Q.Which of the following elements can show covalency greater than 4? (Note: more than one of the given options may be correct.)

Concept: Availability of vacant $d$-orbitals in the valence shell for expansion of octet.

Reasoning:

1. Covalency greater than 4 requires the atom to have vacant $d$-orbitals in its valence shell so it can accommodate more than 8 electrons (expand octet).
2. Be and B belong to the second period (no $d$-orbitals), so their maximum covalency is 4 (Be) and 4 (B) respectively — they cannot exceed 4. …

The key idea is simple: covalency is the number of covalent bonds an atom can form. For most elements, the maximum covalency is limited by the number of valence electrons (the octet rule). But some elements can exceed 4 — they can form 5, 6, or even more bonds. Why? Because they have vacant d-orbitals in the same energy level (the third shell or higher) that can accommodate extra electrons. This is called octet expansion.

Elements in the second period (like Be and B) have only s and p orbitals — no d-orbitals. So they can never have a covalency greater than 4. In fact, Be maxes out at 2 (it's electron-deficient), and B at 3 (though it can form a coordinate bond to reach 4 in species like $\ce{BF4-}$). But 4 is their absolute ceiling.

Elements in the third period and beyond (like P and S) have 3d orbitals. These d-orbitals are empty in the ground state but are close in energy to the 3s and 3p orbitals. When the atom forms many bonds, it can promote electrons into these d-orbitals, allowing it to accommodate more than 8 electrons in its valence shell. That's how P forms $\ce{PCl5}$ (covalency 5) and S forms $\ce{SF6}$ (covalency 6).

Let's check each option.

1. Be (Beryllium) — Atomic number 4, electronic configuration $1s^2 2s^2$. It has only the second shell (n=2), which has no d-orbitals. Its maximum covalency is 2 (as in $\ce{BeCl2}$). It can never exceed 4. So (A) is not correct.

2. P (Phosphorus) — Atomic number 15, configuration $1s^2 2s^2 2p^6 3s^2 3p^3$. It has vacant 3d orbitals. In $\ce{PCl5}$, phosphorus uses $sp^3d$ hybridization, forming 5 bonds. In $\ce{PCl6-}$, it uses $sp^3d^2$ and forms 6 bonds. So covalency can be 5 or 6 — definitely greater than 4. Option (B) is correct. …