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
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.
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