Chemical Bonding & Periodicity - Chemistry MCQ

Species in which ligands are electron deficient (one electron) form 3c–2e^– bonding.

Hence, s-character in lone pair 100-9.6 = 90.4%

Energy required for removing electron from orbitals follows the order ABMO < A.O. < B.M.O.

As the electronegativity of substituted v_P-O also increases

Energy required for removing electron from orbitals follows the order ABMO < A.O. <B.M.O.

Molecular orbital diagram of 

Hence both are gerade.

Lesser the electronegativity of surrounding atom more s-character in that bond. Hence, more bond angle.

Lesser the difference between size of central and surrounding atom lesser will be stereochemical activity of lone pair. Hence, perfect geometry.

Molecular orbital diagram of BeH₂.

More the S-character in C-O bond = more will be strength of bond. Hence, more will be v_c-o value. Hence, more ionization energy and lesser the dipole moment.

The structures of anions in two different salts, [Et₄N]₂[InCl₅] and [NH₃(CH₂)5NH₃][TiCl₅].

For [Et₄N]₂[InCl₅], the anion is [InCl₅]^2-. Indium (In) typically forms a trigonal bipyramidal structure with five chloride ions because it can accommodate five ligands around it.

For [NH₃(CH₂)5NH₃][TiCl₅], the anion is [TiCl₅]^2-. Titanium (Ti) in the +4 oxidation state usually forms a square pyramidal structure with five ligands due to its d^0 configuration.

Therefore, the correct answer is:

Option A: Trigonal bipyramidal and square pyramidal.

Structure of [AI(BH₄)₄]^-

At 298 terminal and bridging hydrogens are involved in dynamic process. Hence, one signal in ¹H NMR. ¹¹B NMR will be a quintet regarding 4 equivalent H.

Hence, ²π_g

Structure of X[Ph₃,SiH₂]

Hence, triplet

 perfect trigonal prismatic geometry

perfect trigonal prismatic geometry has D_3h point group.

2s + 1 = 1, M_l =2 
Hence ¹Δ_g