A molecule with the formula mathrmAX_4mathrmY$\mathrm{AX}_4\mathrm{Y}$ 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 :
### Related Formula
Total valence shell electron pair system equation:
textValence Pairs = textBond Pairs (BP) + textLone Pairs (LP)$$\text{Valence Pairs} = \text{Bond Pairs (BP)} + \text{Lone Pairs (LP)}$$
### 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$\mathrm{F}$) and Oxygen (mathrmO$\mathrm{O}$), matching labels mathrmX$\mathrm{X}$ and mathrmY$\mathrm{Y}$.
* Element mathrmA$\mathrm{A}$ is a rare, monoatomic, non-radioactive p-block element with low ionization energy, identifying it as Xenon (mathrmXe$\mathrm{Xe}$).
* Substituting these components into the target layout formula yields mathrmXeOF_4$\mathrm{XeOF_4}$:
- 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$= 5 \text{ bond regions} + 1 \text{ lone pair} = 6$ (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
### Pattern Recognition
In mathrmsp^3d^2$\mathrm{sp^3d^2}$ architectures containing an explicit lone pair along with an asymmetric double bond (like mathrmXeOF_4$\mathrm{XeOF_4}$), the lone pair always sits directly opposite the double bond to minimize electron repulsion, leaving a clean square pyramidal shape.
### Evaluation Rubric / Model Answer
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### 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 6
### Core Logic
Using VSEPR theory:
(A) BrF_5$BrF_5$: Br has 7 valence electrons. 5 form bonds with F, leaving 1 lone pair. (5 bp + 1 lp) rightarrow$\rightarrow$sp^3d^2$sp^3d^2$ hybridization rightarrow$\rightarrow$ Square pyramidal shape.
(B) H_2O$H_2O$: O has 6 valence electrons. 2 form bonds with H, leaving 2 lone pairs. (2 bp + 2 lp) rightarrow$\rightarrow$sp^3$sp^3$ hybridization rightarrow$\rightarrow$ Bent shape.
(C) ClF_3$ClF_3$: Cl has 7 valence electrons. 3 form bonds with F, leaving 2 lone pairs. (3 bp + 2 lp) rightarrow$\rightarrow$sp^3d$sp^3d$ hybridization rightarrow$\rightarrow$ T-shape.
(D) SF_4$SF_4$: S has 6 valence electrons. 4 form bonds with F, leaving 1 lone pair. (4 bp + 1 lp) rightarrow$\rightarrow$sp^3d$sp^3d$ hybridization rightarrow$\rightarrow$ See-saw shape.
### Step 1: Matching
(A) - (IV)
(B) - (III)
(C) - (I)
(D) - (II)
VSEPR Theory solution diagram for Q71 - JEE Main 2024 MorningVSEPR Theory solution diagram for Q71 - JEE Main 2024 MorningVSEPR Theory solution diagram for Q71 - JEE Main 2024 MorningVSEPR Theory solution diagram for Q71 - JEE Main 2024 Morning
### Evaluation Rubric / Model Answer
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### Chapter Mix
Class 11 Chemistry: Chemical Bonding and Molecular Structure
Q85jee_main_2024_30_jan_morningMolecular Orbital Theory
### Core Logic
According to Molecular Orbital Theory (MOT), the number of molecular orbitals (MOs) formed is equal to the total number of atomic orbitals (AOs) combined.
### Step 1: Counting atomic orbitals
For a single atom in the 2nd period, the valence shell has:
One 2s orbital
Three 2p orbitals (2p_x, 2p_y, 2p_z$2p_x, 2p_y, 2p_z$)
Total = 4 atomic orbitals per atom.
For a diatomic molecule, two such atoms combine. Total atomic orbitals = 4 times 2 = 8$4 \times 2 = 8$.
### Step 2: Forming molecular orbitals
Combining these 8 atomic orbitals yields 8 molecular orbitals:
- From 2s: sigma_2s$\sigma_{2s}$ and sigma^*_2s$\sigma^*_{2s}$ (2 MOs)
- From 2p: sigma_2p_z, pi_2p_x, pi_2p_y, pi^*_2p_x, pi^*_2p_y, sigma^*_2p_z$\sigma_{2p_z}, \pi_{2p_x}, \pi_{2p_y}, \pi^*_{2p_x}, \pi^*_{2p_y}, \sigma^*_{2p_z}$ (6 MOs)
Total MOs = 2 + 6 = 8$2 + 6 = 8$.
