4
$\begingroup$

I was asked the following question in JEE Mains (2nd September, Shift 1) 2020,

If a compound AB4 is polar (having non-zero dipole moment), then it's structure could be

(A) Square planar
(B) Rectangular planar
(C) Square pyramidal
(D) Tetrahedral

My thoughts

  • As this compound has only 4 ligands, therefore it cannot be a square pyramidal.
  • It can neither be a tetrahedral nor a square planar bcoz, it would be non-polar then.

So, we are left only with "rectangular planar", which I'd never heard before. How to move on?

$\endgroup$
7
  • 1
    $\begingroup$ I don't know how one would get to square pyramidal with 4 ligands (I can imagine a weird case...) but that's the only answer that's not overly symmetry so as to eliminate any chance of being polar. I was expecting see-saw. $\endgroup$ – Zhe Sep 3 '20 at 11:28
  • 1
    $\begingroup$ Strictly interpreted none of those geometries would be polar (unless they mean trapezoid not rectangle). Trigonal pyramidal would be but that isn't what they said. $\endgroup$ – matt_black Sep 3 '20 at 12:24
  • 2
    $\begingroup$ My guess is that by square pyramidal they mean that the ligands form the square base of the pyramid and the central atom is the apex, which would make it the correct answer. $\endgroup$ – Andrew Sep 3 '20 at 14:40
  • $\begingroup$ Trigonal bi pyramidal was one option i think.... $\endgroup$ – sai-kartik Sep 3 '20 at 16:08
  • $\begingroup$ @sai-kartik: I don't think so, bcoz if that had been present, I wouldn't ask this question then :P (also, TBP needs AB5 formula) $\endgroup$ – Rahul Verma Sep 3 '20 at 16:48
7
$\begingroup$

Given all $\ce{B}$ are equivalent in terms of AXE notation, for $\ce{AB4}$ the following coordination environments are possible [1, p. 177]:

  • square-planar $(D_\mathrm{4h},$ SP-4);
  • square-pyramidal $(C_\mathrm{4v},$ SPY-4);
  • tetrahedral $(T_\mathrm d,$ T-4);
  • see-saw $(C_\mathrm{2v},$ SS-4).

Table IR-9.3 Polyhedral symbols, geometrical structures and/or polyhedra

Table IR-9.3 Polyhedral symbols, geometrical structures and/or polyhedra

Option (A) square planar implies a centrosymmetrical environment and such molecule cannot be polar. The derivative of (A), option (B) rectangular planar, as well as option (D) tetrahedral also refer to the environments that posses zero dipole moment.

Speaking of (C), I think you are misinterpreting the geometry of square pyramidal coordination. It's a square non-coplanar arrangement with a non-zero dipole moment commonly seen in period 4 transition metal porphyrins $(\ce{MN4}$ fragment) and also lead(II) oxide and tin(II) oxide $(\ce{MO4}$ fragment). According to Brecher [2, p. 1944], the structure of bis(N,N-dimethyldithiocarbamato)lead(II) would be yet another example:

Sample depiction of a compound known to exist in square pyramidal configuration:

bis(N,N-dimethyldithiocarbamato)lead(II)

Since a see-saw option is not present, this leaves only one answer, option (C).

Reference

  1. IUPAC. Nomenclature of Inorganic Chemistry, IUPAC Recommendations 2005 (the “Red Book”), 1st ed.; Connelly, N. G., Damhus, T., Hartshorn, R. M., Hutton, A. T., Eds.; RSC Publishing: Cambridge, UK, 2005.
  2. Brecher, J. Graphical Representation of Stereochemical Configuration (IUPAC Recommendations 2006). Pure and Applied Chemistry 2009, 78 (10), 1897–1970. DOI: 10.1351/pac200678101897.
$\endgroup$
2
0
$\begingroup$

It's square pyramidal. What you miss is there are only four ligands, but the central atom itself forms the apex of the pyramid. In practice, though, such a molecule likely distorts to a seesaw shape which could still be polar.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.