Hybridisation - Free MCQ Practice Test with solutions, NEET Chemistry

Correct answer: A - 1 and diamagnetic.

Each boron atom has 3 valence electrons, so the diatomic species has 6 valence electrons.

Valence molecular orbitals are filled in the order: σ2s, σ2s*, then the π2p set receives the remaining electrons. The occupation (counting valence electrons) is: σ2s (2), σ2s* (2), π2p (2).

Because the question explicitly assumes that Hund's rule is violated, the two electrons in the degenerate π2p orbitals pair in the same π orbital rather than occupying separate π orbitals with parallel spins. This gives no unpaired electrons, so the species is diamagnetic.

Bond order = (number of bonding electrons - number of antibonding electrons)/2.

Bonding electrons = σ2s (2) + π2p (2) = 4. Antibonding electrons = σ2s* (2) = 2.

Bond order = (4 - 2)/2 = 1. Therefore the molecule has bond order 1 and is diamagnetic, which matches option A.

NO₂ molecule has unpaired electrons which are responsible for its brown colour and paramagnetic behavior.

Molecule has zero dipole moment, thus it does not have lone pair. Hence, non-polar.

Thus, (M — X) bonds indicate sp²-hybridisation.

Vinyl acetylene is

bond are at equal position, numbering of C-chains based on IUPAC nomenclature is done from side.

H2​O>HF>NH3​

Strength of hydrogen bonding depends on the size and electronegativity of the atom. 

Smaller the size of the atom, greater is the electronegativity and hence stronger is the H−bonding. Thus, the order of strength of H-bonding is H...F>H...O>H...N. 

But each HF molecule is linked only to two other HF molecules while each H2O molecule is linked to four other H2​O molecules through H−bonding.

Hence, the decreasing order of boiling points is H2​O>HF>NH3​.

Molecules having the same number of hybrid orbitals, have same hybridisation and number of hybrid oebitals.

where,

V= number of valance electrons of central atom

X = number of monovalent atoms

C= charge on cation

A = charge on anion
 

After 1 :1 complex is formed in which NH3 is electron donor and BF3 is electron acceptor

Each atom (N, B) both are sp3 hybridised. Since, lone pairs of N-atoms are also involved in bonding, hence,both are sp3 hybridized and tetrahedral.

Three electron pairs are inclined at 120°.
Two electron-pairs are at 90° with the first three.

Structure is trigonal bipyramidal.

Since acetylene is made up of triple bond. So the hybridization of carbon in acetylene is sp.

(Ip-bp) repulsion make it pyramidal.
Thus, Statement I and II both are correct but Statement II is not the correct explanation of Statement I.

(d) B₂H₆ has following types of bonding

Bridging H-atoms and B-atoms are electron deficient. Each B-atom is however, joined to four H-atoms thus sp³-hybridised.
There is no lone pair or unpaired electron.
Thus, Statement I is incorrect but Statement II is correct.

Size of B is too small to form stable 4-membered ring system. (Also delocalised n-bonding is disturbed). Hence, dimerofBCI3is not formed.
Dimer AI₂CI₆ is stable due to stable ring

Thus, Statements I and II both are correct and Statement II is the correct explanation of Statement 

Due to resonance (C— O) bond length is larger than but smaller than (C— O).

(a) Both N in anion are sp²-hybridised.

(c) Decomposition products are different.

(C— H ) bond length < (C— H ) bond length
Nodal, plane of π-bond is located in molecular plane.

Option B is correct.

In PCl₅, phosphorus uses its five valence electrons to form five P-Cl bonds, so there are 0 lone pairs on P.

In PCl₄⁺, phosphorus is bonded to four chlorines and carries a positive charge; all valence electrons available for bonding are used in the four P-Cl bonds, so there are 0 lone pairs on P.

In PCl₆^-, phosphorus forms six P-Cl bonds (expanded octet for P permitted), and no lone pair remains on the central atom; hence there are 0 lone pairs on P.

Therefore the number of lone pairs on P in I, II and III are 0, 0, 0, which corresponds to option B.

Electronegativity of
sp³ < sp² < sp-hybridised C-atom
(Greater the s-character, greater the electronegativity).
(C— H) electron pair is taken out by most (EN) atom most easily, making it is acidic.