Question

which of the following would you put in box e? * (0.5 points) s block d block p block f block 1 1s²2s²2p⁶3s²3p⁶4s²3d¹⁰4p⁵ 4 xe6s²4f⁴ 2 1s²2s²2p⁴ 5 1s²2s²2p⁶3s²3p⁶4s²3d⁷ 3 1s² 6 ne3s²

Explanation:

Step1: Analyze the electron configuration for p - block elements

The p - block elements have their outermost electrons in p orbitals. The general form of electron configuration for p - block elements (after noble gas core, if applicable) ends with \(ns^{2}np^{x}\) (\(x = 1 - 6\)).

Step2: Check each option

• Option 1: \(1s^{2}2s^{2}2p^{6}3s^{2}3p^{6}4s^{2}3d^{10}4p^{5}\) has the last - filled sub - shell as \(4p\). This is a p - block element (halogen, bromine).
• Option 2: \(1s^{2}2s^{2}2p^{4}\) is for oxygen (p - block, but let's check the position).
• Option 3: \(1s^{2}\) is for helium (s - block in some classifications, but usually noble gas).
• Option 4: \([Xe]6s^{2}4f^{4}\) is an f - block element (lanthanide).
• Option 5: \(1s^{2}2s^{2}2p^{6}3s^{2}3p^{6}4s^{2}3d^{7}\) is a d - block element (cobalt).
• Option 6: \([Ne]3s^{2}\) is a s - block element (magnesium).

Since box E is in the p - block and considering the electron configuration length (higher period for box E compared to box C), the electron configuration \(1s^{2}2s^{2}2p^{6}3s^{2}3p^{6}4s^{2}3d^{10}4p^{5}\) (which has a higher principal quantum number for p - orbital compared to \(1s^{2}2s^{2}2p^{4}\)) is more suitable.

Answer:

1. \(1s^{2}2s^{2}2p^{6}3s^{2}3p^{6}4s^{2}3d^{10}4p^{5}\)