### Evaluation Rubric / Model Answer
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### Chapter Mix
Class 11 Chemistry: Chemical Bonding and Molecular Structure
Q74jee_main_2024_31_jan_eveningIonic Bond and Lattice Energy
Which of the following is least ionic?
A.text(1) BaCl_2$\text{(1) }BaCl_2$
B.text(2) AgCl$\text{(2) }AgCl$
C.text(3) KCl$\text{(3) }KCl$
D.text(4) CoCl_2$\text{(4) }CoCl_2$
Solution
### Core Logic
According to Fajan's rules, covalent character is favored by high charge and small size of the cation, and by cations with a pseudo-noble gas configuration.
Ag^+$Ag^+$ has a pseudo-noble gas configuration (ns^2np^6nd^10$ns^2np^6nd^{10}$), which results in high polarizing power compared to s-block and typical transition elements.
Therefore, AgCl$AgCl$ has the maximum covalent character and is the least ionic among the given options.
Ionic character order: AgCl < CoCl_2 < BaCl_2 < KCl$AgCl < CoCl_2 < BaCl_2 < KCl$
### Step 1: Final Selection
Because AgCl$AgCl$ is the most covalent, it is the least ionic. Hence, option (2) is correct.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 11 Chemistry: Chemical Bonding and Molecular Structure
Q81jee_main_2024_31_jan_eveningDipole Moment and Fractional Charge
A diatomic molecule has a dipole moment of 1.2text D$1.2\text{ D}$. If the bond distance is 1mathrmAA$1\mathrm{\AA}$, then fractional charge on each atom is _________ times 10^-10text esu$\times 10^{-10}\text{ esu}$.
(Given: 1text D = 10^-18text esu cm$1\text{ D} = 10^{-18}\text{ esu cm}$)
Numerical Answer.Answer: 1.2 to 1.2
Solution
### Related Formula
mu = q times d$$\mu = q \times d$$
### Core Logic
Given dipole moment, mu = 1.2text D = 1.2 times 10^-18text esu cm$\mu = 1.2\text{ D} = 1.2 \times 10^{-18}\text{ esu cm}$.
Bond distance, d = 1mathrmAA = 10^-8text cm$d = 1\mathrm{\AA} = 10^{-8}\text{ cm}$.
We need to find the fractional charge q$q$.
### Step 1: Calculation
q = fracmud$$q = \frac{\mu}{d}$$q = frac1.2 times 10^-18text esu cm10^-8text cm$$q = \frac{1.2 \times 10^{-18}\text{ esu cm}}{10^{-8}\text{ cm}}$$q = 1.2 times 10^-10text esu$$q = 1.2 \times 10^{-10}\text{ esu}$$
### Step 2: Final Formatting
The question asks for the fractional charge in the form x times 10^-10text esu$x \times 10^{-10}\text{ esu}$. Therefore, the value is 1.2$1.2$.
*Note: Based on NTA officially accepting 12$12$ (if asked for x times 10^-11$x \times 10^{-11}$) or 1.2$1.2$. We will format it exactly as calculated.*
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 11 Chemistry: Chemical Bonding and Molecular Structure
Q77jee_main_2024_31_jan_morningMolecular Orbital Theory
The linear combination of atomic orbitals to form molecular orbitals takes place only when the combining atomic orbitals
A. have the same energy
B. have the minimum overlap
C. have same symmetry about the molecular axis
D. have different symmetry about the molecular axis
Choose the most appropriate from the options given below:
A.textA, B, C only$\text{A, B, C only}$
B.textA and C only$\text{A and C only}$
C.textB, C, D only$\text{B, C, D only}$
D.textB and D only$\text{B and D only}$
Solution
### Core Logic
Conditions for the linear combination of atomic orbitals (LCAO) to form molecular orbitals:
1. The combining atomic orbitals must have the same or nearly the same energy.
2. The combining atomic orbitals must have the same symmetry about the molecular axis.
3. The combining atomic orbitals must overlap to the maximum extent (not minimum).
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 11 Chemistry: Chemical Bonding and Molecular Structure
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