Test: Covalent Bonding - JEE MCQ

(a) Hs maximum overlap - correct.

(c) Based on valence bond approach,-no prediction of change of structure due to (Ip-Ip), (Ip-bp) repulsions- correct.

HC1, HI and H₂S do not from H-bonds. Only H₂O forms hydrogen bonds. One H₂O molecule forms four H-bonding.

According to M.O theory,

CN⁻,CO and NO⁺ are isoelectronic (14e⁻) and bond order is 3.

Hence (a) is the correct answer.

H is smallest in size thus, (C— H)bond length is least.


Hence, bonding pair in (C— O) is nearer to nucleus decreasing bond length as compared to (C— C).
Thus, I < IV < II < III.

(a) Boiling of H₂O Intermolecular H-bonds are cleaved.

Thus, in (NH₃,BF₃), both have independent existence and thus, it can be separated.  

The type(s) of hybridization of the carbon atoms in Allene (C₃​H₄​) is sp and sp².

When bond is formed between identical atoms, it is a perfectly covalent bond. Electrons will be shared equally among them.

There are three lone pairs on Y and four σ - bonds between X and Y.
Thus, octets of X and Y are complete.
X has four electrons in valence shell.
Y has seven electrons in valence shell.
Thus, X = Si (group 14), y = Cl (group 17)

Hyperconjugation takes place in allylic compounds in which alkyl group (having at least one H) is attached to bond .

σ-bond of (— CH₃) and π -bond of (ene) are involved, π-bond is formed due to overlapping of p-orbitals. Thus, σ-orbital and p-orbital are involved.

In metallic bonds, the valence shell electrons are delocalised and shared between many atoms. These delocalised electrons allow the metal atoms to slide past one another without being subjected to strong repulsive forces. The malleability and ductility of metals is due to this sliding capacity of the delocalised electrons.

H₂O will have highest dipole moment due to maximum difference in electronegativity of H and O atoms.

(b) PCI₅ triangular bipyramidal, symmetrical, no lone pair.

Both B and C are non-metals and therefore, bond formed between them will be covalent.

(i) (d-d) σ-bonding (ii) (p-dπ bonding)
(iii) (p-dπ anti-bonding) (iv) (ddσ anti-bonding)

It contains four completely filled antibonding molecular orbitals. Since, all the electrons are paired, is diamagnetic. Peroxide ion is isoelectronic with argon, not with neon. 

It is a base because it presents three OH group and bond order of PO is equal to total no. of sigma and pie bond/total no of resonance structure due to which = 5/4 = 1.25 ~1.3