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The negativelly charged molecule would involve addition of electrons to anti-bonding orbitals so reducing bond order. The poitivelly charged molecule would involve the loss of electrons in bonding orbials again giving a reduction of bond order.

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Q: Why is the bond order for Nitrogen ion lower than the bond order for nitrogen atom?
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Related questions

Is nitrogen an neutral atom or an ion?

Neutral atom. An atom of nitrogen will also be an isotope of nitrogen.


What particle in a nitrogen atom must remain the same in order for it to stay a nitrogen atom?

The atomic number for nitrogen is 7, meaning that it has 7 protons. In order for it to remain as nitrogen, the number of protons in it must always be 7. If, for example, the number of protons were to increase to 8, it would no longer be nitrogen; it would then be oxygen.


How is a nitrogen atom different from a nitrogen molecule?

A single nitrogen atom can be chemically reactive while a nitrogen molecule is relatively stable and tends to be faf more inert. A nitrogen molecule also has twice the mass of a nitrogen atom.


How many electrons in an atom of nitrogen?

A nitrogen atom has seven electrons.


Which atom is larger nitrogen or carbon?

NITROGEN


What is the increase order of oxidation number of nitrogen atom?

it is three


How big is a nitrogen atom?

The covalent radius of nitrogen atom is 71 pm.


How many electrons does a nitrogen atom need to gain in order to attrain a noble-gas?

Nitrogen has to gain three electrons


Why is nitrogen an atom?

Nitrogen is an element, therefore it is an atom because elements are all atoms.


How do you calculate total nitrogen?

nitrogen is only one atom


What carbon atom forms a covalent bond with nitrogen?

Any carbon atom can form a covalent bond with nitrogen. In hydrogen cyanide, HCN, the carbon atom forms a triple covalent bond with the nitrogen atom. In amino acids, the carbon atom forms a single bond with a nitrogen atom.


Is NO2 made up of 2 nitrogen?

The molecule of nitrogen dioxide has 1 nitrogen atom.