Match the following:
List-I – List-II
1. $NH_4^+$ – (A) $sp^2$,
2. $CO_2$ – (B) $sp^3$,
3. $IF_7$ – (C) $sp$,
4. $AlCl_3^-$ – (D) $sp^3d^3$

Question

Match the following: 
List-I – List-II
1. $NH_4^+$ – (A) $sp^2$, 
2. $CO_2$ –  (B) $sp^3$, 
3. $IF_7$ – (C) $sp$, 
4. $AlCl_3^-$ – (D) $sp^3d^3$

Accepted Answer

The correct option is (1):

Let's break down the hybridization of each species to understand why option (1) is correct. Hybridization is a concept that describes the mixing of atomic orbitals to form new hybrid orbitals suitable for electron pairing to form chemical bonds.

For $NH_4^+$ (Ammonium ion): The central atom is Nitrogen (N). Nitrogen has 5 valence electrons. In $NH_4^+$, nitrogen forms four single bonds with four hydrogen atoms, and it has a +1 charge, meaning it has effectively lost one electron. To determine the hybridization, we count the number of sigma bonds and lone pairs around the central atom. Nitrogen forms 4 sigma bonds and has 0 lone pairs. Therefore, the steric number is 4, which corresponds to $sp^3$ hybridization. This gives a tetrahedral geometry.

For $CO_2$ (Carbon dioxide): The central atom is Carbon (C). Carbon has 4 valence electrons. It forms two double bonds with two oxygen atoms. Each double bond consists of one sigma bond and one pi bond. Therefore, carbon forms 2 sigma bonds and has 0 lone pairs. The steric number is 2, which corresponds to $sp$ hybridization. This results in a linear geometry.

For $IF_7$ (Iodine heptafluoride): The central atom is Iodine (I). Iodine is a halogen and has 7 valence electrons. It forms seven single bonds with seven fluorine atoms and has 0 lone pairs. The steric number is 7, which corresponds to $sp^3d^3$ hybridization. This gives a pentagonal bipyramidal geometry.

For $AlCl_4^-$ (Tetrachloroaluminate ion): The central atom is Aluminum (Al). Aluminum has 3 valence electrons. In $AlCl_4^-$, aluminum forms four single bonds with four chlorine atoms, and it has a -1 charge, meaning it has gained one electron. Therefore, aluminum has 4 bonding pairs and 0 lone pairs. The steric number is 4, which corresponds to $sp^3$ hybridization. This also results in a tetrahedral geometry.

Matching these:1.  $NH\_4^+$: $sp^3$ (Matches with B)
2.  $CO\_2$: $sp$ (Matches with C)
3.  $IF\_7$: $sp^3d^3$ (Matches with D)
4.  $AlCl\_4^-$: $sp^3$ (Matches with A in the options, meaning it is incorrectly matched to $sp^2$)Therefore, option (1) which states "1-B, 2-C, 3-D, 4-A" is the correct match.

The incorrect option is (2): This option suggests "1-C, 2-B, 3-A, 4-D". This is incorrect because, for instance, it incorrectly matches $NH_4^+$ with $sp$ hybridization and $CO_2$ with $sp^3$ hybridization, whereas our analysis shows the opposite.

The incorrect option is (3): This option suggests "1-D, 2-A, 3-C, 4-B". This is incorrect because it misrepresents the hybridization for almost all the given species. For example, it matches $NH_4^+$ with $sp^3d^3$ and $CO_2$ with $sp^2$, which are both wrong.

The incorrect option is (4): This option suggests "1-D, 2-C, 3-B, 4-A". This is incorrect because, while it correctly matches $CO_2$ with $sp$ (C) and $AlCl_4^-$ with $sp^3$ (B, if A was $sp^3$), it incorrectly matches $NH_4^+$ with $sp^3d^3$ (D) and $IF_7$ with $sp^3$ (B).