Question

select the subatomic composition of a vanadium ion (v⁵⁺) with a mass number of 51. refer to a periodic table. 23 protons, 28 neutrons, 23 electrons 28 protons, 23 neutrons, 18 electrons 28 protons, 23 neutrons, 23 electrons 23 protons, 28 neutrons, 18 electrons

Explanation:

Step1: Find atomic number of vanadium

From periodic - table, atomic number of vanadium (V) is 23. Atomic number = number of protons, so number of protons = 23.

Step2: Calculate number of neutrons

Mass number = number of protons+number of neutrons. Given mass number = 51 and number of protons = 23. So, number of neutrons=51 - 23=28.

Step3: Determine number of electrons

Vanadium ion is $V^{5 + }$, which means it has lost 5 electrons. Neutral vanadium has 23 electrons (equal to number of protons). So, number of electrons in $V^{5+}$ is 23 - 5 = 18.

Answer:

23 protons, 28 neutrons, 18 electrons