Match the LIST-I with LIST-II:

LIST-I (Molecules/ion)	LIST-II (Hybridisation of central atom)
(A) PF₅$\mathrm{PF_5}$	(I) dsp²$dsp^2$
(B) SF₆$\mathrm{SF_6}$	(II) sp³d$sp^3d$
(C) Ni(CO)₄$\mathrm{Ni(CO)_4}$	(III) sp³d²$sp^3d^2$
(D) [PtCl₄]²⁻$[\mathrm{PtCl_4}]^{2-}$	(IV) sp³$sp^3$

Choose the correct answer from the options given below:

Core Logic

For CHCl₃$CHCl_3$, the individual dipole vectors do not cancel due to differing electronegativities of H and Cl, leading to a permanent non-zero dipole moment (μ ≠ 0$\mu \neq 0$).

Pattern Recognition

Symmetry yields vector cancellation arrow μ=0$\rightarrow \mu=0$. Asymmetry in CHCl₃$CHCl_3$ prevents cancellation arrow$\rightarrow$ polar.

Chapter Mix

Class 11 Chemistry: Chemical Bonding and Molecular Structure

Step 1: Determine the bond order of CO$\text{CO}$

Carbon monoxide (CO$\text{CO}$) contains 6 + 8 = 14$6 + 8 = 14$ total electrons. Its structural representation is C$\text{C}\equiv\text{O}$, matching a bond order value of 3.

Step 2: Determine the bond order of NO^+$\text{NO}^+$

The nitrosonium ion (NO^+$\text{NO}^+$) contains 7 + 8 - 1 = 14$7 + 8 - 1 = 14$ total electrons. Since it is isoelectronic with N₂$\text{N}_2$ and CO$\text{CO}$ (14 electrons$14\text{ electrons}$), its corresponding bond order value is also 3.

Step 3: Sum the results

Pattern Recognition

Isoelectronic species possessing 14 total electrons consistently demonstrate a bond order value of 3.

Chapter Mix

Class 11 Chemistry: Chemical Bonding and Molecular Structure

Core Logic

Let us determine the steric number (Z$Z$) and number of lone pairs (LP$LP$) for the central atom in each given molecule. Formula: Z = (1)/(2) (V + M - C + A)$Z = \frac{1}{2} (V + M - C + A)$ Where V$V$ = valence electrons on central atom, M$M$ = number of monovalent atoms, C$C$ = cationic charge, A$A$ = anionic charge. LP = Z - Bond Pairs (B.P.)$LP = Z - \text{Bond Pairs (B.P.)}$

• O₃$O_3$: Central atom O (V=6$V=6$). It forms one double bond and one dative bond. It has 1 lone pair remaining.
• H₂O$H_2O$: Central atom O (V=6$V=6$). Z = (1)/(2)(6 + 2) = 4$Z = \frac{1}{2}(6 + 2) = 4$. LP = 4 - 2 = 2$LP = 4 - 2 = 2$.
• SF₄$SF_4$: Central atom S (V=6$V=6$). Z = (1)/(2)(6 + 4) = 5$Z = \frac{1}{2}(6 + 4) = 5$. LP = 5 - 4 = 1$LP = 5 - 4 = 1$ (See-saw shape).
• ClF₃$ClF_3$: Central atom Cl (V=7$V=7$). Z = (1)/(2)(7 + 3) = 5$Z = \frac{1}{2}(7 + 3) = 5$. LP = 5 - 3 = 2$LP = 5 - 3 = 2$ (T-shape).
• NH₃$NH_3$: Central atom N (V=5$V=5$). Z = (1)/(2)(5 + 3) = 4$Z = \frac{1}{2}(5 + 3) = 4$. LP = 4 - 3 = 1$LP = 4 - 3 = 1$ (Pyramidal).
• BrF₅$BrF_5$: Central atom Br (V=7$V=7$). Z = (1)/(2)(7 + 5) = 6$Z = \frac{1}{2}(7 + 5) = 6$. LP = 6 - 5 = 1$LP = 6 - 5 = 1$ (Square Pyramidal).
• XeF₄$XeF_4$: Central atom Xe (V=8$V=8$). Z = (1)/(2)(8 + 4) = 6$Z = \frac{1}{2}(8 + 4) = 6$. LP = 6 - 4 = 2$LP = 6 - 4 = 2$ (Square Planar).

Step 1: Final Counting

The compounds containing exactly ONE lone pair on the central atom are O₃$O_3$, SF₄$SF_4$, NH₃$NH_3$, and BrF₅$BrF_5$.

Total count = 4.

Chapter Mix

Class 11 Chemistry: Chemical Bonding and Molecular Structure

Core Logic

Using Molecular Orbital (MO) Theory, we evaluate the bond order (BO = (Nb - Nₐ)/(2)$BO = \frac{N_b - N_a}{2}$) and magnetic nature (unpaired electrons = paramagnetic, all paired = diamagnetic) for each species:

Step 1: Final Selection

We need the species that satisfies BOTH conditions:

• Paramagnetic
• Bond Order = 1

Looking at the table, B₂$B_2$ is the only molecule that is paramagnetic (it has 2 unpaired electrons in degenerate π₂ₚ$\pi_{2p}$ orbitals) and has a bond order of 1.

Total number of such species = 1.

Pattern Recognition

B₂$B_2$ (10 electrons) and O₂$O_2$ (16 electrons) are the classic exceptions in MO theory that are paramagnetic despite having an even number of electrons. B₂$B_2$ has BO = 1, and O₂$O_2$ has BO = 2.

Chapter Mix

Class 11 Chemistry: Chemical Bonding and Molecular Structure

Core Logic

According to VSEPR theory:

• [Ni(CN)₄]²⁻$[Ni(CN)_4]^{2-}$: dsp²$dsp^2$ hybridization arrow$\rightarrow$ Square Planar.
• PCl₅$PCl_5$: sp³d$sp^3d$ hybridization with 0 lone pairs arrow$\rightarrow$ Trigonal Bipyramidal.
• BrF₅$BrF_5$: Bromine has 7 valence electrons. It forms 5 single bonds with Fluorine, leaving 1 lone pair. sp³d²$sp^3d^2$ hybridization arrow$\rightarrow$ geometry is octahedral, but shape is Square Pyramidal.
• PF₅$PF_5$: sp³d$sp^3d$ hybridization with 0 lone pairs arrow$\rightarrow$ Trigonal Bipyramidal.

Pattern Recognition

AX₅E₁$AX_5E_1$ configuration (5 bond pairs + 1 lone pair) typically yields a Square Pyramidal shape.

Chapter Mix

Class 11 Chemistry: Chemical Bonding and Molecular Structure Class 12 Chemistry: Coordination Compounds