal to the atomic number, so \(p^{+}=35\). For a neutral atom, the number of electrons (\(e^{-}\)) is equal to the number of protons. But since it's not specified as an ion here (assuming neutral for the first part of the table), \(e^{-}=35\). The number of neutrons (\(n^{0}\)) is calculated using the formula \(n^{0}=A - Z\). So \(n^{0}=80 - 35=45\).

Step2: Atomic Notation for \(^{65}\text{Zn}\)

The atomic number (\(Z\)) of zinc (\(Zn\)) is \(30\) (from the periodic table). The mass number (\(A\)) is given as \(65\). The number of protons (\(p^{+}\)) is \(Z = 30\). For a neutral atom, \(e^{-}=p^{+}=30\). The number of neutrons (\(n^{0}\)) is \(n^{0}=A - Z=65 - 30 = 35\).

Step3: Charge of Nitrogen (\(N\))

Nitrogen is in group \(15\) (or \(5A\)) of the periodic table. Elements in group \(15\) tend to gain \(3\) electrons to achieve a full - outer shell. When an atom gains electrons, it forms a negatively - charged ion. The charge is \( - 3\).

Step4: Charge of Aluminium (\(Al\))

Aluminium is in group \(13\) (or \(3A\)) of the periodic table. Elements in group \(13\) tend to lose \(3\) electrons to achieve a full - outer shell. When an atom loses electrons, it forms a positively - charged ion. The charge is \(+3\).

Step5: Atomic Notation for \(^{63}\text{Cu}^{+}\)

The atomic number (\(Z\)) of copper (\(Cu\)) is \(29\) (from the periodic table). The mass number (\(A\)) is given as \(63\). The number of protons (\(p^{+}\)) is \(Z = 29\). For the ion \(Cu^{+}\), it has lost \(1\) electron. So the number of electrons (\(e^{-}\)) is \(e^{-}=29 - 1=28\). The number of neutrons (\(n^{0}\)) is \(n^{0}=A - Z=63 - 29 = 34\).

Step6: Atomic Notation for \(^{79}\text{Se}^{2 -}\)

The atomic number (\(Z\)) of selenium (\(Se\)) is \(34\) (from the periodic table). The mass number (\(A\)) is given as \(79\). The number of protons (\(p^{+}\)) is \(Z = 34\). For the ion \(Se^{2 -}\), it has gained \(2\) electrons. So the number of electrons (\(e^{-}\)) is \(e^{-}=34+2 = 36\). The number of neutrons (\(n^{0}\)) is \(n^{0}=A - Z=79 - 34 = 45\).

Step7: Valence Electrons of Carbon (\(C\))

Carbon has an electron configuration of \(1s^{2}2s^{2}2p^{2}\). The valence electrons are in the \(n = 2\) shell. The number of valence electrons is \(4\) (sum of electrons in \(2s\) and \(2p\) sub - shells: \(2 + 2\)).

Step8: Valence Electrons of Neon (\(Ne\))

Neon has an electron configuration of \(1s^{2}2s^{2}2p^{6}\). The valence electrons are in the \(n = 2\) shell. The number of valence electrons is \(8\) (sum of electrons in \(2s\) and \(2p\) sub - shells: \(2+6\)).

Answer:

1. Atomic \#: \(35\), \(\#\) of protons: \(35\), \(\#\) of electrons: \(35\), \(\#\) of neutrons: \(45\)
2. Atomic \#: \(30\), Mass \#: \(65\), \(\#\) of protons: \(30\), \(\#\) of electrons: \(30\), \(\#\) of neutrons: \(35\)
3. d. \(-3\)
4. d. \(+3\)
5. Atomic \#: \(29\), Mass \#: \(63\), \(\#\) of electrons: \(28\), \(\#\) of neutrons: \(34\)
6. Mass \#: \(79\), \(\#\) of electrons: \(36\), \(\#\) of neutrons: \(45\)
7. b. \(4\)
8. a. \(8\)