Question

1. the elements c, be, na, mg, ne, o listed in order of increasing first ionization energy are: a. ne, o, c, mg, be, na b. be, na, mg, c, o, ne c. na, be, c, o, mg, ne d. ne, na, mg, o, c, be e. na, mg, be, c, o, ne

Explanation:

Step1: Recall ionization energy trends

Ionization energy (IE) generally increases across a period (left to right) and decreases down a group (top to bottom). Exceptions: filled/half - filled subshells (e.g., Be has \(2s^2\), O has \(2p^4\); Be > B trend, N > O trend).

Step2: Analyze each element's position

• Na (sodium): Group 1, Period 3. Low IE (easiest to lose e⁻).
• Mg (magnesium): Group 2, Period 3. IE > Na (same period, right of Na; \(3s^2\) filled).
• Be (beryllium): Group 2, Period 2. IE > Mg? Wait, Period 2 is above Period 3. Be is in Period 2, Mg in Period 3. But Be has \(2s^2\), Mg has \(3s^2\). Also, across Period 2: Be (Group 2), C (Group 14), O (Group 16), Ne (Group 18).
• C (carbon): Group 14, Period 2. IE: Be < C? Wait, Be (\(2s^2\)) vs C (\(2s^22p^2\)). Be has filled \(s\) - subshell, but C is to the right of Be in Period 2. Wait, actually, in Period 2: Li < B < Be < C < O < F < Ne? No, correction: The order in Period 2 (from left to right, with exceptions) is: Li (lowest), then B (lower than Be because Be has \(2s^2\) filled, B has \(2s^22p^1\)), then Be, then C, then O (lower than N because N has \(2p^3\) half - filled, O has \(2p^4\)), then F, then Ne (highest in Period 2).
• Now, comparing across periods: Na (Period 3, Group 1) has lower IE than Mg (Period 3, Group 2). Mg (Period 3, Group 2) has lower IE than Be (Period 2, Group 2) because Be is in a higher period (smaller atomic radius, stronger nuclear attraction). Then Be < C (Period 2, Group 14), C < O (Period 2, Group 16), O < Ne (Period 2, Group 18, noble gas, highest IE).

So the order from lowest to highest (increasing IE): Na (lowest), then Mg, then Be, then C, then O, then Ne.

Answer:

E. Na, Mg, Be, C, O, Ne