How many valence electrons does your compound have? sodium citrate
The correct answer and explanation is:
To determine the number of valence electrons in sodium citrate, let’s break down the compound. Sodium citrate has the formula Na₃C₆H₅O₇. It is composed of 3 sodium (Na) atoms, 6 carbon (C) atoms, 5 hydrogen (H) atoms, and 7 oxygen (O) atoms.
Sodium (Na):
Sodium is an alkali metal in Group 1 of the periodic table, and elements in this group have 1 valence electron. Since there are 3 sodium atoms in sodium citrate, they contribute a total of: 3×1=3 valence electrons3 \times 1 = 3 \text{ valence electrons}
Carbon (C):
Carbon is in Group 14 of the periodic table and has 4 valence electrons. Since there are 6 carbon atoms in sodium citrate, they contribute a total of: 6×4=24 valence electrons6 \times 4 = 24 \text{ valence electrons}
Hydrogen (H):
Hydrogen, although in Group 1, has 1 valence electron. There are 5 hydrogen atoms in sodium citrate, contributing: 5×1=5 valence electrons5 \times 1 = 5 \text{ valence electrons}
Oxygen (O):
Oxygen is in Group 16 and has 6 valence electrons. With 7 oxygen atoms in sodium citrate, they contribute: 7×6=42 valence electrons7 \times 6 = 42 \text{ valence electrons}
Total Valence Electrons:
Now, adding the valence electrons from each element: 3(Na)+24(C)+5(H)+42(O)=74 valence electrons3 (\text{Na}) + 24 (\text{C}) + 5 (\text{H}) + 42 (\text{O}) = 74 \text{ valence electrons}
Thus, sodium citrate has a total of 74 valence electrons.
Explanation:
Valence electrons are important because they determine how atoms interact and bond with other atoms. In sodium citrate, the sodium ions (Na⁺) will donate their single valence electron to form ionic bonds with the citrate ions (C₆H₅O₇³⁻), which results in the formation of a stable ionic compound. The electron distribution in sodium citrate plays a role in its ability to act as a buffer in various chemical processes, including in biological systems where it helps maintain pH balance.