Midterm 2

General Chemistry 1

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Question 1

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 r_Mg2+ = 72 pm r_O2- = 140 pm

Solution

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^- → O^2- (g) Energy = EA₁ + EA₂ = 1.07 aJ

Third step: Mg²⁺ (g) + O^2- (g) → MgO (g) Energy = E_coulomb = 231 x

\frac{(+ 2) (- 2)}{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.

Question 2

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.

Solution

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

Question 3

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

Data:

m_electron = 9.109 x 10^-25 mg

h = 6.626 x 10^-34 kg.m².s^-1

Solution

λ = $\frac{h}{mv}$

v = $\frac{h}{λm}$

v = $\frac{6.626 \times 10^{- 34}}{(271.2 \times 10^{- 12}) (9.109 \times 10^{- 31})}$ = 2.682 x 10⁶ m.s^-1

Question 4

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 m_electron = 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.

Solution

1) c = λν ⇒ ν₀ = $\frac{c}{λ_{0}}$ ⇒ ν₀ = $\frac{3.00 \times 10^{8}}{253 \times 10^{- 9}}$ = 1.19 x 10¹⁵ s^-1

2) ν= $\frac{c}{λ}$ ⇒ ν = $\frac{3.00 \times 10^{8}}{260 \times 10^{- 9}}$ = 1.15 x 10¹⁵ s^-1

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

Question 5

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

Solution

1) Mo: Z = 42 ⇒ 1s²2s²2p⁶3s²3p⁶4s²3d¹⁰4p⁶5s¹4d⁵

2) outermost shell of Mo ⇒ n = 5 ⇒ 5s

5s: n = 5, l = 0, m_l = 0, m_s = + or – ½

3) +4 ⇒ Mo⁴⁺ ⇒ 1s²2s²2p⁶3s²3p⁶4s²3d¹⁰4p⁶4d²

+6 ⇒ Mo⁶⁺ ⇒ 1s²2s²2p⁶3s²3p⁶4s²3d¹⁰4p⁶

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