Test: Formal Charge & Resonance (May 27) - JEE MCQ

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.

Due to delocalisation of π-electrons, (C—Cl) bond is stable and it does not show S_N reactions; Cl directly attached (C=C) bond, i.e. vinyl group.

Formal charge (F)

where, v = valence electrons = 6
s = shared electrons = 4
u = unshared electron = 4

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

S_N1 means unimolecular (1) nucleophilic (N) substitution (S). Rate is independent of molar concentration of H2_O but is dependend on molar concentration of

(¹⁴C at changed position due to resonance).

(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.

Strength of H-bonding depends on the electronegativity of the atom which follows the order: F > O > N .
Strength of H-bond is in the order:
H……. F > H…….. O > H…….. N
But each H₂O molecule is linked to 4 other H₂O molecules through H-bonds whereas each HF molecule is linked only to two other HF molecules.
Hence, correct decreasing order of the boiling points is H_zO > HF > NH₃.

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.