Why is PCl₅ possible but NCl₅ is not — explain with d orbitals

Question

Why does phosphorus form $\text{PCl}_5$ but nitrogen cannot form $\text{NCl}_5$ ? Explain with reference to d orbitals.

Solution — Step by Step

Nitrogen (Z = 7): $1s^2\ 2s^2\ 2p^3$

Ground state: 3 unpaired electrons in 2p → can form 3 bonds → $\text{NCl}_3$ is possible.

Phosphorus (Z = 15): $1s^2\ 2s^2\ 2p^6\ 3s^2\ 3p^3$

Ground state: 3 unpaired electrons in 3p → can form 3 bonds → $\text{PCl}_3$ is possible.

So far, both behave similarly.

Phosphorus has empty 3d orbitals in its valence shell (n = 3 has s, p, and d subshells).

When excited, one electron from 3s can be promoted to an empty 3d orbital:

3s^2\ 3p^3\ 3d^0 \xrightarrow{\text{excitation}} 3s^1\ 3p^3\ 3d^1

Now P has 5 unpaired electrons → can form 5 bonds → $\text{PCl}_5$ is possible!

The energy needed for this promotion is compensated by the energy released in forming 2 additional P–Cl bonds.

Nitrogen is in the second period (n = 2). Its valence shell has only 2s and 2p subshells.

There are no 2d orbitals — d orbitals start from n = 3 only. So nitrogen cannot expand its valence shell beyond 4 electron pairs (octet).

Even if we try to promote a 2s electron to some higher orbital, the energy cost would be enormous and the bond formation energy cannot compensate. More fundamentally: there is no 2d subshell for any element.

Therefore, nitrogen is limited to a maximum of 4 bonds (in $\text{NH}_4^+$ , for example), and $\text{NCl}_5$ cannot exist.

The ability to expand octet (form more than 4 bonds) requires vacant d orbitals in the same valence shell. These are available from the 3rd period onwards (3d, 4d, 5d…). Elements in the 2nd period (C, N, O, F) have only s and p orbitals in their valence shell — no d orbitals — so they cannot exceed the octet.

This explains many comparisons:

$\text{PCl}_5$ exists, $\text{NCl}_5$ does not
$\text{SF}_6$ exists, $\text{OF}_6$ does not
$\text{H}_2\text{SO}_4$ is stable, $\text{H}_2\text{CO}_4$ would require carbon to expand octet

Why This Works

The octet rule works perfectly for 2nd period elements because they simply don’t have anywhere to put extra electrons — the 2p orbitals are the highest, and there’s no 2d. It’s not a rigid chemical law; it’s a consequence of quantum mechanics.

For 3rd period and beyond, the d orbitals are energetically accessible (not too far in energy from 3s and 3p). Promoting an electron there costs some energy, but forming two extra bonds releases more. The net result is exothermic — $\text{PCl}_5$ is thermodynamically stable.

Alternative Method — Compare Electronegativity and Size

There’s a secondary reason beyond d orbitals: nitrogen is much smaller and more electronegative than phosphorus. Five chlorine atoms around tiny, electronegative nitrogen would cause severe steric repulsion and unfavourable electron-electron repulsion. Even if d orbitals were available, the geometry would be unstable. Phosphorus is larger, reducing steric strain.

Common Mistake

Saying “nitrogen doesn’t have d orbitals at all.” Nitrogen does have 3d, 4d, etc. — all elements have infinite orbitals. The correct statement is: nitrogen has no d orbitals in its valence shell (n = 2 shell). Only n = 3 and higher have d subshells. This is a precise but important distinction.

Question

Solution — Step by Step

Why This Works

Alternative Method — Compare Electronegativity and Size

Common Mistake

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