1. which of the following atoms would you expect to release less energy than carbon when an electron is added to its valence?
oxygen (o)
fluorine (f)
chlorine (cl)
neon (ne)
2. suppose a laboratory wants to identify an unknown pure substance. the valence electrons of the substance's atoms feel an effective nuclear charge of +4. if the unknown substance has a higher electronegativity value than silicon (si), what other characteristics would you expect the substance to exhibit?
it would have larger atomic radii than si and higher ionization energies than si.
it would have larger atomic radii than si and lower ionization energies than si.
it would have smaller atomic radii than si and higher ionization energies than si.
it would have smaller atomic radii than si and lower ionization energies than si.
3. the trend for ionization energy is a general increase from left to right across a period. however, nitrogen (n) is found to have a higher first ionization energy value than oxygen (o). explain this exception to the general trend in terms of electron arrangements and attraction/repulsion.
4. a hydrate of na2co3 has a mass of 4.31 g before heating. after heating, the mass of the anhydrous compound is found to be 3.22 g. determine the formula of the hydrate and then write out the name of the hydrate. show all your work for the calculations for full credit.

1) Answer is: neon (Ne).

Noble gases are in group 18: helium (He), neon (Ne), argon (Ar), krypton (Kr), xenon (Xe) and radon (Rn). They have very low chemical reactivity.

Noble gases have very stable electron configuration and does not need to gain electrons, only when they gain energy.

Fluorine has the greatest affinity for electrons and tendency to be reduced, than oxygen and chlorine.

2) Answer is: It would have smaller atomic radii than Si and higher ionization energies than Si.

The atomic radius varies with increasing atomic number, but usually increases because of increasing of number of electrons. Because unknown metal has hifher electronegativity, it means that electrons are closer to nucleus, so it has amaller radius.

The ionization energy (Ei) is the minimum amount of energy required to remove the valence electron, when element lose electrons, oxidation number of element grows (oxidation process). Because valence electrons in unknown elements are closer to nucleus, the ionization energy is higher.

3) Answer is: nitrogen has more stable electron configuration (half fill 2p orbital) than oxygen atom, so it needs more energy to remove the valence electron.

Atomic number of nitrogen is 7, it has 7 protons and 7 electrons.

Electron configuration of nitrogen atom: ₇N 1s² 2s² 2p³.

Oxygen atom has atomic number 8, it means it has eight protons and eight electrons, so atom has neutral charge. Oxygen is a nonmetal.

Electron configuration of oxygen atom: ₈O 1s² 2s² 2p⁴.

Oxygen atom has six valence electrons, so it is in 16 group of Periodic table of elements.

4) Answer is: formula is Na₂CO₃ × 2H₂O.

m(Na₂CO₃ × nH₂O) = 4.31 g; mass of hydrate of sodium carbonate.

m(Na₂CO₃) = 3.22 g; mass of anhydrous sodium carbonate.

n(Na₂CO₃) = m(Na₂CO₃) ÷ M(Na₂CO₃).

n(Na₂CO₃) = 3.22 g ÷ 106 g/mol.

n(Na₂CO₃) = 0.0303 mol; amount of sodium carbonate.

m(H₂O) = 4.31 g - 3.22 g.

m(H₂O) = 1.09 g; mass of water.

n(H₂O) = 1.09 g ÷ 18 g/mol.

n(H₂O) = 0.0606 mol; amount of water.

n(Na₂CO₃) : n(H₂O) = 0.0303 mol : 0.0606 mol.

n(Na₂CO₃) : n(H₂O) = 1 : 2.

since the atomic number is the number of protons in the nucleus, as we move to the left on the periodic table, the electrostatic pull between the nucleus and the valence electron shell decreases. as we move down a group (vertical column), the valence shell energy level gets further and further from the nucleus.