Which of the following compounds is LEAST likely to exist? — Bonding Chemistry Question

STEPS:

1. Analyze the positions of the central atoms on the periodic table: Identify the period of each central atom in the options. Nitrogen ($\text{N}$) is a second-period element. Phosphorus ($\text{P}$) is a third-period element, and antimony ($\text{Sb}$) is a fifth-period element.
2. Apply the Octet Rule and shell capacity principles: Elements in the second period have valence shells consisting of only $2s$ and $2p$ subshells, which can accommodate a maximum of 8 valence electrons (four pairs). Because they lack energetically accessible $d$-orbitals, second-period elements cannot expand their octet to form more than four covalent bonds.
3. Understand expanded octets in heavier elements: Elements in the third period or below possess vacant, low-energy $d$-orbitals in their valence shell (such as $3d$ for $\text{P}$ and $5d$ for $\text{Sb}$). This allows them to expand their valence shell to accommodate 10 or more electrons, enabling them to form five covalent bonds in hypervalent compounds.
4. Evaluate the bonding in $\text{NI}_5$: For $\text{NI}_5$ to exist, the central nitrogen atom must form five single covalent bonds with five iodine atoms, which would require nitrogen to share 10 valence electrons (an expanded octet). Since nitrogen is in the second period and cannot expand its octet, it cannot form five bonds, making $\text{NI}_5$ least likely to exist.
5. Consider steric hindrance as a compounding factor: Nitrogen is a very small atom, whereas iodine is a very large halogen atom. Trying to pack five bulky iodine atoms around a single tiny nitrogen atom creates severe steric strain and electron-cloud repulsion, further preventing $\text{NI}_5$ from forming.

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WHY_OTHERS_WRONG:

• A is incorrect: $\text{PCl}_5$ readily exists. Since phosphorus is in the third period, it has accessible $3d$ orbitals that allow it to expand its octet and share 10 valence electrons with five chlorine atoms.
• B is incorrect: $\text{PBr}_3$ readily exists. Phosphorus forms three single bonds with bromine and retains one lone pair, which perfectly satisfies the standard octet rule.
• C is incorrect: $\text{NF}_3$ readily exists. The nitrogen atom forms three single bonds with fluorine and retains one lone pair, satisfying the octet rule without requiring valence shell expansion.
• E is incorrect: $\text{SbF}_5$ readily exists. Antimony is a heavy element in the fifth period and can easily expand its octet to form five stable covalent bonds with small, highly electronegative fluorine atoms.