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2 . Determine the ground state electronic configuration of Nz bond order, and all possible terms for molecular electronic states....

Question

2 . Determine the ground state electronic configuration of Nz bond order, and all possible terms for molecular electronic states.

2 . Determine the ground state electronic configuration of Nz bond order, and all possible terms for molecular electronic states.



Answers

Describe the electronic structure of each of the following, using molecular orbital theory. Calculate the bond order of each and decide whether it should be stable. For each, state whether the substance is diamagnetic or paramagnetic. a. $\mathrm{Be}_{2}$ b. $\mathrm{He}_{2}^{+}$ c. $\mathrm{N}_{2}^{-}$

Here we are taking a look at the molecule be too fast. And so the total number of electrons is 10. And so we can calculate the bond order where we have two electrons in the sigma, Two s two electrons in the sigma start to us and two electrons in the pie to pee. We have one heart for bonding electrons. Takeaway to anti bonding electrons gives us a bond order of one. So this molecule is stable and so we have two electrons in the bonding to pay that unpaid. So it is power magnetic. So the next example what we've got is B two plus and so we have nine electrons in total. So a Bondo is now one of three take away too, which is equal to one half. So again this molecule is stable, slightly less stable and again it is power magnetic. And so finally what we've got is both to minors. So our bonding order is equal to 1/2. Multiplied by eight bonding electrons take away five anti bonding electrons gives us 3/2. So again this is even less stable. And the first example. And so what we have is pro magnetic. Again,

In this problem We want to write Electronique configuration for ground Start off carbon s we hill A Tommy number off carbon is equal to six So the short dead but a number of other towns in carbon Adam is equal to six. So in oneness Exceptional dear r Two electrons so in toe exceptional. There are tutor turns. So do the trans are left. So these address will be in Don't be social So they're turning confusion off carbon ese one is too dress toe on a group e six.

The atomic number of nitrogen. So that is seven. So the total number of electrons in end two is 14 electrons. Where the configuration in the molecular orbital is sigma to us to sigma. Start to us too pi two P four sigma to P two where the bonding order is eight. Take away two divided by two to get three as the bonding order. So we have a triple bond. So we have n triple bond nitrogen. Next we can look at the first excited electronic configuration event too. Got into with a star. So that is sigma two S two sigma star to us too, pi two P four sigma to P one high star to P one. The bond order here 70 quay three divided by two. We got a bond order of two. So therefore the transition decreases the bond order from three. In the fast calculations to in the second calculation where the ground state we can see is die magnetic. All of the electrons are paired. However, in the excited state we have para magnetic species because we have unpaid electrons.

So the atomic number for nitrogen is seven. And so the total number of electrons in end to We seven multiplied by two. So that gives us 14 electrons. So the molecular orbital configuration we feel sigma two, S two sigma star to us too. Hi to people Sigma to P two, where the bond order is equal to one half of the bonding electrons. Eight subtract to the anti bonding electrons. What we get is a bond order of three. And so the fast excited state of the electron configuration Is as follows, where we have σ two past two sigma star to us too. Hi to P4, sigma two people one and high star to P one where our bond order is now 1/2, seven. Subtract three because we've put an electron into an anti bonding over till we get a bond order too. And so therefore the transition decreases the bond order from 3-2. And so the ground state is dire magnetic and the excited state is power magnet.


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