Test: Molecular Orbital Theory (May 30) - NEET MCQ

(C—H) bond lengths are different from (B—H) and (N—H) bond lengths. Both are planar.

sp² hybridisation gives three sp² hybrid orbitals which are planar triangular forming an angle of 120° with each other.
The electronic configurations of three elements A, B and C are given below.
Answer the questions from 14 to 17 on the basis of these configurations.
A ls²2s²2p⁶
B ls²2s²2p⁶3s²3p³
C ls²2s²2p⁶3s²3p^s 

Structure is decided based on formal charge.

where, v = valence electrons
s = shared electrons
u = unshared electrons

Structure with at least one neutral atom is favoured.

Structure with least formal charge on each atom, makes greater contributions. Also, structures with at least one neutral atom is also favoured II and III identical.

Only structure a is able to show the valence electrons along with charge distributions. So, ith the Lewis structure of HNO₃ .

(C—Cl) bond has (C=Cl) double bond character, thus it is stable.

Benzyl carbocation is stable due to resonance, hence (C—Cl) bond is cleaved easily.

Plan Oxygen is more electronegative than oxygen, hence the structure with a negative formal charge on oxygen atom is probably lower in energy than the structure that has a negative formal charge on N-atom is more accurate.
Thus (I) is more accurate than (II).

(a) All (O—O) bonds in O₃ are equivalent, thus it is resonance hybrid of I and II- True.

(d) No unpaired electron in O₃ -diamagnetic.
Thus, (d) is false.

double bond single bond character character

Lone pair on N-atom is not delocalised in pyridine.

Thus total = 6π-electrons.