Which of the two lons from the list given have the geometry that is explained by the same hybridization of orbitals `NO_(2)^(-),NO_(3)^(-) ,NH_(2)^(-) NH_(4)^(+) SCN^(-)`?

Options

A. `NH_4^+` and `NO_3^-`
B. `SCN^-` and `NH_2^-`
C. `NO_2^-` and `NH_2^-`
D. `NO_2^-` and `NO_3^-`

Explanation

To determine which two ions from the given list have geometries explained by the same hybridization of orbitals, we will analyze each ion step by step.

### Step 1: Analyze NH₄⁺
- **Valence Electrons**: Nitrogen has 5 valence electrons. In NH₄⁺, one electron is removed (due to the positive charge), leaving nitrogen with 4 valence electrons.
- **Bonding**: Nitrogen forms 4 bond pairs with 4 hydrogen atoms.
- **Hybridization**: The hybridization is calculated as the number of bond pairs plus the number of lone pairs. Here, there are 4 bond pairs and 0 lone pairs.
- **Hybridization Type**: \(sp^3\)

### Step 2: Analyze NO₃⁻
- **Valence Electrons**: Nitrogen has 5 valence electrons. In NO₃⁻, one additional electron is added (due to the negative charge), giving nitrogen 6 valence electrons.
- **Bonding**: Nitrogen forms 3 bond pairs with 3 oxygen atoms.
- **Hybridization**: There are 3 bond pairs and 0 lone pairs.
- **Hybridization Type**: \(sp^2\)

### Step 3: Analyze SCN⁻
- **Valence Electrons**: Carbon has 4 valence electrons. In SCN⁻, one additional electron is added, giving carbon 5 valence electrons.
- **Bonding**: Carbon forms 2 bond pairs (double bonds) with sulfur and nitrogen.
- **Hybridization**: There are 2 bond pairs and 0 lone pairs.
- **Hybridization Type**: \(sp\)

### Step 4: Analyze NH₂⁻
- **Valence Electrons**: Nitrogen has 5 valence electrons. In NH₂⁻, one additional electron is added, giving nitrogen 6 valence electrons.
- **Bonding**: Nitrogen forms 2 bond pairs with 2 hydrogen atoms and has 2 lone pairs.
- **Hybridization**: There are 2 bond pairs and 2 lone pairs.
- **Hybridization Type**: \(sp^3\)

### Step 5: Analyze NO₂⁻
- **Valence Electrons**: Nitrogen has 5 valence electrons. In NO₂⁻, one additional electron is added, giving nitrogen 6 valence electrons.
- **Bonding**: Nitrogen forms 2 bond pairs (double bonds) with 2 oxygen atoms and has 1 lone pair.
- **Hybridization**: There are 2 bond pairs and 1 lone pair.
- **Hybridization Type**: \(sp^2\)

### Summary of Hybridizations:
- NH₄⁺: \(sp^3\)
- NO₃⁻: \(sp^2\)
- SCN⁻: \(sp\)
- NH₂⁻: \(sp^3\)
- NO₂⁻: \(sp^2\)

### Conclusion:
The ions that have the same hybridization are:
- NH₄⁺ and NH₂⁻ (both \(sp^3\))
- NO₃⁻ and NO₂⁻ (both \(sp^2\))

Thus, the pairs of ions that have geometries explained by the same hybridization of orbitals are:
- **NH₄⁺ and NH₂⁻** (both \(sp^3\))
- **NO₃⁻ and NO₂⁻** (both \(sp^2\))