Question

1. fill in the following table:

atom	# of valence electrons	# of unpaired electrons in the valence shell	# of covalent bonds the atom forms to reach stability	the atom’s valence*
₆c
₇n
₈o

• the atom’s valence is defined as the bonding capacity of an atom, that is the number of covalent bonds an atom can form usually equals the number of unpaired electrons in its valence (outermost) shell.

Explanation:

Step1: Analyze \( _1\text{H} \)

Hydrogen has an atomic number of 1. Its electron configuration is \( 1s^1 \), so the number of valence electrons is 1. The number of unpaired electrons in the valence shell is 1 (since the \( 1s \) orbital has 1 electron, unpaired). To reach stability (like helium's electron configuration, \( 1s^2 \)), it needs 1 more electron, so it forms 1 covalent bond. The valence is equal to the number of covalent bonds, so valence is 1.

Step2: Analyze \( _6\text{C} \)

Carbon has an atomic number of 6. Electron configuration: \( 1s^2 2s^2 2p^2 \). Valence electrons are in the \( n = 2 \) shell, so \( 2s^2 + 2p^2 = 4 \) valence electrons. In the \( 2p \) subshell, there are 2 unpaired electrons (Hund's rule: electrons fill orbitals singly first). But carbon can hybridize, and in general, to reach a stable octet, it forms 4 covalent bonds (since it needs 4 more electrons to complete the octet: \( 8 - 4 = 4 \)). The number of unpaired electrons in the valence shell (considering bonding) is 4 (as it can form 4 bonds), and the valence is 4.

Step3: Analyze \( _7\text{N} \)

Nitrogen has an atomic number of 7. Electron configuration: \( 1s^2 2s^2 2p^3 \). Valence electrons: \( 2s^2 + 2p^3 = 5 \). In the \( 2p \) subshell, there are 3 unpaired electrons (Hund's rule: 3 electrons in 3 \( p \) orbitals, each unpaired). To reach an octet, it needs \( 8 - 5 = 3 \) more electrons, so it forms 3 covalent bonds. The valence is 3 (equal to the number of covalent bonds, which equals the number of unpaired electrons).

Step4: Analyze \( _8\text{O} \)

Oxygen has an atomic number of 8. Electron configuration: \( 1s^2 2s^2 2p^4 \). Valence electrons: \( 2s^2 + 2p^4 = 6 \). In the \( 2p \) subshell, there are 2 unpaired electrons (since \( p \) orbitals can hold 6 electrons; \( 2p^4 \) means 2 orbitals have paired electrons and 2 have unpaired? Wait, no: \( p \) has 3 orbitals. \( 2p^4 \): first 3 orbitals get 1 each (unpaired), then the fourth pairs with one, so number of unpaired electrons is 2. To reach an octet, it needs \( 8 - 6 = 2 \) more electrons, so it forms 2 covalent bonds. The valence is 2.

Atom	# of valence electrons	# of unpaired electrons in the valence shell	# of covalent bonds the atom forms to reach stability	The atom’s valence*
\( _6\text{C} \)	4	4	4	4
\( _7\text{N} \)	5	3	3	3
\( _8\text{O} \)	6	2	2	2

Answer:

The filled table is as shown above.