How many protons, electrons and Neutrons does the Vanadium +5 ion (V) have?
a) P_____
b) n______
c) e______
The correct answer and explanation is :
The vanadium ( V^{5+} ) ion has the following subatomic particles:
- Protons (P): 23
- Neutrons (n): 28 (for the most common isotope, Vanadium-51)
- Electrons (e): 18
Explanation:
Vanadium (V) is a transition metal with an atomic number of 23, meaning that in its neutral state, it has 23 protons and 23 electrons. The number of neutrons in an atom is determined by subtracting the atomic number from the atomic mass. The most common isotope of vanadium is Vanadium-51, which has an atomic mass of 51. Since neutrons and protons contribute to the atomic mass, we calculate:
[
\text{Neutrons} = \text{Atomic Mass} – \text{Protons} = 51 – 23 = 28
]
For the electrons, we need to consider the charge of the ion. A neutral vanadium atom has 23 electrons. When vanadium loses five electrons to form the (V^{5+}) ion, the electron count is:
[
\text{Electrons} = 23 – 5 = 18
]
Thus, the final count for the (V^{5+}) ion is:
- Protons: 23 (unchanged because protons define the element)
- Neutrons: 28 (for the V-51 isotope)
- Electrons: 18 (due to the +5 charge from electron loss)
Vanadium commonly forms multiple oxidation states, but the +5 oxidation state is one of its most stable and commonly occurring states, found in compounds like vanadium pentoxide (V₂O₅). Losing electrons affects the chemical reactivity and bonding properties of the ion, making (V^{5+}) a strong oxidizing agent in redox reactions.