Question:
Species that are both isoelectronic and isostructural are:
Species that are both isoelectronic and isostructural are:
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Species that have the same central atom group and identical ligand arrangement (like $\text{CH}_4$, $\text{NH}_4^+$, and $\text{BH}_4^-$) are generally isoelectronic and isostructural with each other.
Updated On: Jun 16, 2026
- $\text{NH}_4^+$ and $\text{BH}_4^-$
- $\text{O}_3$ and $\text{NO}_2^+$
- $\text{N}_2\text{O}$ and $\text{NO}_2$
- $\text{NH}_2^-$ and $\text{BH}_4^-$
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The Correct Option is A
Solution and Explanation
Step 1: Understanding the Question:
The question asks us to identify a pair of chemical species that are both isoelectronic (having the same number of electrons) and isostructural (having the same spatial molecular geometry).
Step 2: Key Formula or Approach:
1. Two species are isoelectronic if they have the same total number of electrons (or valence electrons).
2. Two species are isostructural if they have the same molecular shape and geometry as predicted by VSEPR theory.
Step 3: Detailed Explanation:
• Let us analyze Option A containing the ammonium ion ($\text{NH}_4^+$) and the tetrahydroborate ion ($\text{BH}_4^-$).
• Let us calculate the total number of electrons in each species:
itemize
• For $\text{NH}_4^+$: Nitrogen has $7$ electrons, and each of the $4$ Hydrogens has $1$ electron. Subtracting $1$ electron for the positive charge:
\[ \text{Total Electrons} = 7 + (4 \times 1) - 1 = 10 \text{ electrons} \]
• For $\text{BH}_4^-$: Boron has $5$ electrons, and each of the $4$ Hydrogens has $1$ electron. Adding $1$ electron for the negative charge:
\[ \text{Total Electrons} = 5 + (4 \times 1) + 1 = 10 \text{ electrons} \]
Since both species have $10$ electrons, they are isoelectronic.
Now let us evaluate their molecular geometry using VSEPR theory:
• The central nitrogen in $\text{NH}_4^+$ has $4$ bonding pairs of electrons and $0$ lone pairs, giving a steric number of $4$, which corresponds to $\text{sp}^3$ hybridization and a tetrahedral shape.
• The central boron in $\text{BH}_4^-$ has $4$ bonding pairs of electrons and $0$ lone pairs, also giving a steric number of $4$, corresponding to $\text{sp}^3$ hybridization and a tetrahedral shape.
Since both have a tetrahedral geometry, they are isostructural.
Let us briefly check the other options to verify:
• In Option B, $\text{O}_3$ is bent ($18$ valence electrons), while $\text{NO}_2^+$ is linear ($16$ valence electrons). They are neither isoelectronic nor isostructural.
• In Option C, $\text{N}_2\text{O}$ is linear ($16$ valence electrons), while $\text{NO}_2$ is bent ($17$ valence electrons).
• In Option D, $\text{NH}_2^-$ is bent, while $\text{BH}_4^-$ is tetrahedral.
itemize
Step 4: Final Answer:
Therefore, $\text{NH}_4^+$ and $\text{BH}_4^-$ are both isoelectronic and isostructural.
The question asks us to identify a pair of chemical species that are both isoelectronic (having the same number of electrons) and isostructural (having the same spatial molecular geometry).
Step 2: Key Formula or Approach:
1. Two species are isoelectronic if they have the same total number of electrons (or valence electrons).
2. Two species are isostructural if they have the same molecular shape and geometry as predicted by VSEPR theory.
Step 3: Detailed Explanation:
• Let us analyze Option A containing the ammonium ion ($\text{NH}_4^+$) and the tetrahydroborate ion ($\text{BH}_4^-$).
• Let us calculate the total number of electrons in each species:
itemize
• For $\text{NH}_4^+$: Nitrogen has $7$ electrons, and each of the $4$ Hydrogens has $1$ electron. Subtracting $1$ electron for the positive charge:
\[ \text{Total Electrons} = 7 + (4 \times 1) - 1 = 10 \text{ electrons} \]
• For $\text{BH}_4^-$: Boron has $5$ electrons, and each of the $4$ Hydrogens has $1$ electron. Adding $1$ electron for the negative charge:
\[ \text{Total Electrons} = 5 + (4 \times 1) + 1 = 10 \text{ electrons} \]
Since both species have $10$ electrons, they are isoelectronic.
Now let us evaluate their molecular geometry using VSEPR theory:
• The central nitrogen in $\text{NH}_4^+$ has $4$ bonding pairs of electrons and $0$ lone pairs, giving a steric number of $4$, which corresponds to $\text{sp}^3$ hybridization and a tetrahedral shape.
• The central boron in $\text{BH}_4^-$ has $4$ bonding pairs of electrons and $0$ lone pairs, also giving a steric number of $4$, corresponding to $\text{sp}^3$ hybridization and a tetrahedral shape.
Since both have a tetrahedral geometry, they are isostructural.
Let us briefly check the other options to verify:
• In Option B, $\text{O}_3$ is bent ($18$ valence electrons), while $\text{NO}_2^+$ is linear ($16$ valence electrons). They are neither isoelectronic nor isostructural.
• In Option C, $\text{N}_2\text{O}$ is linear ($16$ valence electrons), while $\text{NO}_2$ is bent ($17$ valence electrons).
• In Option D, $\text{NH}_2^-$ is bent, while $\text{BH}_4^-$ is tetrahedral.
itemize
Step 4: Final Answer:
Therefore, $\text{NH}_4^+$ and $\text{BH}_4^-$ are both isoelectronic and isostructural.
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