Covalent & Coordinate Bonding - Free MCQ Practice Test with solutions,

To complete octet, N shares 2 electrons and oxygen shares 3 electrons.

Thus, (a) is correct.

The maximum covalency is total no. of  unpaired electron in ground state or in excited state which can be used for bond formation.

N(7) = 1s²2s²2p³
Nitrogen does not have (2d) orbitals. Thus, (more than 8) electrons cannot be accommodated in the second orbit.

Based on structure:


 bond has one electron; bond order y = 0.5.
 bond has two electrons; bond order x = 1.0. 

Nonbonding electron pairs means lonepair of electrons.​

• SF₄​ molecule has one non-bonding electron pair on the central atom Sulfur.
• ICl₃​ molecule has two non-bonding electron pairs on the central atom Iodine. 
• SO₂​ molecule has one non-bonding electron pair on the central atom Sulfur.

So the answer is D.

)bond enthalpy > )bond enthalpy hence, is stable.

Bond enthalpy of six (P— P) bond = 6 x 215 = 1290 k j mol^-1

Bond enthalpy of six (N — N) bond = 6 x 160
= 960 kJ mol^-1
Thus, P₄ is more stable than N₄.

In structure I, octet of oxygen is complete but octet of carbon is incomplete. Hence, carbon is an acceptor of an electron pair from oxygen. Thus, covalency of carbon is three in CO.

(a) Octet of Cl expanded to (9).
(b) Octet Xe expanded to (10).
(c) Octet of P expanded to (10).
 

Due to unpaired electron paramagnetic, N-atom in NO₂ is electron deficie nt thus, to complete octet, dimer is formed.

In N₂O₄ formation, each N-atom gets charge. Dipole-dipole repulsion increases (N— N) bond length which is larger than (N— N)covalent bond.

Thus, (a), (b), (c) and (d) are correct. 

Incorrect (b)

two sp-hybrid orbitals are along x-axis.

Thus Option C is correct

Option B is correct.

AlCl3 is a Lewis acid because the central Al atom has an incomplete octet and can accept an electron pair; therefore it behaves as an electrophile.

SO3 shows resonance structures in which the sulphur atom carries partial positive character; this makes sulphur electron-deficient and able to accept electron density, so SO3 acts as an electrophile in many reactions (for example, electrophilic aromatic sulfonation).

NO2+ (the nitronium ion) is a classical strong electrophile; nitrogen in NO2+ is electron-deficient and readily accepts electron pairs from nucleophiles.

All three species in option B are therefore electron-seeking and qualify as electrophiles; hence the correct choice is B.

Tetracyanomethane, where σ bond =8, ratio = 1, π bond = 8.

 Carbon dioxide, O=C=O

σ = 2, π = 2 

Ratio = 1.

Greater the electronegativity of carbon atom
(sp³ < sp² < sp)
Greater the attraction for (C— H) bonding pair hence, shorter the (C— H)bond.

Both Ionic and Covalent bond arise from the tendency of atoms to attain stable configuration of electrons.

α and π bonds exist.
Oxidation number of N = + 3 it can be increased to +5 (maximum) and decreased to -3 (minimum) HNO₂ is also a ’ligand.

HNO₂ is also the complexing agent 

When H₂O₂ is added to an acidified solution of a dichromate, Cr₂O₇^2-, aO deep blue coloured complex, chromic peroxide, CrO₅ [ or CrO(O₂)₂] is formed.
Cr₂O₇^2- +2H⁺+4H₂O₂ ---> 2CrO(O₂)₂ +5H₂O
This deep blue coloured complex.

Oxidation state of Cr is +6 due to the presence of two peroxide linkages which can be calculated as Cr peroxide normal
x+(-1)4+(-2) =0
x-6 =0
x=+6

There are 4 atoms that are directly linked to each Cr atom.