Question

given the elements cl, ge, and k and the values 418, 1255, and 784 kj/mol of possible first ionization energies, match the atoms with their first ionization energies.
cl: 1255 kj/mol; ge: 784 kj/mol; and k: 418 kj/mol
cl: 1255 kj/mol; ge: 418 kj/mol; and k: 784 kj/mol
cl: 418 kj/mol; ge: 784 kj/mol; and k: 1255 kj/mol
cl: 784 kj/mol; ge: 1255 kj/mol; and k: 418 kj/mol
question 6
1 pts
rank the following from least to greatest ionization energy: carbon (c), nitrogen (n), oxygen (o).
c < o < n
c < n < o
o < n < c
n < c < o

Explanation:

• First question:
• Ionization energy generally increases across a period and decreases down a group.
• \(K\) (potassium) is an alkali metal (Group 1), \(Ge\) (germanium) is a metalloid (Group 14), and \(Cl\) (chlorine) is a non - metal (Group 17).
• Alkali metals have relatively low ionization energies as they easily lose one electron to achieve a noble gas configuration. Non - metals have higher ionization energies as they tend to gain electrons.
• So, \(K\) has the lowest ionization energy (\(418\ kJ/mol\)), \(Ge\) has an intermediate value (\(784\ kJ/mol\)), and \(Cl\) has the highest (\(1255\ kJ/mol\)).
• Second question:
• Ionization energy generally increases across a period. But there is an exception between \(N\) and \(O\).
• \(N\) has a half - filled \(2p\) sub - shell (\(2p^{3}\)) which gives it extra stability. Removing an electron from \(O\) (\(2p^{4}\)) is easier than from \(N\) because after losing one electron, \(O^{+}\) has a half - filled \(2p\) sub - shell.
• So, the order from least to greatest ionization energy is \(C

Answer:

• First question: \(Cl: 1255\ kJ/mol; Ge: 784\ kJ/mol; and K: 418\ kJ/mol\)
• Second question: \(C < O < N\)