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The hybridization state of the central atom in HgCl2 is –
(1) sp
(2) sp2
(3) sp3

Answer
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Hint: The structure of HgCl2 is linear and the hybridization state of the central atom can be predicted by using VSEPR theory. This theory defines the hybridization and geometry of a compound by determining the number of electrons involved in bonding.

Complete answer:
To find out the hybridization of the central atom, we need to calculate the steric number. The steric number is the number of atoms bonded to a central atom of a molecule plus the no. of lone pairs attached to it. It is used alongside VSEPR theory to determine the molecular geometry and thus the hybridization of the central atom.
The atomic number of Hg is 80 and thus its electronic configuration is –
Hg[Xe]4f145d106s2
Here, the outermost orbital is 6s-orbital. So, it has 2 valence electrons. Since it is bonded to two chlorine atoms, therefore the two electrons will be shared with these two chlorine atoms and it will form HgCl2. Thus, it has no lone pairs of electrons. Therefore, the steric number of HgCl2 comes out to be 2.
Steric number=No. of atoms bonded to central atom+No. of lone pairsSteric number(HgCl2)=2+0=2
This suggests that the molecule possesses linear geometry in which the hybridization of the central atom is sp-hybridized. The Lewis structure of HgCl2 is drawn below.
seo images

Hence, the correct answer is (1) sp.

Note:
The steric number defines the shape and geometry of the molecule by assigning the position of atoms in such a way that the repulsion between electron pairs will be minimal. For example, a steric number of 2 provides a linear shape in which the bonding electrons are 180 degrees apart.
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