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Chemical Bonding and Molecular Structure question

2022 · 27 Jul · Shift 2 · Q19

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Chemical Bonding and Molecular Structure
2022 · 27 Jul · Shift 2 · Q19

Chemical Bonding and Molecular Structure question

2022 · 27 Jul · Shift 2 · Q19

JEE MainChemistryChemical Bonding and Molecular StructureNumerical+4 / −1

The number of molecule(s) or ion(s) from the following having non-planar structure is

\underline{\hspace{2cm}}

.

\mathrm{NO}_{3}^{-}, \mathrm{H}_{2} \mathrm{O}_{2}, \mathrm{BF}_{3}, \mathrm{PCl}_{3}, \mathrm{XeF}_{4}, \mathrm{SF}_{4}, \mathrm{XeO}_{3}, \mathrm{PH}_{4}^{+}, \mathrm{SO}_{3},\left[\mathrm{Al}(\mathrm{OH})_{4}\right]^{-}

Numerical answer

View written solutionFree

Correct answer: 6

We determine the geometry of each species using VSEPR theory and check whether the whole molecule/ion is planar or non-planar.

Analyze each species:

(i) $\mathrm{NO_3^-}$

Central atom: N
Three bond pairs, no lone pair on N
Geometry: trigonal planar
Hence planar

(ii) $\mathrm{H_2O_2}$

Structure: $\mathrm{H-O-O-H}$
Due to lone pair repulsions on each O, the molecule has a non-planar (skew/open book) structure
Hence non-planar

(iii) $\mathrm{BF_3}$

Central atom: B
Three bond pairs, no lone pair
Geometry: trigonal planar
Hence planar

(iv) $\mathrm{PCl_3}$

Central atom: P
Three bond pairs and one lone pair
Electron pair geometry: tetrahedral
Molecular shape: trigonal pyramidal
Hence non-planar

(v) $\mathrm{XeF_4}$

Central atom: Xe
Four bond pairs and two lone pairs
Electron pair geometry: octahedral
Molecular shape: square planar
Hence planar

(vi) $\mathrm{SF_4}$

Central atom: S
Four bond pairs and one lone pair
Electron pair geometry: trigonal bipyramidal
Molecular shape: see-saw
Hence non-planar

(vii) $\mathrm{XeO_3}$

Central atom: Xe
Three bond pairs and one lone pair
Molecular shape: trigonal pyramidal
Hence non-planar

(viii) $\mathrm{PH_4^+}$

Central atom: P
Four bond pairs, no lone pair
Geometry: tetrahedral
Hence non-planar

(ix) $\mathrm{SO_3}$

Central atom: S
Three bond pairs, no lone pair
Geometry: trigonal planar
Hence planar

(x) $\left[\mathrm{Al(OH)_4}\right]^{-}$

Central atom: Al
Four bond pairs, no lone pair
Geometry: tetrahedral
Hence non-planar

Count the non-planar species:

Non-planar are:

$\mathrm{H_2O_2}$
$\mathrm{PCl_3}$
$\mathrm{SF_4}$
$\mathrm{XeO_3}$
$\mathrm{PH_4^+}$
$\left[\mathrm{Al(OH)_4}\right]^{-}$

So total number is $6$

Comparison with stored answer:

Derived answer = $6$
Stored correct answer = $6$
They agree.

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