Question

3 look at the following electron configurations. which of the atoms will behave like krypton?
a (1s^2 2s^2 2p^6)

b (1s^2 2s^2 2p^6 3s^2 3p^6 4s^2 3d^{10} 4p^4)

c (1s^2 2s^2 2p^6 3s^2 3p^6 4s^2 3d^{10})

d (1s^2 2s^2 2p^6 3s^2 3p^4)

Explanation:

To solve this, we first recall that Krypton (Kr) is a noble gas with a full valence shell (stable electron configuration). Noble gases have electron configurations ending with a full p - orbital (ns²np⁶ for the outermost shell, in general for main - group noble gases). Let's analyze each option:

Option A

The electron configuration is \(1s^{2}2s^{2}2p^{6}\). This is the electron configuration of Neon (Ne), which is a noble gas, but not Krypton. Krypton is in the 4th period, while Neon is in the 2nd period.

Option B

The electron configuration is \(1s^{2}2s^{2}2p^{6}3s^{2}3p^{6}4s^{2}3d^{10}4p^{4}\). The outermost shell here is the 4th shell, and the p - orbital has 4 electrons, which is not a full p - orbital (a full p - orbital has 6 electrons). So this atom will not have the same behavior as Krypton.

Option C

Wait, there is a mistake in the original option C. The correct electron configuration for an atom with a configuration that would be related to Krypton - like behavior (a noble gas - like configuration) should have a full p - orbital in the outermost shell. Let's re - evaluate. Wait, maybe there was a typo. If we assume that the correct configuration for an atom that would be isoelectronic or have the same valence shell configuration as Krypton (which has electron configuration \([Ar]4s^{2}3d^{10}4p^{6}\)), let's check again.

Wait, the original option C is \(1s^{2}2s^{2}2p^{6}3s^{2}3p^{6}4s^{2}3d^{10}\). Wait, no, that's the configuration of Zinc (Zn). Wait, maybe there is a mistake in the options. Wait, perhaps the intended option was one with a full 4p orbital. Wait, maybe the user made a typo. But among the given options, let's re - check:

Wait, Krypton has electron configuration \(1s^{2}2s^{2}2p^{6}3s^{2}3p^{6}4s^{2}3d^{10}4p^{6}\). So an atom that has a similar (noble gas) electron configuration should have a full valence p - orbital.

Wait, let's re - examine the options:

Option A: \(1s^{2}2s^{2}2p^{6}\) (Neon, noble gas, period 2)

Option B: \(1s^{2}2s^{2}2p^{6}3s^{2}3p^{6}4s^{2}3d^{10}4p^{4}\) (Selenium, Se, has 6 valence electrons in the 4th shell but p⁴, not noble gas)

Option C: \(1s^{2}2s^{2}2p^{6}3s^{2}3p^{6}4s^{2}3d^{10}\) (Zinc, Zn, d - block element, not noble gas)

Option D: \(1s^{2}2s^{2}2p^{6}3s^{2}3p^{4}\) (Sulfur, S, p⁴ in 3rd shell, not noble gas)

Wait, there must be a mistake in the options. But if we assume that option C was supposed to be \(1s^{2}2s^{2}2p^{6}3s^{2}3p^{6}4s^{2}3d^{10}4p^{6}\) (which is Krypton itself), but among the given options, the closest in terms of having a filled inner shell and a configuration that is a noble gas - like (full valence shell) is Option A is Neon, but none of the options seem to be Krypton - like. Wait, maybe the question is about which atom has a noble gas electron configuration (like Krypton, which is a noble gas). So noble gases have full valence shells.

Neon (\(1s^{2}2s^{2}2p^{6}\)) is a noble gas, Krypton is also a noble gas. So if we consider "behave like Krypton" as having noble gas properties (inert, full valence shell), then Option A has a full valence shell (2nd shell, p⁶), so it behaves like a noble gas (like Krypton). But Krypton is in the 4th period. Wait, maybe the question has a typo. But among the given options, Option A is a noble gas (Neon) with a full valence shell, so it will behave like Krypton (a noble gas) in terms of chemical inertness.

Answer:

A. \(1s^{2}2s^{2}2p^{6}\)