Question

1. order the following from lowest to highest electronegativity: o, na, f, ne, c, li
2. order the following from lowest to highest ionization energy: o, na, f, ne, c, li
3. order the following from smallest to largest atomic radius: o, na, f, ne, c, li

Explanation:

8. Ordering by electronegativity

Step1: Recall electronegativity trends

Electronegativity increases across a period (left - to - right) and decreases down a group. Noble gases (like Ne) have very low electronegativity as they have a full valence shell and do not tend to gain electrons easily.

Step2: Analyze each element

• \(Li\) is in group 1, period 2. \(Na\) is in group 1, period 3 (lower electronegativity than \(Li\) as it is down the group).
• \(C\) and \(O\) are in period 2 ( \(C\) is to the left of \(O\), so \(C
• \(F\) is the most electronegative element. \(Ne\) is a noble gas.

So the order is \(Na < Li < C < O < F\) ( \(Ne\) is not considered in a normal electronegativity comparison as it is a noble gas and the values are not well - defined in the same context as the other elements for chemical bonding purposes)

Step1: Recall ionization energy trends

Ionization energy increases across a period (left - to - right) and decreases down a group. Noble gases (\(Ne\)) have very high ionization energies because of their stable full - valence - shell configuration.

Step2: Analyze each element

• \(Na\) and \(Li\) are group 1 elements (\(Na\) has a lower ionization energy than \(Li\) as it is down the group).
• \(C\), \(O\), \(F\) are in period 2. The order within the period is \(C < O
• \(Ne\) has the highest ionization energy among these.

So the order is \(Na < Li < C < O < F < Ne\)

Step1: Recall atomic radius trends

Atomic radius decreases across a period (left - to - right) and increases down a group.

Step2: Analyze each element

• \(Na\) is in period 3, so it has the largest radius among these.
• \(Li\) is in period 2, group 1. \(C\), \(O\), \(F\) are in period 2 ( \(Li\) has a larger radius than \(C\), \(C\) has a larger radius than \(O\), \(O\) has a larger radius than \(F\)).
• \(Ne\) has a “van der Waals” radius (which is larger than covalent radii of non - noble gases in the same period, but still smaller than \(Na\) and \(Li\) in terms of atomic size comparison for the purpose of this ordering).

So the order is \(F < O < C < Ne < Li < Na\)

Answer:

\(Na < Li < C < O < F\)

9. Ordering by ionization energy