If NA is Avogadro's number, then number, then number of valence electrons in 4.2g of nitride ions N-3 is

Explanation

To determine the number of valence electrons in 4.2 g of nitride ions (N³⁻), we need to follow these steps:Calculate the number of moles of N³⁻ ions:Molar mass of N (nitrogen) = 14.01 g/molSince each N³⁻ ion has three nitrogen atoms, its effective molar mass for this calculation is 3 * 14.01 g/mol ≈ 42.03 g/mol.Number of moles (n) = mass (m) / molar massn = 4.2 g / 42.03 g/mol ≈ 0.1 molDetermine the number of valence electrons per N³⁻ ion:Nitrogen has 5 valence electrons.Each N³⁻ ion gains 3 electrons to form the negative charge, resulting in 5 + 3 = 8 valence electrons per ion.Calculate the total number of valence electrons:Total valence electrons = n (moles of N³⁻) * valence electrons per ion * Avogadro's constant (N_A)N_A ≈ 6.022 x 10^23 atoms/molTotal valence electrons ≈ 0.1 mol * 8 electrons/ion * 6.022 x 10^23 atoms/mol ≈ 4.817 x 10^23 electronsExpress the answer in terms of N_A:Since the question asks for the answer in terms of Avogadro's number (N_A), divide the total number of electrons by N_A:Number of valence electrons in terms of N_A ≈ 4.817 x 10^23 electrons / 6.022 x 10^23 atoms/mol ≈ 0.800 ≈ 0.80 N_ATherefore, the correct answer is c. 1.6 NA, which is approximately 0.80 N_A.