IChO 1975 · Theory problem 8

Molecular properties table and isoelectronic species

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a) Mark by the "+" in a corresponding window those molecules for which the assertions written on the left side of the Table, are true.

Assertion Molecule
\(\ce{C2H4}\) \(\ce{N2H4}\) \(\ce{H2O2}\) \(\ce{H2F2}\)
There is a covalent bond between two equal atoms
The molecule contains a double bond
The molecule is planar
The molecule is polar
There is also a hydrogen bond in the molecule
It has basic properties in relation to water

b) The following assertion and Table are incomplete. Fill in the dotted places by a missing word and the missing formulas, respectively.

Assertion: The electronic structures of the molecules (ions) being placed in the Table one under the other are . . . . . . . . . . . . . . . . . . . . . . . . .

\(\ce{CH4}\) \(\ce{C2H6}\) \(\ce{CO3^2-}\) . . . . . . \(\ce{C2O4^2-}\) . . . . . .
\(\ce{NH4+}\) \(\ce{N2H6^2+}\) . . . . . . \(\ce{NO2+}\) . . . . . . \(\ce{N2}\)

Official English text of the problem.

Answers

  • a) Place the “+” signs as follows. Empty cells indicate false assertions.
Assertion \(\ce{C2H4}\) \(\ce{N2H4}\) \(\ce{H2O2}\) \(\ce{H2F2}\)
There is a covalent bond between two equal atoms + + +
The molecule contains a double bond +
The molecule is planar + +
The molecule is polar + + +
There is also a hydrogen bond in the molecule +
It has basic properties in relation to water +
  • b) Missing word: identical, in the intended sense that the corresponding species are isoelectronic, with analogous Lewis-electron structures.

One consistent completion is:

\(\ce{CH4}\) \(\ce{C2H6}\) \(\ce{CO3^2-}\) \(\ce{CO2}\) \(\ce{C2O4^2-}\) \(\ce{C2^2-}\)
\(\ce{NH4+}\) \(\ce{N2H6^2+}\) \(\ce{NO3-}\) \(\ce{NO2+}\) \(\ce{N2O4}\) \(\ce{N2}\)

Carbon monoxide, \(\ce{CO}\), is also a valid answer for the final blank.

Solution

a)

The structures represented by the four formulas are:

Species Name SMILES Relevant structure
\(\ce{C2H4}\) Ethene (ethylene) C=C Two trigonal-planar carbon atoms joined by a double bond; all six atoms are coplanar.
\(\ce{N2H4}\) Hydrazine NN A single nitrogen–nitrogen bond; each nitrogen is pyramidal. The equilibrium conformation is nonplanar and gauche.
\(\ce{H2O2}\) Hydrogen peroxide OO A single oxygen–oxygen bond and a nonplanar, skew equilibrium conformation.
\(\ce{H2F2}\) Hydrogen fluoride dimer [H]F.[H]F Two hydrogen fluoride units joined by a hydrogen bond in a planar equilibrium arrangement.

Ordinary SMILES does not specify these conformations or encode the hydrogen bond between the two components of the hydrogen fluoride dimer.

  1. Covalent bond between two equal atoms.
    Ethene contains a carbon–carbon bond, hydrazine a nitrogen–nitrogen bond, and hydrogen peroxide an oxygen–oxygen bond. The hydrogen fluoride dimer contains covalent hydrogen–fluorine bonds and a hydrogen bond; it has no covalent fluorine–fluorine or hydrogen–hydrogen bond.

  2. Double bond.
    Only ethene contains a double bond: its carbon–carbon bond consists of one \(\sigma\) bond and one \(\pi\) bond.

  3. Planarity.
    Ethene is planar because the carbon atoms are trigonal planar and their parallel \(p\) orbitals form the \(\pi\) bond. The hydrogen fluoride dimer is also planar, although it need not be represented as a straight chain. Hydrazine and hydrogen peroxide have nonplanar equilibrium structures.

  4. Polarity.
    Ethene’s symmetry gives a zero permanent molecular dipole moment. The bond dipoles and electron distributions in hydrazine, hydrogen peroxide and the hydrogen fluoride dimer do not cancel, so these three species are polar.

  5. Hydrogen bonding within the listed species.
    Only the hydrogen fluoride dimer contains a hydrogen bond as part of the specified aggregate: a hydrogen covalently attached to one fluorine interacts with the other fluorine. Individual hydrazine and hydrogen peroxide molecules contain no intramolecular hydrogen bond. Their ability to form hydrogen bonds between separate molecules does not warrant a “+” in this row.

  6. Basic properties toward water.
    Hydrazine accepts a proton from water through a nitrogen lone pair:

\[ \ce{N2H4 + H2O <=> N2H5+ + OH-}. \]

Here water is O, the hydrazinium ion is N[NH3+], and the hydroxide ion is [OH-]. Formation of hydroxide accounts for the basic solution. Ethene does not appreciably act as a base toward water; hydrogen peroxide and hydrogen fluoride exhibit acidic behavior in water.

b)

For a species with signed charge \(q\), expressed in units of the elementary charge, the total electron count is

\[ N_{\mathrm e}=\sum_{\text{atoms}} Z-q. \]

Thus a positive charge removes electrons and a negative charge adds electrons.

Replacing a carbon atom by nitrogen increases the nuclear charge by one. Increasing the species’ positive charge by one at the same time preserves its electron count. Applying this consistently gives all six pairs in the completed table.

Upper species: name and SMILES Lower species: name and SMILES Total-electron check
Methane, \(\ce{CH4}\): C Ammonium ion, \(\ce{NH4+}\): [NH4+] \(6+4=7+4-1=10\)
Ethane, \(\ce{C2H6}\): CC Hydrazine-1,2-diium ion, \(\ce{N2H6^2+}\): [NH3+][NH3+] \(2(6)+6=2(7)+6-2=18\)
Carbonate ion, \(\ce{CO3^2-}\): [O-]C(=O)[O-] Nitrate ion, \(\ce{NO3-}\): [O-][N+](=O)[O-] \(6+3(8)+2=7+3(8)+1=32\)
Carbon dioxide, \(\ce{CO2}\): O=C=O Nitronium ion, \(\ce{NO2+}\): O=[N+]=O \(6+2(8)=7+2(8)-1=22\)
Oxalate ion, \(\ce{C2O4^2-}\): [O-]C(=O)C(=O)[O-] Dinitrogen tetroxide, \(\ce{N2O4}\): [O-][N+](=O)[N+](=O)[O-] \(2(6)+4(8)+2=2(7)+4(8)=46\)
Acetylide dianion (ethynediide), \(\ce{C2^2-}\): [C-]#[C-] Dinitrogen, \(\ce{N2}\): N#N \(2(6)+2=2(7)=14\)

The SMILES show representative Lewis structures; they do not encode resonance delocalization in the oxygen-containing groups.

The corresponding pairs have the same valence-electron counts and analogous bonding arrangements. In particular, carbon dioxide and the nitronium ion are linear, while the acetylide dianion and dinitrogen both have a triple bond and one lone pair on each atom.

Carbon monoxide, \(\ce{CO}\), SMILES [C-]#[O+], also has a triple bond, one lone pair on each atom, and

\[ N_{\mathrm e}=6+8=14. \]

It therefore also qualifies for the final blank.

“Identical” here describes the intended electron-count and Lewis-structure correspondence; it does not imply identical orbital energies, bond strengths or other physical properties.

The answers were checked against the official answers published by the organisers.