Question

what is the effective nuclear charge experienced by the 2p and 3p electrons of a chlorine atom (cl), respectively?
15, 7
11, 7
17, 7
17, 10
question 2
1 pts
rank the following atoms in terms of decreasing atomic radius.
f, mg, na, o, n
f, o, n, mg, na
na, n, o, mg, f
na, mg, n, o, f

Explanation:

Step1: Determine the effective nuclear charge for 2p electrons in Cl

• Chlorine (Cl) has an atomic number \(Z = 17\).
• For 2p electrons, the shielding electrons (\(S\)): Inner electrons are \(1s^{2}2s^{2}\). Using the shielding rules (Slater's rules: electrons in the same shell as the valence electron contribute \(0.35\), electrons in the shell below contribute \(0.85\), and electrons in shells further below contribute \(1.00\)). For 2p electrons in Cl (\(n = 2\)), the shielding electrons are \(1s^{2}2s^{2}\). \(S=(2\times1.00)+(2\times0.85) = 2 + 1.7=3.7\) (but a simpler way: for \(n = 2\) electrons in Cl, the number of non - valence electrons is \(2\) (from \(1s^{2}\)) and \(2\) (from \(2s^{2}\)). Using the formula \(Z_{eff}=Z - S\), where \(Z = 17\) and \(S=(2\times1)+(2\times1)= 6\) (another approximation: for \(n = 2\) electrons, the core electrons are \(K - shell(2)\) and \(2s\) electrons (\(2\)). \(Z_{eff}=17-(2 + 2)=11\)

Step2: Determine the effective nuclear charge for 3p electrons in Cl

• For 3p electrons in Cl (\(n = 3\)), the non - valence electrons are \(1s^{2}2s^{2}2p^{6}\). So \(S=(2\times1)+(8\times1)=10\)
• Using the formula \(Z_{eff}=Z - S\), with \(Z = 17\) and \(S = 10\), \(Z_{eff}=17 - 10=7\)

for atomic radius ranking:

• Atomic radius trends:
• Down a group: atomic radius increases due to the addition of new electron shells.
• Across a period: atomic radius decreases (from left to right) because of increasing nuclear charge (more protons in the nucleus) which pulls the electrons closer.
• \(Na\) and \(Mg\) are in the third period (\(n = 3\)), \(N\), \(O\), \(F\) are in the second period (\(n = 2\)). Since \(n = 3>n = 2\), \(Na\) and \(Mg\) are larger than \(N\), \(O\), \(F\).
• Among \(Na\) and \(Mg\) (third period), \(Na\) is larger than \(Mg\) (as we move from left to right in a period, atomic radius decreases).
• Among \(N\), \(O\), \(F\) (second period), \(N>O>F\) (atomic radius decreases from left to right in a period)

Answer:

11,7