How many d electrons are in the valence shell of the Mo4+ cation? How many unpaired electron spins
The Correct Answer and Explanation is:
To determine the number of d electrons in the valence shell of the Mo4+ cation, we need to follow these steps:
Step 1: Determine the Electron Configuration of Mo
Molybdenum (Mo) has an atomic number of 42, meaning its neutral atom has 42 electrons. The electron configuration for a neutral Mo atom is:Mo:[Kr] 4d5 5s1\text{Mo}: [Kr]\, 4d^5\, 5s^1Mo:[Kr]4d55s1
This configuration shows that Mo has 5 electrons in the 4d orbital and 1 electron in the 5s orbital.
Step 2: Determine the Electron Configuration of Mo4+
When Mo loses four electrons to form the Mo4+ cation, these electrons will first be removed from the outermost orbitals, which are the 5s and 4d orbitals.
Thus, Mo4+ will have:Mo4+:[Kr] 4d2\text{Mo}^{4+}: [Kr]\, 4d^2Mo4+:[Kr]4d2
This means that Mo4+ has 2 electrons in the 4d orbital.
Step 3: Determine the Number of Unpaired Electrons
The 4d subshell can hold up to 10 electrons, but in this case, Mo4+ has only 2 electrons in the 4d orbital. These two electrons will occupy different 4d orbitals to minimize electron-electron repulsion. Since each electron in the 4d orbitals will be unpaired, Mo4+ has 2 unpaired electrons.
Conclusion
- The number of d electrons in the valence shell of the Mo4+ cation is 2.
- The number of unpaired electron spins in Mo4+ is 2.
This is because the electron configuration of Mo4+ has 2 electrons in the 4d orbitals, and each of these electrons is unpaired.
