# Drag the “orbitals with electrons” box to the energy scale shown.

Question: Drag the “orbitals with electrons” box to the energy scale shown. Your goal in this exercise is to show nitrogen in an sp3-hybridized state which is capable of forming 3 bonds with three other atoms. Fill answer fields from left to right. Place orbitals with two electrons on the left (if any).

Solution: Concepts and reason

An orbital is the distribution of the probability density that is associated with the given electron centered at a given point. If the point is the nucleus of an atom, an atomic orbital is the resultant. If the point is the center of mass of multiple nuclei within a molecule, a molecular orbital is the resultant.

When an atom is required to bond with the other atoms to form a molecule, the valence atomic orbitals with almost similar energies ‘hybridize’ to form hybridized orbitals. These orbitals of identical shapes have identical associated energies.

The number of hybrid orbitals formed is equal to the number of atomic orbital hybridizing. Fundamentals

The hybridization of an atom in a molecule could be correlated with the number of bonds formed by it. This is tabulated as follows:

The sum of the principal quantum number and the azimuthal quantum number informs about the energy associated with a given orbital.

$(n+l)\left( {{\rm{n + l}}} \right)(n+l)$

Here

$n−{\rm{n}}\,{\rm{ - }}\,n$− Principal quantum number

$l−{\rm{l}}\,{\rm{ - }}l$− Azimuthal quantum number

The electronic configuration of nitrogen is $1s22s22p3{\rm{1}}{{\rm{s}}^{\rm{2}}}{\rm{2}}{{\rm{s}}^{\rm{2}}}{\rm{2}}{{\rm{p}}^{\rm{3}}}1s22s22p3$.

The $(n+l)\left( {{\rm{n + l}}} \right)(n+l)$ values associated with the atomic orbitals involved are as follows:

$1s−12s−22p−3\begin{array}{l}\\{\rm{1s}}\,{\rm{ - }}\,{\rm{1}}\\\\{\rm{2s}}\,{\rm{ - }}\,2\\\\{\rm{2p}}\,{\rm{ - }}\,{\rm{3}}\\\end{array}1s−12s−22p−3​$

The atomic orbitals with lower $(n+l)\left( {{\rm{n + l}}} \right)(n+l)​$ values have lower associated energies. Therefore, they are filled first. In nitrogen, the orbital with the lowest associated energy, $1s{\rm{1s}}1s​$ , is filled foremost, followed by $2s{\rm{2s}}2s​$ , and finally ​$2p{\rm{2p}}2p ​$.

The hybridized orbitals possess a blend of characteristics associated with their constituent atomic orbitals. Therefore, they could be placed at an intermediate position between the respective atomic orbitals (here, $2sand2p{\rm{2s}}\,{\rm{and}}\,{\rm{2p}}2sand2p$ ).

Hence, the arrangement is given below:

The arrangement is given below:

The $sp3{\rm{s}}{{\rm{p}}^{\rm{3}}}sp3$ orbitals possess $25%{\rm{25\% }}25%$ ‘s’ character and $75%{\rm{75\% }}75%$ ‘p’ character. Correspondingly, the stabilization or the effective energy of the hybridized orbitals is a little lesser than the un-hybridized $′2p′{\rm{'2p'}}′2p′$ orbitals, but quite greater than the un-hybridized $′2s′{\rm{'2s'}}′2s′$orbitals. The electronic population of the un-hybridized, valence orbitals is 666 . Correspondingly, the hybridized orbitals also have the same population.