Test: Formal Charge & Resonance - JEE MCQ

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

In structure (a), octet of N-atom is not complete. In all the other structures, octets of N and all the O-atoms are complete.

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 

BH₄– has 4 bond pairs and 0 lone pairs, so it is tetrahedral. CO₃(2^-) has 3 bond pairs and 0 lone pairs, so it is trigonal planar. NH₂- has 4 bond pairs and 2 lone pairs, so it is bent or angular. H₃O^+​ has 4 bond pairs and 1 lone pair, so it is a trigonal pyramid.

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.

Greater the resonating structures, greater the stability.
I : (Y)with negative charge has greater delocalisation of n-electrons than HNO₃(X).

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.

Due to complete-delocalisation of π-electrons, benzene exists as resonance structures.

Thus, (C=C) bonds and (C —C) bonds are of equal length and stability.
All (C— H) bonds are equivalent.
C-atom is also sp²-hybridised.

double bond single bond character character

double bond single bond character character

double bond single bond character character

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

Thus total = 6π-electrons.

Addition in conjugated diene can be 1 ,2 - or 1 , 4-typ

At position A,B it is called 1 , 4-addition.