Draw the molecular orbital diagram of N2 ^+2. Calculate it's bond orbital (BO), state it's magnetic effect, suggest it's Lewis structure, write down it's electronic configuration and assign four quantum numbers.
To draw the molecular orbital diagram of N2^+2, we need to consider that N2^+2 has a total of 13 valence electrons (5 from each nitrogen atom plus 3 from the positive charge).
The molecular orbital diagram for N2^+2 would look as follows:
1. σ2p bonding
2. σ*2p antibonding
3. π2p bonding
4. π*2p antibonding
5. σ2s bonding
6. σ*2s antibonding
The bond order (BO) can be calculated as
BO = (number of bonding electrons - number of antibonding electrons) / 2
BO = (8 - 5) / 2 = 1.5
N2^+2 has a bond order of 1.5.
N2^+2 is paramagnetic, as it has unpaired electrons.
The Lewis structure for N2^+2 would be N≡N^+.
The electronic configuration for N2^+2 is (σ2s)^2(σ*2s)^2 (π2p)^4(π*2p)^3.
Assigning four quantum numbers for one of the electrons in N2^+2:
1. Principal quantum number (n) = 2
2. Azimuthal quantum number (l) = 1
3. Magnetic quantum number (m_l) = 0
4. Spin quantum number (m_s) = +1/2