Question

Electron Configuration of [Cr(H₂O)₆]₂

A. d₂s₄
B. d₄s₂
C. d₃s₃
D. d7s-1

Answer 1

The electron configuration of [Cr(H₂O)₆]²+ is expected to be [Ar]3d³ based on chromium having a d³ ion configuration after losing two electrons; the complex ion is typically high-spin with three unpaired electrons. The options provided do not match this configuration, suggesting an error in the question or its options.

Step-by-step explanation:

The question asks for the electron configuration of the complex ion [Cr(H₂O)₆]²+. Chromium, in its neutral state, has the electron configuration [Ar]3d⁵ 4s¹. When we form the [Cr(H₂O)₆]²+ ion, chromium loses two electrons. Since electrons are first removed from the higher energy 4s orbital, chromium becomes a d³ ion. This results in the electron configuration [Ar]3d³ as further removal of electrons occurs from the 3d orbitals. Factoring in the nature of the ligands in the complex, which are water molecules in this case, the complex is typically a high-spin complex because H₂O is a weak field ligand and does not induce a large splitting in the d orbital energies that would favor a low-spin configuration. This results in three unpaired electrons. Hence, the most accurate option for the electron configuration of [Cr(H₂O)₆]²+ would be d³s⁰, which corresponds to [Ar]3d³, not present in the options provided. Since none of the options provided fully match the expected electron configuration, a direct match cannot be made, implying a possible error in the options or the question itself.

Answer 2

The electron configuration of [Cr(H₂O)₆]²⁺ complex is a d₄ high-spin configuration, due to water being a weak field ligand, which does not cause significant splitting of the d orbitals.

This results in the configuration d4s0 with four unpaired d electrons. The correct answer is option B. d₄s₂, in the choices provided.

Step-by-step explanation:

The electron configuration of [Cr(H₂O)₆]²⁺ needs to be determined. Chromium in its neutral state has the electron configuration of [Ar]₃d₅₄s₁.

When it forms the [Cr(H₂O)₆]²⁺ complex, it loses 2 electrons from its 4s orbital and 1 electron from the 3d orbital, resulting in a d₄ configuration.

Since water is a weak field ligand, it causes a small splitting energy (Δo) in the d orbitals.

This small splitting energy is less than the spin-pairing energy; therefore, the complex will adopt a high-spin configuration, with electrons filling the higher energy eg orbitals before pairing up in the lower energy t₂g orbitals.

As a result, the d4 high-spin electron configuration of Cr²⁺ in this complex is d₄s₀, with four unpaired d electrons and no s electrons, which is option B in the choices provided.

The correct answer is option B. d₄s₂, in the choices provided.