Chemistry · Periodic Classification Of Properties · NEET
NCERT gives three reasons. (1) They are very SMALL in size. (2) They have a large charge-to-radius ratio (high polarising power). (3) They have high electronegativity. On top of this, the first member of a group has only FOUR valence orbitals for bonding (one 2s + three 2p). The elements below it also have empty 3d orbitals. This single fact — no d-orbitals — causes most of the 'anomalies' NEET tests.
Boron is in the second period. Its valence shell (n=2) has only 2s and 2p orbitals — there is NO 2d orbital in nature. So boron can hold at most 4 electron pairs, giving a maximum covalency of 4 (it forms BF4-). Aluminium is in the third period and has empty 3d orbitals, so it can expand its shell to hold 6 pairs and form AlF6 3-. This is a direct NEET 2018 question.
Covalency = number of bonds an atom can make. The number of bonds is limited by the number of usable orbitals. A period-2 atom has just 4 valence orbitals (2s + three 2p), so it can form at most 4 bonds. Period-3 atoms and below have extra empty d-orbitals, so they can go beyond 4. Remember: 4 orbitals means covalency 4.
The first element of a group is so small that its properties match the element ONE step down and ONE step to the right — its diagonal neighbour. Li is like Mg, Be is like Al, B is like Si. This happens because moving right increases charge but moving down increases size, and these two effects cancel to give a similar charge/size ratio (polarising power). This is called the diagonal relationship.
Be2+ and Li+ are very small ions with a high charge/size ratio, so they have strong polarising power (they pull electron clouds of the anion). By Fajans' rules, high polarising power means MORE covalent character. That is why BeCl2 and LiCl are more covalent than the compounds of the elements below them. Careful: in the NEET 2023 Assertion, the claim that Li/Be form 'pronounced IONIC' compounds is FALSE — they form pronounced COVALENT compounds.
Second period atoms are small, so their p-orbitals are compact and close together. This lets them overlap sideways (pπ-pπ overlap) and form strong double and triple bonds — e.g. C=C, C≡C, N=N, N≡N, C=O, C≡N. Larger atoms below them (like Si, P) are too big for good pπ-pπ overlap, so they prefer single bonds. This is why N2 is a gas (triple bond) but P exists as P4.
Which one of the following elements is unable to form MF6 3- ion?
Assertion (A): Lithium and beryllium, unlike their other respective group members, form compounds with pronounced ionic character. Reason (R): Lithium and magnesium have similar properties due to the diagonal relationship.
Among CaH2, BeH2, BaH2, the order of ionic character is
Try the real previous-year questions from this chapter — each with the answer and a full solution.
They are Li, Be, B, C, N, O, F and Ne — the 8 elements in the second row of the periodic table (n=2). In NEET, the 'anomalous first member' idea mainly means Li, Be, B and C, each differing from the group below it.
It is 4. A period-2 atom has only 4 valence orbitals (one 2s + three 2p) and no d-orbitals, so it can form at most 4 bonds. That is why boron makes only BF4- and cannot make BF6 3-.
Three NCERT reasons: (1) small size, (2) large charge/radius ratio (high polarising power), (3) high electronegativity — plus only 4 valence orbitals (no d-orbitals). These give it covalency limited to 4, more covalent compounds, strong pπ-pπ multiple bonds, and a diagonal relationship with the element to its lower-right.
More COVALENT than expected. Their tiny ions have high charge/size ratio (high polarising power), so by Fajans' rules their compounds show pronounced covalent character. This is a favourite NEET Assertion-Reason trap.
Nitrogen is a small second-period atom, so its 2p orbitals overlap well sideways (pπ-pπ) to make a strong N≡N triple bond. Phosphorus is larger; its 3p orbitals overlap poorly for π bonds, so it prefers single bonds and exists as P4. This is a direct effect of anomalous second-period behaviour.
