# Midterm 2

General Chemistry 1

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Consider the formation of gaseous magnesium oxide from elemental gaseous magnesium and oxygen.

1) Write the chemical equation of this reaction.

2) Determine the overall change in energy for the formation of gaseous magnesium oxide.

3) Does this formation release or absorb energy? Explain.

4) Would it require more energy to melt CO (s) or MgO (s)? Explain.

Data:

I1 (Mg) = 1.23 aJ

I2 (Mg) = 2.41 aJ

EA1 (O) = -0.234 aJ

EA2 (O) = 1.30 aJ

k = 231 aJ.pm

rMg2+ = 72 pm

rO2- = 140 pm

1) Mg (g) + O (g) → MgO (g)

2)

First step:

Mg (g) → Mg+ (g) + 1 e-

Mg+ (g) → Mg2+ (g) + 1 e-

Energy = I1 + I2 = 1.23 + 2.41 = 3.64 aJ

Second step:

O (g) + 1 e- → O- (g)

O- (g) + 1 e- → O2- (g)

Energy = EA1 + EA2 = 1.07 aJ

Third step:

Mg2+ (g) + O2- (g) → MgO (g)

Energy = Ecoulomb = 231 x $\frac{\left(+2\right)\left(-2\right)}{212}$= -4.36 aJ

Energy of reaction: E = 3.64 + 1.07 – 4.36 = 0.35 aJ

3) E > 0 ⇒ absorb energy

4) CO(s) ⇒ covalent bonding

MgO (s) ⇒ ionic bonding

ionic compound has higher melting point than covalent compound.

1) Draw the complete photoelectron spectra for beryllium.

2) What is the difference between the photoelectron spectra for beryllium and that of Be2+?

3) Is the radius of a neutral atom of helium smaller or greater than the radius of Be2+ ion? Explain.

1) Be: Z = 4 ⇒ 1s22s2 ⇒ 2 subshells: 1s and 2s ⇒ 2 peaks on the complete photoelectron spectra.

1s = core orbital; 2s = valence orbital:

The core orbitals have higher binding energies than valence orbitals ⇒ 1s is on the left, 2s is on the right.

The two orbitals are 2 electrons ⇒ same height

2) Be2+: 1s2 ⇒ only one peak on the complete photoelectron spectra.

No more peak around 4 MJ.mol-1

3) Be: Z = 4 ⇒ 4 protons

He: Z = 2 ⇒ 2 protons

Be has higher nuclear charge ⇒ electrons in Be2+ are closer to the nucleus than those of He

⇒ radius of Be2+ < radius of He

What is the velocity of an electron which has a De Broglie wavelength of 271.2 pm?

Data:

melectron = 9.109 x 10-25 mg

h = 6.626 x 10-34 kg.m2.s-1

λ = $\frac{\mathrm{h}}{\mathrm{mv}}$

v = $\frac{\mathrm{h}}{\mathrm{\lambda m}}$

v =  = 2.682 x 106 m.s-1

1) The longest wavelength of light that causes electrons to be ejected from the surface of a copper plate is 253 nm. What is the threshold frequency of the copper plate knowing that melectron = 9.109 x 10-25 mg?

2) If light with a wavelength of 260 nm shines on the copper plate, will electrons be ejected from the copper atoms? Explain.

1) c = λν ⇒ ν0 = $\frac{\mathrm{c}}{{\mathrm{\lambda }}_{0}}$ ⇒ ν0 =  = 1.19 x 1015 s-1

2) ν = $\frac{\mathrm{c}}{\mathrm{\lambda }}$ ⇒ ν =  = 1.15 x 1015 s-1

ν < ν0 ⇒ electrons won’t be ejected from the copper atoms.

Molybdenum (Z = 42) has several oxidation states.

1) Write the electronic configuration of Mo knowing it has half-filled valence orbitals.

2) Give a set of quantum numbers for an electron on the outermost shell of Mo.

3) The most stable oxidation states of Mo are +4 and +6. Give the corresponding ions and their electronic configuration

1) Mo: Z = 42 ⇒ 1s22s22p63s23p64s23d104p65s14d5

2) outermost shell of Mo ⇒ n = 5 ⇒ 5s

5s: n = 5, l = 0, ml = 0, ms = + or – ½

3) +4 ⇒ Mo4+ ⇒ 1s22s22p63s23p64s23d104p64d2

+6 ⇒ Mo6+ ⇒ 1s22s22p63s23p64s23d104p6