A molecule with the formula mathrmAX_4mathrmY has all its elements from p-block. Element A is rarest, monoatomic, non-radioactive from its group and has the lowest ionization enthalpy value among A, X and Y. Elements X and Y have first and second highest electronegativity values respectively among all the known elements. The shape of the molecule is :

Solution & Explanation

### Related Formula Total valence shell electron pair system equation: textValence Pairs = textBond Pairs (BP) + textLone Pairs (LP) ### Core Logic Let's decode individual identities based on the descriptive properties: * The elements with the first and second highest electronegativity values across the entire periodic table are Fluorine (mathrmF) and Oxygen (mathrmO), matching labels mathrmX and mathrmY. * Element mathrmA is a rare, monoatomic, non-radioactive p-block element with low ionization energy, identifying it as Xenon (mathrmXe). * Substituting these components into the target layout formula yields mathrmXeOF_4: - Xenon brings 8 valence electrons. It forms 4 single bonds with F and 1 double bond with O, consuming 6 electrons and leaving 1 lone pair on the central atom. - Steric Number = 5 text bond regions + 1 text lone pair = 6 (Octahedral electronic arrangement). ### Step 1: Geometry Determination Placing the double-bonded oxygen and lone pair along vertical spatial axes yields a stable square pyramidal molecular shape layout:
XeOF4 square pyramidal spatial geometry diagram for Q45
XeOF4 square pyramidal spatial geometry diagram for Q45
### Pattern Recognition In mathrmsp^3d^2 architectures containing an explicit lone pair along with an asymmetric double bond (like mathrmXeOF_4), the lone pair always sits directly opposite the double bond to minimize electron repulsion, leaving a clean square pyramidal shape. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Chemistry: Chemical Bonding and Molecular Structure Class 12 Chemistry: The p-Block Elements

More Chemical Bonding and Molecular Structure Previous-Year Questions — Page 7

Q85 jee_main_2024_31_jan_morning Hybridization
The number of species from the following in which the central atom uses sp^3 hybrid orbitals in its bonding is NH_3, SO_2, SiO_2, BeCl_2, CO_2, H_2O, CH_4, BF_3
Numerical Answer. Answer: 4 to 4

Solution

### Core Logic Analyzing the hybridization of the central atom in each species: - NH_3: 3 bp + 1 lp = 4 electron domains rightarrow sp^3 - SO_2: 2 bp + 1 lp = 3 electron domains rightarrow sp^2 - SiO_2: A giant covalent network where each Si is bonded to 4 oxygens tetrahedrally rightarrow sp^3 - BeCl_2: 2 bp + 0 lp = 2 electron domains rightarrow sp - CO_2: 2 bp + 0 lp = 2 electron domains rightarrow sp - H_2O: 2 bp + 2 lp = 4 electron domains rightarrow sp^3 - CH_4: 4 bp + 0 lp = 4 electron domains rightarrow sp^3 - BF_3: 3 bp + 0 lp = 3 electron domains rightarrow sp^2 Total species with sp^3 hybridization: NH_3, SiO_2, H_2O, CH_4. Total count = 4. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Chemistry: Chemical Bonding and Molecular Structure
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JEE Physics: Waves (+15.5%) | Electrostatics: Concentric Shells (-29.7%) | Modern Physics: Photoelectric Clones (+34.2%) | Mathematics: Definite Integrals (+18.1%) | Chemistry: Coordination Splitting (-11.4%) | JEE Physics: Waves (+15.5%) | Electrostatics: Concentric Shells (-29.7%) | Modern Physics: Photoelectric Clones (+34.2%) | Mathematics: Definite Integrals (+18.1%) | Chemistry: Coordination Splitting (-11.4%)