Chemical Bonding and Molecular Structure Test -12

(i) Carbon and nitrogen molecules are diamagnetic as they contain all paired electrons.

(ii) Superoxide ion and oxygen molecule are paramagnetic as they contain 1 and 2 unpaired electrons respectively.

The molecule \(Li_2\) is a stable molecule in the gas phase, with a bond order of one.

Bond Order = 2(bonding electrons) - 0(anti-bonding \(e^-\))2

= 1

The paramagnetic nature of oxygen molecules is best explained on the basis of molecular orbital theory.

Molecular orbital theory was developed, in the years after valence bond theory had been established (1927), primarily through the efforts of Friedrich Hund, Robert Mulliken, John C. Slater, and John Lennard-Jones.

MO theory was originally called the Hund-Mullikan theory.

2 outer most electrons because sulphur has \(1s^2 , 2s^2 , 2px^2 , 2py^2 , 2pz^2 , \) \(3s^2 , 3px^2 , 3py^1 , 3pz^1 . \)

So, it has 1 unpair electron in 3py and 1 unpair electron in 3pz.

Due to electrostatic force of attraction, the melting points and boiling points of ionic compounds are high.

(i) The bond order of CO is three.

(ii) Bond order is the number of covalent bonds between any pair of atoms in a molecule.

Intermolecular hydrogen bonding is strongest in methylamine [\(CH_3-NH_2\)] because N has more electron affinity.

One lone pair of electrons is present on Xe in \(XeOF_4\). Xe has eight valence electrons out of which four are shared with 4 electrons of 4F atoms.

2 electrons of Xe are shared with 2 electrons of O atom.

Thus, only one lone pair of electrons is remaining on Xe atom.

We know that ionic characters

= \(16[E_A - E_B]+3.5 \times [E_A - E_B]^2\)

or ionic characters = 72.24%

A co-ordinate bond is a union of one electrovalent and one covalent bond, the volatility of these compounds lies between that of covalent and ionic compounds. Thus their melting and boiling points are higher than covalent compounds and lower than ionic compounds.

Electrovalent compounds have high melting point and high boiling point.

Electronegativities are used to determine the polarity of covalent bonds. The polarity of a covalent bond can be judged by determining the difference between the Electronegativities of the two atoms involved in the covalent bond.

(i) Number of electrons in the valence orbit of nitrogen in an ammonia molecule is 5.

(ii) It has 5 electrons in its valence shell, hence in ammonia molecule, it completes its octet by sharing of three electrons with three H atom,

Hence, it has 8 electrons in its valence shell in ammonia.

\(\because\) NaH is an ionic hydride (\(Na^+H^-\))

\(\therefore\) Covalent bond is absent.

(i) Van der Waals, the weakest is London dispersion. Then if you have a molecule with a more electronegative atom, then you start having a dipole, where you have one side where molecule becomes polar and you have the interaction between the positive and the negative side of the pole.

(ii) It gets a dipole-dipole interaction.

The valency of noble gases is zero because their octet is complete. Octet means that outermost orbit has complete eight electrons to stay stable.

The noble gases are already stable, so their tendency to gain or lose electrons is zero.

Hence, their valency is also zero.

Configuration of \(O_2\) molecule is

[\(\sigma(1s)^2 \sigma^*(1s)^2\sigma(2s)^2 \sigma *(2s)^2 \pi(2p_x)^2\pi(2p_y)^2\) \(\sigma(2p_z)^2 \pi ^*(2p_x)^1\pi^*(2p_y)^1]\)

Bond order in \(N_2, O^{2+}_2\) is 3 (calculate by energy level diagram).