Which of the following has $$dsp^2$$ hybridization A. $$NiCl_4^{2-}$$ B. $$SCl_4$$C. $$NH_4^{+}$$ D. $$PtCl_4^{2-}$$

Question

Which of the following has $$dsp^2$$ hybridization A. $$NiCl_4^{2-}$$   B. $$SCl_4$$C. $$NH_4^{+}$$ D. $$PtCl_4^{2-}$$

Vedantu Content Team · Accepted Answer

Hint: We know that molecules and ions have shapes that are predicted by the valence shell electron pair repulsion (VSEPR) model. In $$dsp^2$$ hybridization, one d-orbital, s-orbital, and two p-orbital are hybridized forming a square planar structure. Complete step by step solution:Hybridization as we know is the mixing of two atomic orbitals of identical or nearly identical energies to get unique orbitals of comparable energy. The central atom orbitals have hybridization.When one d, s, and two p- orbitals of identical atomic orbitals blend to provide new hybrid orbitals of equivalent energy, $$dsp^2$$ hybridization.A. $$NiCl_4^{2-}$$-        It is a complex compound. Ni has a +2 charge in this compound. So, the total no.of electrons on the central metal atom is 8. All these electrons lie in the 3d of the central metal with six electrons paired up and two electrons remaining unpaired. Here $$Cl^-$$ is a weak field ligand. So it will not pair up the single electrons. So, the one 4s and three 4p orbitals hybridize to form four $$sp^3$$ hybrid orbitals. These hybrid orbitals adjust four pairs of electrons from chloride ions. So, A is incorrect. B.  $$SCl_4$$S is the central atom here. It has 6 electrons in its valence shell. Four electrons take part in bond formation. The remaining two electrons form a lone pair. So, there are four bond pairs and one lone pair. So, there is $$sp^3$$ hybridization for four bond pairs. So, B is incorrect. C. $$NH_4^{+}$$  Ammonia undergoes a reaction with hydrogen ions and forms ammonium cation. In this, a nitrogen atom carries four sigma bonds with four hydrogen atoms and no lone pair. Accordingly, the hybridization of Nitrogen (N) in the ammonium ion is $$sp^3$$. So, C is incorrect. D. $$PtCl_4^{2-}$$It is a complex compound. The oxidation state of Pt is +2.There are eight electrons in the 4d orbital. Here the ligands work as strong ligands therefore pairing happens. There is one vacant orbital. Hence, one orbital of 4d, 5s, and two orbitals of 5p take part in $$dsp^2$$ hybridization and square planar geometry. So, D is correct.  So, option D is correct.Note: Elements usually integrate with other elements either by losing, gaining, or sharing electrons known as chemical bonding. Usually, elements integrate with others with the support of valence electrons. Valence electrons are present in the outermost shell of an atom.

Which of the following has \[dsp^2\] hybridization A. \[NiCl_4^{2-}\] B. \[SCl_4\]C. \[NH_4^{+}\] D. \[PtCl_4^{2-}\]

Which of the following has \[dsp^2\] hybridization
A. \[NiCl_4^{2-}\]
B. \[SCl_4\]
C. \[NH_4^{+}\]
D. \[PtCl_4^{2-}\]