Question

One of the following having square planar structure is:
$\mathrm{A}) \mathrm{NH}_{4}^{+}$
$B) B F_{4}^{-}$
$\mathrm{C}) \mathrm{XeF}_{4}$
$\mathrm{D}) \mathrm{SCl}_{4}$

Answer 1

In order to find out the geometry, we first have to find out the hybridisation of the given compounds. In order to find the hybridisation, let's find out the lone pairs on each compound given.

Complete step-by-step answer:
$\mathrm{NH}_{4}^{+}$ is ammonium cation ,the lone pair on the nitrogen atom (N) in ammonia, represented as a line above the N, forms the bond with a proton (H $^{+}$ ). Thereafter, all four $\mathrm{N}-\mathrm{H}$ bonds are equivalent, being polar covalent bonds.The ion has a tetrahedral structure and is isoelectronic with methane and borohydride.

Boron has 3 valence electrons, and each of the four fluorides contributes one electron to each covalent bond. The overall negative charge of the molecule contributes another electron, so overall we have 3 + 4 + 1 = 8 valence electrons which form 4 electron pairs.
The pairs arrange themselves in space to maximise the distance between them, thereby reducing the electronic repulsion between them to a minimum. Thus it would be tetrahedral geometry.

$\mathrm{SCl}_{4}$ has seesaw molecular geometry because you must take into account the effect that the lone pair on S has on the shape. If no lone pair was there on S, then it would have been tetrahedral. But it has a seesaw geometry.

Now, in $\mathrm{XeF}_{4}$ the hybridisation is $\mathrm{sp}^{2} \mathrm{d}^{2} .$ Thus it gives us a square planar structure. The central atom has 6 electron groups out of which 2 pairs are lone pairs.

The answer is C)

Note: The hybridisation is the key factor in deciding the shape and geometry of the compound. Another factor is the distribution of the lone pairs in the compound.