Question

The number of electrons, protons and neutrons in a species are equal to \(10\), \(11\) and \(12\) respectively. The proper symbol of the species is

• A. \({}^{23}_{11}Na\)
• B. \({}^{23}_{11}Na^+\)
• C. \({}^{23}_{10}Ne\)
• D. \({}^{23}_{11}Na^-\)
• E. \({}^{23}_{11}Na^{2+}\)

Hint:

Charge on an ion is found from: \[ \text{charge} = \text{protons} - \text{electrons} \]

Answer 1

The Correct Option is B

Given: - protons \(=11\), so atomic number \(Z=11\), element is sodium - neutrons \(=12\), so mass number: \[ A=11+12=23 \] Electrons \(=10\), but neutral sodium has \(11\) electrons. So one electron is lost: \[ Na^+ \] Thus the species is: \[ \boxed{{}^{23}_{11}Na^+} \] Hence, \[ \boxed{(B)} \]