Chemical Bonding and Atomic Structure MCQs for NEET — Chemistry Questions with Answers

As the number of lone pair of electrons on central atom increases the bond angle decreases.

${\mathrm{NH}}_4^{+}>{\mathrm{NH}}_3 >{\mathrm{NH}}_2^{-}$

NH2-, NH3, and NH4+ have H-N-H bond angles of 105, 107 and 109 repectively

Smaller is the size of ion, higher is the lattice energy.

For a T-shaped molecule like MX3, the central atom M has 3 bonding pairs and 2 non-bonding pairs of electrons. The presence of 2 non-bonding electron pairs gives the molecule a T-shaped geometry, as the bonding pairs occupy more space and push the non-bonding pairs towards the corners.

In PCl₃, the hybridised state of P is sp³ while in PCl₅, the hybridised state of P is sp³d

NH₃ - 107 ,

NF₃ ,102,

PH₃ - 93,

PF₃- 97

H₂O- 104.5

Use MOT

B₂ is most paramagnetic due to presence of two unpaired electrons.

In ${\mathrm{XeF}}_2$ & ${\mathrm{I}}_3^{-}$, the hibridisation state of central atom is sp³d but shape is linear.

XeOF4 is an oxyfluoride of xenon, with xenon in the +6 oxidation state. The molecule has a square pyramidal shape, with the xenon atom at the center, surrounded by 4 fluorine atoms in a square planar arrangement and 1 oxygen atom at the apex.

CO is not a homo-nuclear atom like C₂, N₂ or even O₃, O₃ (both these categories are different: with and without 2s-2p mixing). So there is a large discrepancy in the atomic energy levels of 2s, 2p e^- of C and O.As a result, 2p(pi)x, 2p(pi)y and 2p(sigma)z have lower energy than 2s(sigma)*. So the e^- lost is from 2s(sigma)*2 and not 2p(sigma)z. And hence the bond order increases from 3 to 3.5 and not decreases to 2.5 . (The typical school textbook formula doesn't work for species like CO,CO⁺ and even NO,NO⁺ in many situations)