Question

name: pd: chemistry u3l12_q2 ct module day 1 hw
the first five ionization energies of an unknown element are listed in the table above.
**4. which of the following statements correctly identifies the element and cites the evidence supporting the identification?
(a) na, because of the large difference between the first and the second ionization energies
(b) al, because of the large difference between the third and fourth ionization energies
(c) si, because the fifth ionization energy has the greatest value
(d) p, because a neutral atom of p has five valence electrons
**5.
the complete photoelectron spectrum of an element is given above. which of the following electron configurations is consistent with the spectrum?
(a) ( 1 s^{2} 2 s^{2} 2 p^{1} )
(b) ( 1 s^{2} 2 s^{2} 2 p^{6} 3 s^{2} 3 p^{3} )
(c) ( 1 s^{2} 2 s^{2} 2 p^{6} 3 s^{2} 3 p^{6} )
(d) ( 1 s^{2} 2 s^{2} 2 p^{6} 3 s^{2} 3 p^{6} 4 s^{2} 3 d^{5} )

1. which of the following lists mg, p, and cl in order of increasing atomic radius?

(a) ( mathrm{cl}<mathrm{p}<mathrm{mg} )
(b) ( mathrm{cl}<mathrm{mg}<mathrm{p} )
(c) ( mathrm{mg}<mathrm{p}<mathrm{cl} )
(d) ( mathrm{mg}<mathrm{cl}<mathrm{p} )

Explanation:

Step1: Analyze ionization energy for question 4

Ionization energy is the energy required to remove an electron from an atom. A large jump in ionization energy occurs when removing an electron from a noble - gas - like electron configuration. For \(Al\) (\(1s^{2}2s^{2}2p^{6}3s^{2}3p^{1}\)), after removing 3 electrons (\(Al^{3+}:1s^{2}2s^{2}2p^{6}\)), removing the fourth electron is very difficult as it is from a noble - gas - like shell. The third ionization energy is \(2745\ kJ/mol\) and the fourth is \(11577\ kJ/mol\), a large jump.

Step2: Analyze photoelectron spectrum for question 5

In a photoelectron spectrum, the number of peaks corresponds to the number of different electron environments. The areas under the peaks correspond to the number of electrons in each environment. For \(1s^{2}2s^{2}2p^{6}3s^{2}3p^{3}\), we have different electron shells and sub - shells. The \(1s\) (2 electrons), \(2s\) (2 electrons), \(2p\) (6 electrons), \(3s\) (2 electrons), \(3p\) (3 electrons) which can match the relative number of electrons in the spectrum.

Step3: Analyze atomic radius for question 6

Atomic radius decreases across a period from left to right (due to increasing nuclear charge). \(Mg\), \(P\), and \(Cl\) are in the same period (\(Mg\) is in group 2, \(P\) in group 15, \(Cl\) in group 17). As we move from \(Mg\) to \(P\) to \(Cl\), the nuclear charge increases, pulling the electrons closer. So \(Cl\) has the smallest radius among them, followed by \(P\), and \(Mg\) has the largest.

Answer:

1. B. \(Al\), because of the large difference between the third and fourth ionization energies
2. B. \(1s^{2}2s^{2}2p^{6}3s^{2}3p^{3}\)
3. A. \(Cl < P < Mg\)