Before You Begin

Question #44296

Difficulty: 8/10 | 🟢 Free

The Question

Elements $X$ and $Y$ belong to the same group of the periodic table, with $Y$ located two periods below $X$ . They form monovalent radicals that act as ligands. The first ionization enthalpies ( $IE_1$ ) and electron affinities ( $EA$ ) of $X$ and $Y$ are given as follows: For $X$ , $IE_1 = 7.8$ eV/atom and $EA = 1.2$ eV/atom. For $Y$ , $IE_1 = 9.4$ eV/atom and $EA = 2.6$ eV/atom. A central atom $M$ (Group 15) forms a neutral, pentacoordinate complex \ce{MX2Y3}. Assuming the thermodynamic stability of the isomers is governed by Bent's rule based on the Mulliken electronegativity scale, identify the correct statement regarding the most stable isomer of \ce{MX2Y3}.

Options

A.
Both $X$ ligands occupy the axial positions, giving the molecule $D_{3h}$ symmetry and a net dipole moment $\mu = 0$ .
B.
Both $X$ ligands occupy the equatorial positions, giving the molecule $C_{2v}$ symmetry and a net dipole moment $\mu \neq 0$ .
C.
One $X$ ligand occupies an axial position and the other an equatorial position, giving the molecule $C_s$ symmetry and $\mu \neq 0$ .
D.
Two $Y$ ligands occupy the axial positions, giving the molecule $D_{3h}$ symmetry and a net dipole moment $\mu = 0$ .

Check Your Answer

Correct Answer

Step-by-Step Solution

Explanation

How the Examiner Thinks

Mechanism_Trap

Don't get tricked by the process! Learn the mechanical traps used in competitive exam questions.

Concepts Behind This Question

Ionization Enthalpy (IE) Anomalies|Electronegativity Scales|Dipole Moment Vector Addition

Why This Question Is Tricky

Common Mistakes

The problem tests the 'Hidden Variable' of anomalous electronegativity trends caused by the lanthanide/d-block contraction.

Students often blindly apply the 'down-the-group decrease' heuristic.

If they do, they will incorrectly assume $EN(X) > EN(Y)$ , place the two $X$ ligands in the axial positions, and deduce a highly symmetric $D_{3h}$ molecule with a calculated zero dipole moment (Option A).

By forcing the calculation via the Mulliken scale, the problem reveals that $Y$ is actually more electronegative, leading to a $C_{2v}$ geometry where the dipole moment is non-zero.

Quick Facts About This Question

Question Information

Subject	Chemistry
Chapter	Classification of Elements and Periodicity in Properties - Class 11 Part 1
Concept	Ionization Enthalpy (IE) Anomalies
Difficulty	8
Cognitive Load	7
Target Solve Time	160 sec
Access	Free