Question

when a certain weak - field ligand forms an octahedral complex with $ce{zn^{2+}}$ cation, the energies of the valence $d$ orbitals on the zinc atom are split according to this electron box diagram: (diagram description: energy in kj/mol, with orbitals $z^2$, $x^2 - y^2$ at around - 400 kj/mol level, and $xy$, $yz$, $xz$ at around - 600 kj/mol level) using this diagram, answer the following questions. how many unpaired $d$ electron spins does the zinc atom have? (blank for answer) is the complex paramagnetic or diamagnetic? (options: paramagnetic, diamagnetic) predict the color of the complex. (options: blue or green, red or violet, orange or yellow, white or transparent)

Explanation:

Step1: Determine Zn²⁺ electron configuration

Zinc (Zn) has an atomic number of 30, so its electron configuration is $[Ar] 3d^{10} 4s^2$. When it forms $Zn^{2+}$, it loses the 4s electrons, so the configuration becomes $[Ar] 3d^{10}$. This means the 3d orbitals have 10 electrons.

Step2: Analyze electron filling in d orbitals

In the octahedral complex, the d orbitals split into two sets. But with 10 electrons, all d orbitals will be completely filled (each of the 5 d orbitals has 2 electrons, paired). So there are no unpaired electrons.

Step3: Determine magnetic property

A complex with no unpaired electrons is diamagnetic (paramagnetic has unpaired electrons).

Step4: Predict color

If a complex has no unpaired electrons (diamagnetic, and in this case, the d orbitals are fully filled, so there's no d - d electron transition that absorbs visible light), it will appear white or transparent (since no visible light is absorbed, it reflects or transmits all, appearing white/transparent).

Answer:

s:
How many unpaired d electron spins does the zinc atom have? $\boldsymbol{0}$
Is the complex paramagnetic or diamagnetic? $\boldsymbol{\text{diamagnetic}}$
Predict the color of the complex. $\boldsymbol{\text{white or transparent}}$