A molecule with the formula \( AX_4Y \) 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:
A molecule with the formula \( AX_4Y \) 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:
Show Hint
- Square pyramidal
- Octahedral
- Planar
- Tetrahedral
The Correct Option is A
Approach Solution - 1
The problem requires us to determine the shape of a molecule with the formula AX\(_4\)Y, where the elements A, X, and Y are identified based on a series of descriptive clues.
Concept Used:
The solution involves two main concepts:
- Identification of Elements using Periodic Trends: We will use fundamental knowledge of the periodic table, including trends in electronegativity, ionization enthalpy, and elemental properties (like natural state and abundance) to identify A, X, and Y.
- Valence Shell Electron Pair Repulsion (VSEPR) Theory: Once the molecule is identified, we will use VSEPR theory to predict its shape. This theory states that the geometry of a molecule is determined by minimizing the electrostatic repulsion between electron pairs (both bonding pairs and lone pairs) around the central atom. The steps are:
- Identify the central atom.
- Calculate the steric number (SN): \( \text{SN} = (\text{Number of atoms bonded to central atom}) + (\text{Number of lone pairs on central atom}) \).
- Determine the electron geometry from the SN (e.g., SN=6 \(\rightarrow\) Octahedral).
- Determine the final molecular shape from the arrangement of atoms (VSEPR type AX\(_m\)E\(_n\)).
Step-by-Step Solution:
Step 1: Identify Element X.
The clue for X is that it has the "first and highest electronegativity value... among all the known elements." The most electronegative element in the periodic table is Fluorine.
\[ \text{Element X = Fluorine (F)} \]
Step 2: Identify Element Y.
The clue for Y is that it has the "second highest electronegativity value... among all the known elements." The second most electronegative element is Oxygen.
\[ \text{Element Y = Oxygen (O)} \]
Step 3: Identify Element A.
The clues for element A are:
- It is from the p-block.
- It is the "rarest, monatomic, non-radioactive from its group." The property of being monatomic points to the noble gases (Group 18). Among the non-radioactive noble gases (He, Ne, Ar, Kr, Xe), Xenon is the rarest.
- It has the "lowest ionization enthalpy value among A, X and Y." Let's check this: IE\(_1\)(F) \(\approx\) 1681 kJ/mol; IE\(_1\)(O) \(\approx\) 1314 kJ/mol; IE\(_1\)(Xe) \(\approx\) 1170 kJ/mol. Xenon indeed has the lowest ionization enthalpy of the three.
Therefore, element A is Xenon.
\[ \text{Element A = Xenon (Xe)} \]
The molecular formula is thus XeOF\(_4\).
Step 4: Apply VSEPR Theory to XeOF\(_4\).
- Central Atom: Xenon (Xe) is the central atom as it is the least electronegative.
- Valence Electrons of Xe: Xenon is a noble gas, so it has 8 valence electrons.
- Bonding and Lone Pairs:
- Xe forms four single bonds with the four fluorine atoms.
- Xe forms one double bond with the oxygen atom (as oxygen is divalent).
- Total electrons from Xe used in bonding = 4 (for 4 F atoms) + 2 (for 1 O atom) = 6 electrons.
- Remaining non-bonding electrons on Xe = 8 - 6 = 2 electrons.
- This corresponds to 1 lone pair of electrons on the central Xe atom.
Step 5: Calculate the Steric Number (SN) and determine the shape.
- The number of atoms bonded to the central atom is 4 (F) + 1 (O) = 5.
- The number of lone pairs on the central atom is 1.
- Steric Number (SN) = (Number of bonded atoms) + (Number of lone pairs) = 5 + 1 = 6.
- A steric number of 6 corresponds to an octahedral electron geometry.
- The VSEPR type of the molecule is AX\(_5\)E\(_1\) (5 bonding pairs, 1 lone pair).
- According to VSEPR theory, a molecule with an AX\(_5\)E\(_1\) arrangement has a square pyramidal shape. The lone pair occupies one of the axial positions to minimize repulsion, and the five atoms occupy the remaining five vertices of the octahedron.
The shape of the molecule XeOF\(_4\) is Square pyramidal.
Approach Solution -2
The molecule \( \text{A} \text{X}_2 \text{Y}_2 \) follows the square pyramidal structure based on the given criteria. The electronegativity and ionization energy of element A explain its rarest behavior. 
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