Molecular Geometry

Step 1: Pt(II) is a d⁸ metal ion.
Step 2: d⁸ complexes commonly show square planar geometry.

Step 1: PH₃ exists but BiCl₃ does not as a stable molecule.
Step 2: SO₂ has pπ-pπ bonding, not π-π.
Step 3: I₃^+ has bent geometry.
Step 4: SeF₄ is seesaw while CH₄ is tetrahedral.

Step 1: VSEPR theory. T-shape corresponds to AX₃E₂ geometry.
Step 2: Lone pairs count. Two lone pairs occupy equatorial positions.

Step 1: Apply Molecular Orbital Theory (MOT).
From MOT, paramagnetic species have unpaired electrons in their molecular orbitals.
Step 2: Conclusion.
Both and have unpaired electrons, making them paramagnetic. Final Answer:

Step 1: BF and NH both have tetrahedral geometry.
Step 2: Both involve hybridization.

Step 1: is tetrahedral, giving bond angle .
Step 2: has puckered ring structure with bond angle .

Step 1: Boron forms only three bonds.
Step 2: BF has only six electrons around boron.

Step 1: Total charge on PO₄³- is -3, distributed equally over 4 oxygen atoms.
Step 2: Formal charge on each oxygen: (-3)/(4)=-0.75
Step 3: Total P–O bonds =5 (one double bond + three single bonds): Average bond order=(5)/(4)=1.25

Bond length is inversely proportional to bond order. Bond order decreases in the order: O₂⁺ > O₂ > O₂⁻ > O₂²- Hence bond length increases in the reverse order.

Step 1: Lone pairs: IF₇:0, IF₅:1, ClF₃:2, XeF₂:3.

P₄ has a tetrahedral structure with bond angle 60^∘. S₈ has a puckered ring structure with bond angle 107^∘.

Step 1: The phosphate ion PO₄³- has four equivalent resonance structures. In each structure, there is: • one P=O double bond • three P-O⁻ single bonds
Step 2: Total bond order of all P–O bonds: 2 + 1 + 1 + 1 = 5
Step 3: Average P–O bond order: Bond order = (5)/(4) = 1.25
Step 4: Total negative charge on oxygen atoms is -3, shared equally among 4 oxygen atoms: Formal charge on each O = (-3)/(4) = -0.75

Step 1: In gaseous phase, water molecule exists as an individual molecule. Step 2: Oxygen has two lone pairs and two bond pairs. Step 3: According to VSEPR theory, lone pair–lone pair and lone pair–bond pair repulsions cause bending of the molecule. ∠ HOH = 104.5^∘

Step 1: Directional bonds are characteristic of covalent compounds. Step 2: \beginitemize • NaCl, BaO, CsCl – ionic (non-directional) • CO₂ – covalent with fixed bond angles (directional) \enditemize

Step 1: Bond order decreases as electrons are added to antibonding orbitals. Bond order: O₂⁺ (2.5) > O₂ (2) > O₂⁻ (1.5) > O₂²- (1) Step 2: Bond length is inversely proportional to bond order. ⟹ Shortest → Longest O₂⁺ < O₂ < O₂⁻ < O₂²-

Water (H₂O) is a simple molecule, but its geometry can be determined using VSEPR theory (Valence Shell Electron Pair Repulsion theory). Here’s how we analyze it:
1. Lewis Structure: - Oxygen has 6 valence electrons, and each hydrogen has 1 valence electron. In H₂O, oxygen forms two single bonds with hydrogen atoms, and it has two lone pairs of electrons remaining. 2. Electron Geometry: - The oxygen atom has 4 electron pairs around it: 2 bonding pairs (with H) and 2 lone pairs. According to VSEPR theory, these 4 electron pairs are arranged in a tetrahedral geometry to minimize repulsion. 3. Molecular Geometry: - However, the lone pairs are not visible in the molecule’s structure. The lone pairs cause the bonding pairs to repel each other, resulting in a bent molecular shape. - The bond angle between the hydrogen-oxygen-hydrogen atoms is approximately 104.5°, which is less than the ideal tetrahedral angle (109.5°) due to the lone pair repulsion. 4. Conclusion: - The molecular geometry of water is bent.