(a)What is the bond order of the diatomic molecule BN?
(b) Is BN paramagnetic?
(c) Rank the following compounds in order of increasing bond energy: B2, N2, BN.
(d) Rank the following compounds in order of increasing bond length: B2, N2, BN.
Bond energy B2<BN<N2
Bond Length N2<BN<B2
not paramagnetic
(a) To determine the bond order of a diatomic molecule, we need to calculate the number of bonding electrons and the number of antibonding electrons.
For BN, the electron configuration is:
B: 1s2 2s2 2p1
N: 1s2 2s2 2p3
To form a bond, B donates one of its 2s electrons to the empty 2p orbital of N. This forms a sigma bond.
The molecular orbital diagram for BN is as follows:
B N
σ(2p) π*(2p) σ*(2p)
N has 5 valence electrons, and B has 3 valence electrons.
In the bonding molecular orbital (σ(2p)), there are 2 electrons, and in the antibonding molecular orbital (σ*(2p)), there are 0 electrons.
The bond order is calculated as:
Bond Order = (Number of bonding electrons - Number of antibonding electrons) / 2
Bond Order = (2 - 0) / 2 = 1
Therefore, the bond order of the diatomic molecule BN is 1.
(b) To determine if BN is paramagnetic, we need to consider the presence of unpaired electrons.
In BN, there is one unpaired electron in the π*(2p) antibonding orbital of nitrogen.
Since there is an unpaired electron, BN is paramagnetic.
(c) To rank the compounds in order of increasing bond energy (strongest bond to weakest bond), we need to compare their bond orders.
The bond order of B2 is 1.
The bond order of N2 is 3.
The bond order of BN is 1.
Ranking the compounds from increasing bond energy: B2 < BN < N2.
(d) To rank the compounds in order of increasing bond length (shortest bond to longest bond), we need to compare their bond orders.
The bond order of B2 is 1.
The bond order of N2 is 3.
The bond order of BN is 1.
Ranking the compounds from increasing bond length: N2 < BN < B2.
(a) To determine the bond order of the diatomic molecule BN, we need to know the number of bonding electrons and the number of anti-bonding electrons. To calculate this, we can use molecular orbital theory.
The molecular orbital diagram for BN can be constructed by combining the atomic orbitals of boron (B) and nitrogen (N).
Boron has electron configuration: 1s2 2s2 2p1
Nitrogen has electron configuration: 1s2 2s2 2p3
When these atomic orbitals combine, they form molecular orbitals. By filling up these molecular orbitals with electrons according to the Aufbau principle and Pauli's exclusion principle, we get the following molecular orbital diagram:
sigma2p*(antibonding)
sigma2p(antibonding)
pi2p*(antibonding)
pi2p(bonding)
sigma2p(bonding)
Counting the electrons in the bonding orbitals (sigma2p and pi2p), we have a total of 2 bonding electrons. Counting the electrons in the antibonding orbitals (sigma2p* and pi2p*), we have a total of 3 antibonding electrons.
The bond order is then given by the formula: (number of bonding electrons - number of antibonding electrons) / 2
In this case, the bond order of BN can be calculated as follows:
(2 - 3) / 2 = -0.5
Therefore, the bond order of the diatomic molecule BN is -0.5.
(b) To determine if BN is paramagnetic or not, we need to examine the molecular orbital diagram once again. If a molecule has any unpaired electrons, it is considered paramagnetic.
Looking at the molecular orbital diagram for BN, we can see that there are no unpaired electrons because all bonding and antibonding orbitals are filled in pairs. Therefore, BN is diamagnetic, not paramagnetic.
(c) To rank the compounds in order of increasing bond energy (or bond strength), we need to consider their bond lengths. Generally, shorter bond lengths correspond to stronger bonds and higher bond energies.
Experimentally, the bond lengths for B2, N2, and BN are as follows:
B2 - 1.54 angstroms
N2 - 1.09 angstroms
BN - 1.45 angstroms
Based on the bond lengths, we can rank the compounds in order of increasing bond energy as follows:
N2 < BN < B2
(d) To rank the compounds in order of increasing bond length, we can use the bond lengths provided in the previous question.
Based on the given bond lengths:
B2 - 1.54 angstroms
N2 - 1.09 angstroms
BN - 1.45 angstroms
We can determine the order of increasing bond length as follows:
N2 < BN < B2